Some Properties of the Number 2023

For my one hundredth post of the year 2022, I thought I'd create a post about the forthcoming year before it arrives. Firstly, looking at Numbers Aplenty, we find that a 4×4 square can be tiled in 2023 ways using the L triomino and the monomino. Figure 1 shows four examples and click here to see all 2023 possible arrangements. See OEIS A220054 and A127870.

Figure 1

Another interesting property that Numbers Aplenty brings up is the following:

2023 = (2+0+2+3)(2²+0²+2²+3²)²

It also turns out that the prime factors, when concatenated, form a palindrome.

2023 = 7 x 17 x 17 --> 71717

This as it turns out is not a common property (see OEIS A046447). In the range of composite numbers up to 40,000, there are only 193 such numbers. In terms of years, it has been 320 years since this last happened and next occurrence will be rather sooner, in 2048. Here are the numbers (permalink):

4, 8, 9, 16, 25, 27, 32, 39, 49, 64, 69, 81, 119, 121, 125, 128, 129, 159, 219, 243, 249, 256, 259, 329, 339, 343, 403, 429, 469, 507, 512, 625, 669, 679, 729, 795, 1024, 1207, 1309, 1329, 1331, 1533, 1547, 1587, 1589, 1703, 2023, 2048, 2097, 2187, 2319, 2321, 2359, 2401, 2649, 2701, 3039, 3125, 3421, 3503, 3629, 3633, 3639, 3729, 3897, 3899, 3975, 4096, 4303, 4607, 4719, 4839, 5289, 5295, 5547, 5565, 5603, 5739, 6209, 6401, 6561, 6591, 6597, 6979, 7269, 7663, 7819, 7833, 8029, 8192, 8203, 8427, 8519, 8745, 8921, 9329, 9583, 9703, 9707, 9795, 10001, 10021, 10201, 10299, 10599, 10731, 11199, 11499, 12207, 12729, 12795, 13129, 13489, 13573, 13801, 13929, 14133, 14147, 14189, 14295, 14399, 14641, 14829, 15459, 15463, 15549, 15563, 15625, 15987, 16233, 16384, 16507, 16807, 16863, 16959, 17161, 17295, 17363, 17859, 18489, 18789, 18829, 19003, 19683, 19689, 19875, 19907, 20111, 20289, 20489, 20589, 20667, 21259, 21959, 22139, 22507, 22719, 22801, 23019, 23619, 24601, 24901, 25459, 26349, 27879, 28273, 28601, 28609, 28847, 28951, 29729, 29829, 31129, 31363, 31423, 31829, 32357, 32761, 32768, 33173, 33339, 34391, 35143, 36481, 36501, 36763, 36849, 38307, 38899, 38955, 39599, 39939, 39963

On the topic of palindromes, the sum of 2023 and its reverse 3202 is a palindrome. Of all the numbers from 1 to 40,000, there are 9456 (or almost 25%) with this property.

2023 + 3202 = 5225

2023 is also palindromic in base 16 since it equals 7e7. In the range of numbers up to 40,000, this occurs 409 times.

2023 = 7e7

The numbers are:

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 34, 51, 68, 85, 102, 119, 136, 153, 170, 187, 204, 221, 238, 255, 257, 273, 289, 305, 321, 337, 353, 369, 385, 401, 417, 433, 449, 465, 481, 497, 514, 530, 546, 562, 578, 594, 610, 626, 642, 658, 674, 690, 706, 722, 738, 754, 771, 787, 803, 819, 835, 851, 867, 883, 899, 915, 931, 947, 963, 979, 995, 1011, 1028, 1044, 1060, 1076, 1092, 1108, 1124, 1140, 1156, 1172, 1188, 1204, 1220, 1236, 1252, 1268, 1285, 1301, 1317, 1333, 1349, 1365, 1381, 1397, 1413, 1429, 1445, 1461, 1477, 1493, 1509, 1525, 1542, 1558, 1574, 1590, 1606, 1622, 1638, 1654, 1670, 1686, 1702, 1718, 1734, 1750, 1766, 1782, 1799, 1815, 1831, 1847, 1863, 1879, 1895, 1911, 1927, 1943, 1959, 1975, 1991, 2007, 2023, 2039, 2056, 2072, 2088, 2104, 2120, 2136, 2152, 2168, 2184, 2200, 2216, 2232, 2248, 2264, 2280, 2296, 2313, 2329, 2345, 2361, 2377, 2393, 2409, 2425, 2441, 2457, 2473, 2489, 2505, 2521, 2537, 2553, 2570, 2586, 2602, 2618, 2634, 2650, 2666, 2682, 2698, 2714, 2730, 2746, 2762, 2778, 2794, 2810, 2827, 2843, 2859, 2875, 2891, 2907, 2923, 2939, 2955, 2971, 2987, 3003, 3019, 3035, 3051, 3067, 3084, 3100, 3116, 3132, 3148, 3164, 3180, 3196, 3212, 3228, 3244, 3260, 3276, 3292, 3308, 3324, 3341, 3357, 3373, 3389, 3405, 3421, 3437, 3453, 3469, 3485, 3501, 3517, 3533, 3549, 3565, 3581, 3598, 3614, 3630, 3646, 3662, 3678, 3694, 3710, 3726, 3742, 3758, 3774, 3790, 3806, 3822, 3838, 3855, 3871, 3887, 3903, 3919, 3935, 3951, 3967, 3983, 3999, 4015, 4031, 4047, 4063, 4079, 4095, 4097, 4369, 4641, 4913, 5185, 5457, 5729, 6001, 6273, 6545, 6817, 7089, 7361, 7633, 7905, 8177, 8194, 8466, 8738, 9010, 9282, 9554, 9826, 10098, 10370, 10642, 10914, 11186, 11458, 11730, 12002, 12274, 12291, 12563, 12835, 13107, 13379, 13651, 13923, 14195, 14467, 14739, 15011, 15283, 15555, 15827, 16099, 16371, 16388, 16660, 16932, 17204, 17476, 17748, 18020, 18292, 18564, 18836, 19108, 19380, 19652, 19924, 20196, 20468, 20485, 20757, 21029, 21301, 21573, 21845, 22117, 22389, 22661, 22933, 23205, 23477, 23749, 24021, 24293, 24565, 24582, 24854, 25126, 25398, 25670, 25942, 26214, 26486, 26758, 27030, 27302, 27574, 27846, 28118, 28390, 28662, 28679, 28951, 29223, 29495, 29767, 30039, 30311, 30583, 30855, 31127, 31399, 31671, 31943, 32215, 32487, 32759, 32776, 33048, 33320, 33592, 33864, 34136, 34408, 34680, 34952, 35224, 35496, 35768, 36040, 36312, 36584, 36856, 36873, 37145, 37417, 37689, 37961, 38233, 38505, 38777, 39049, 39321, 39593, 39865

2023 is also a:

• Duffinian number: a composite number that has no factors in common with its sum of divisors \( \sigma(\tau)) \).
  
  
• Harshad number: a number that is divisible by the sum of its digits. In fact, 2022 was a Harshad number and so are 2024 and 2025 we have a run of four consecutive Harshad numbers.
  
  
• Lucky number: a number that survives a sieving process which is similar to the Eratosthenes sieve that can be used to compute the prime numbers.
  
  
• Junction number: a number that can be formed in two ways by adding a number to its sum of digits. Here we have 1997 + sod(1997) = 2015 + sod(2015) = 2023 where sod means sum of digits.

Turning to the Online Encyclopedia of Integer Sequences (OEIS), we find that it is a member of OEIS A063416:

A063416

Multiples of 7 whose sum of digits is equal to 7.

The sequence begins:

7, 70, 133, 322, 511, 700, 1015, 1141, 1204, 1330, 2023, 2212, 2401, 3031, 3220, 4102, 5110, 7000, 10024, 10150, 10213, 10402, 11032, 11221, 11410, 12040, 12103, 13111, 13300, 15001, 20041, 20104, 20230, 21112, 21301, 22120, 23002, 24010 

There is a strong connection between 2023 and the digit 7:

• the sum of its digits is 7
• it is divisible by 7
• both of its factors (7 and 17) contain the digit 7
• in octal, it has the representation 3747
• in hexadecimal, it has the representation 7e7
• in base 17, it has the representation 700
• its Fibonacci seed numbers are 17 and 49 = 7 x 7

By "Fibonacci seed numbers" I mean that by starting with 17 and 49 and proceeding Fibonacci-style, we get to 2023 thus:

17, 49, 66, 115, 181, 296, 477, 773, 1250, 2023

It is a member of OEIS A070258:

A070258

Smallest of 3 consecutive numbers each divisible by a square.

• \(2023 = 7 \times 17^2\)
• \(2024 = 2^3 \times 11 \times 23\)
• \(2025 = 3^4 \times 5^2\)

The initial members of the sequence are:

48, 98, 124, 242, 243, 342, 350, 423, 475, 548, 603, 724, 774, 844, 845, 846, 1024, 1250, 1274, 1323, 1375, 1420, 1448, 1519, 1664, 1674, 1680, 1681, 1682, 1848, 1862, 1924, 2007, 2023, 2056, 2106, 2150, 2223, 2275, 2348, 2366, 2523, 2527, 2574, 2644

Well, that's largely what I decided to present from Numbers Aplenty and the OEIS. What else is there of interest about the number 2023? I thought I'd ask ChatGPT the following question: what are some interesting facts, mathematical or otherwise, about the number 2023? Figure 2 is a screenshot of what it spat out.

Figure 2

As usual with ChatGPT, when it comes to Mathematics, it can spew out some nonsense. In case, it claims 2023 is a prime number! Similarly, it's claim that "2023 is the sum of three consecutive prime numbers: 661 + 673 + 677 is false. These are indeed consecutive prime numbers but they add up to 2011. Can it get any worse? Well, yes. The claim is made that "2023 is the hypotenuse of a Pythagorean triple: (683, 992, 2023)" and testing this out we find that \(683^2+992^2=1,450,553\) while \(2023^2=4,092,529\).  We're not done yet! Another bold claim is made that "2023 can be written as the sum of two cubes in two different ways: \(2023 = 13^3 + 2^3 = 11^3 + 10^3\) ". Nope. \(13^3+2^3=2205\) and \(11^3+10^3=2331\). It turns out that 2023 cannot be expressed as a sum of two cubes. A final claim is made that 2023 has the representation 11111011111 in binary. No. The correct representation is 11111100111.

Clearly time to give up on ChatGPT as a source of information about Mathematics. I've written about this AI bot's deficiencies in a post titled ChatGPT and a Tragi-Comedy of Integration. Let's not leave off without a final laugh. After the bot's spectacular failures, I asked it: 

I'm impressed in a negative way. Except for points 2 and 6, every mathematical statement that you made was  wrong. 2023 is not a prime number and so on. How can you get things so wrong? 

Its response was: 

I apologize for any errors in my previous response. As a language model, I can provide information and answer questions to the best of my ability based on the knowledge and information that I have been trained on. However, I am not infallible and it is possible for me to make mistakes or provide incorrect information.

In fact, 2023 is not a prime number. It can be divided evenly by 3, 11, and 61, so it is a composite number. I apologize for any confusion this may have caused.

I hope this clears up any misunderstandings and I apologize for any errors in my previous response. Please feel free to ask me any other questions you may have.

Actually, looking at the atomic number of Calcium, I checked and found that it was 20 not 2023. This perhaps gives an insight into what's going wrong with the bot. It's looking at the first two digits (20) and ignoring the last two digits (23). As for the Mathematics, I simply don't know how its coming up with its answers.

All triprimes can be envisaged as "bricks", 3-D solid objects. If all three primes are distinct, we have a sphenic number and an associated sphenic brick. The volume of the brick is the number itself but there is always an associated number: the surface area of the brick. In the case of 2023, the associated number is 1054 (square units). This number is sphenic and the associated sphenic brick has a surface area of 1246 square units. This number in turn is sphenic and the associated area is 1630 square units. Once again, 1630 is sphenic with associated surface area of 2302 square units. At this point, the chain is broken because 2302 is biprime. This is summarised below:$$ \begin{align} 2023 &= 7 \times 17 \times 17 &\rightarrow 1054\\1054 &= 2 \times 17 \times 31 &\rightarrow 1246\\1246 &= 2 \times 7 \times 89 &\rightarrow 1630\\1630&= 2 \times 5 \times 163 &\rightarrow 2302\\2302 &= 2 \times 1151 \end{align}$$Figure 3 shows a list of the triprime numbers between 2000 and 2100. Thus there is another twelve years before another triprime number, 2035, is encountered. 2035 is also a sphenic number because it has three distinct factors

Figure 3

Numbers Within Numbers

As sometimes happens, the number associated with my diurnal age seems to contain nothing much of interest. Today the number was 26930 and I was temporarily stuck. However, I noticed something in Numbers Aplenty. See Figure 1.

Figure 1

The hexadecimal representation of 26930 is 6932. All the digits in the hexadecimal representation are contained within the decimal representation, each in the same proportion. I thought that this would be an interesting criterion to apply to the numbers from 10 to 40000. The numbers from 0 to 9 trivially satisfy the criterion. 

It turns out that there are only 90 decimal numbers that satisfy. This is the list (Permalink):

Hex     Decimal

35 --- 53

173 --- 371

391 --- 913

411 --- 1041

412 --- 1042

413 --- 1043

414 --- 1044

415 --- 1045

416 --- 1046

417 --- 1047

418 --- 1048

419 --- 1049

541 --- 1345

791 --- 1937

821 --- 2081

822 --- 2082

823 --- 2083

824 --- 2084

825 --- 2085

826 --- 2086

827 --- 2087

828 --- 2088

829 --- 2089

1004 --- 4100

1415 --- 5141

1524 --- 5412

1826 --- 6182

2008 --- 8200

2419 --- 9241

3012 --- 12306

4010 --- 16400

4011 --- 16401

4012 --- 16402

4013 --- 16403

4014 --- 16404

4015 --- 16405

4017 --- 16407

4018 --- 16408

4019 --- 16409

4167 --- 16743

4815 --- 18453

5021 --- 20513

5042 --- 20546

5221 --- 21025

6251 --- 25169

6528 --- 25896

6582 --- 25986

6702 --- 26370

6872 --- 26738

6921 --- 26913

6932 --- 26930

7221 --- 29217

7412 --- 29714

7830 --- 30768

7840 --- 30784

8020 --- 32800

8021 --- 32801

8022 --- 32802

8024 --- 32804

8025 --- 32805

8026 --- 32806

8027 --- 32807

8028 --- 32808

8029 --- 32809

8052 --- 32850

8335 --- 33589

8834 --- 34868

8840 --- 34880

8841 --- 34881

8842 --- 34882

8844 --- 34884

8845 --- 34885

8846 --- 34886

8847 --- 34887

8848 --- 34888

8849 --- 34889

9026 --- 36902

9056 --- 36950

9274 --- 37492

9830 --- 38960

9831 --- 38961

9832 --- 38962

9833 --- 38963

9834 --- 38964

9835 --- 38965

9837 --- 38967

9838 --- 38968

9839 --- 38969

9848 --- 38984

9982 --- 39298

The hexadecimal numbers will have the same or fewer digits than the decimal ones. Here is the list of decimal numbers on their own:

53, 371, 913, 1041, 1042, 1043, 1044, 1045, 1046, 1047, 1048, 1049, 1345, 1937, 2081, 2082, 2083, 2084, 2085, 2086, 2087, 2088, 2089, 4100, 5141, 5412, 6182, 8200, 9241, 12306, 16400, 16401, 16402, 16403, 16404, 16405, 16407, 16408, 16409, 16743, 18453, 20513, 20546, 21025, 25169, 25896, 25986, 26370, 26738, 26913, 26930, 29217, 29714, 30768, 30784, 32800, 32801, 32802, 32804, 32805, 32806, 32807, 32808, 32809, 32850, 33589, 34868, 34880, 34881, 34882, 34884, 34885, 34886, 34887, 34888, 34889, 36902, 36950, 37492, 38960, 38961, 38962, 38963, 38964, 38965, 38967, 38968, 38969, 38984, 39298

The algorithm can  be easily modified to accommodate bases up to 36 and the range can be extended, let's say to 100,000. Here are the results for base 36 in the range up to 100,000 (Permalink):

Base 36 Decimal

41 --- 145

75 --- 257

185 --- 1589

213 --- 2631

247 --- 2743

268 --- 2816

291 --- 2917

340 --- 4032

573 --- 6735

659 --- 7965

680 --- 8064

814 --- 10408

850 --- 10548

896 --- 10698

1357 --- 50731

1504 --- 53140

1875 --- 57281

There are only 17 decimal numbers that satisfy in this range. These are listed below in decimal form only:

145, 257, 1589, 2631, 2743, 2816, 2917, 4032, 6735, 7965, 8064, 10408, 10548, 10698, 50731, 53140, 57281

If we wish to look at bases below 10 then the numbers will have the same or more digits and so the algorithm will need to be modified. We will require all the digits in base 10 format to be contained within the number in the lower base format and these digits need to be in the same proportion. Here are the results for base 8 in range from 8 to 40,000 (Permalink):

Base 8 Decimal

107 --- 71

1347 --- 743

1357 --- 751

2371 --- 1273

3165 --- 1653

4152 --- 2154

6130 --- 3160

6131 --- 3161

6132 --- 3162

6133 --- 3163

6134 --- 3164

6135 --- 3165

6136 --- 3166

6137 --- 3167

6232 --- 3226

12765 --- 5621

13620 --- 6032

14562 --- 6514

15713 --- 7115

15720 --- 7120

15721 --- 7121

15722 --- 7122

15723 --- 7123

15724 --- 7124

15726 --- 7126

15727 --- 7127

15764 --- 7156

17074 --- 7740

37621 --- 16273

52371 --- 21753

73560 --- 30576

75341 --- 31457

102345 --- 34021

102473 --- 34107

103462 --- 34610

104365 --- 35061

105436 --- 35614

106273 --- 36027

106347 --- 36071

There are only 39 numbers that satisfy. Here is the list in decimal format only:

71, 743, 751, 1273, 1653, 2154, 3160, 3161, 3162, 3163, 3164, 3165, 3166, 3167, 3226, 5621, 6032, 6514, 7115, 7120, 7121, 7122, 7123, 7124, 7126, 7127, 7156, 7740, 16273, 21753, 30576, 31457, 34021, 34107, 34610, 35061, 35614, 36027, 36071

Thus it can be seen that whenever a number seems boring, there is always something of interest that remains to be discovered and that property can often be generalised, as was done here, to find interesting properties of other numbers.

Getting back to comparing decimal numbers to their hexadecimal equivalents we can make conditions more stringent by requiring that the lengths of both numbers be the same so that the digits of both are simply permutations of the other. In the range up to 100,000, there are 24 numbers that qualify. These are (Permalink):

Hex Decimal

35 --- 53

173 --- 371

391 --- 913

1004 --- 4100

1415 --- 5141

1524 --- 5412

1826 --- 6182

2008 --- 8200

2419 --- 9241

12570 --- 75120

12571 --- 75121

12572 --- 75122

12573 --- 75123

12574 --- 75124

12575 --- 75125

12576 --- 75126

12577 --- 75127

12578 --- 75128

12579 --- 75129

12765 --- 75621

15086 --- 86150

16549 --- 91465

18197 --- 98711

18499 --- 99481

The numbers in decimal format only are:

53, 371, 913, 4100, 5141, 5412, 6182, 8200, 9241, 75120, 75121, 75122, 75123, 75124, 75125, 75126, 75127, 75128, 75129, 75621, 86150, 91465, 98711, 99481

Similarly in the range up to 100,000, there are only 16 numbers in base 8 format that have the same digits as their decimal equivalants (Permalink):

Base 8 Decimal

2371 --- 1273

3165 --- 1653

4152 --- 2154

6130 --- 3160

6131 --- 3161

6132 --- 3162

6133 --- 3163

6134 --- 3164

6135 --- 3165

6136 --- 3166

6137 --- 3167

6232 --- 3226

37621 --- 16273

52371 --- 21753

73560 --- 30576

75341 --- 31457

Here are the numbers in decimal format only:

1273, 1653, 2154, 3160, 3161, 3162, 3163, 3164, 3165, 3166, 3167, 3226, 16273, 21753, 30576, 31457

Lightweight mathematics of course but sometimes it's good just to have fun with numbers. The table below shows the number of decimal numbers whose value in another base is an anagram of its digits. The bases range from 4 to 28. For bases 2 and 3 and for 29 and beyond, there are no such numbers.

Base 26 is interesting because the ten decimal numbers that satisfy are consecutive and the hexadecimal numbers formed from them are similarly consecutive and formed simply by moving the leading 9 of the decimal number one place to the right. Thus 910 in decimal becomes 190 in hexadecimal, 911 becomes 191 and so on until 919 that becomes 199.

Base 26 Decimal

190 --- 910

191 --- 911

192 --- 912

193 --- 913

194 --- 914

195 --- 915

196 --- 916

197 --- 917

198 --- 918

199 --- 919

All this can be explored via this permalink.

A Special Class of Semiprimes

Let's recall that a semiprime is a number with two, not necessarily distinct, prime factors. The very first semiprime is 4 = 2 x 2 with only one distinct prime factor. The next is 6 = 2 x 3 with two distinct prime factors. Concatenation involves combining the two factors together so that 2 x 2 becomes 22 and 2 x 3 becomes 23. Concatenating the factors of a semiprime with only one distinct prime factor can never produce a prime because of the repetition of digits. Thus 22 is not prime. However, concatenating the factors of a semiprime with two distinct prime factors can produce a prime. 23 is an example. 

However, the factors need not be written is ascending order. We could just as well write 6 = 3 x 2 and in this case concatenating the digits produces 32 which is not a prime number. The first example of a semiprime whose factors can be concatenated either way to produce a prime is 21 because 21 = 3 x 7 giving 37 and 21 = 7 x 3 giving 73. So this is clear enough. Now let's turn our attention to the so-called emiprimes, a semiprime that remains a semiprime when its digits are reversed. The first example of a semiprime that is an emirpimes is 15 because 15 = 3 x 5 and 51 = 3 x 17.

What I want to find is a list of semiprimes with the following properties:

• the semiprime is also an emirpimes
• the semiprime has two distinct prime factors
• the concatenation of the prime factors of the semiprime in ascending order is a prime
• the concatenation of the prime factors of the semiprime in descending order is a prime
• the emirpimes has two distinct prime factors
• the concatenation of the prime factors of the emirpimes in ascending order is a prime
• the concatenation of the prime factors of the emirpimes in descending order is a prime

This is a demanding list of properties for any semiprime and not surprisingly very few satisfy. Here is a list of such numbers, with factorisation, up to 100,000 (Permalink):

3099 = 3 * 1033

9903 = 3 * 3301

10519 = 67 * 157

11707 = 23 * 509

13993 = 7 * 1999

16387 = 7 * 2341

18247 = 71 * 257

19039 = 79 * 241

30607 = 127 * 241

32667 = 3 * 10889

36367 = 41 * 887

38697 = 3 * 12899

39487 = 7 * 5641

39931 = 73 * 547

70603 = 13 * 5431

70711 = 31 * 2281

72247 = 7 * 10321

73099 = 13 * 5623

74227 = 199 * 373

74281 = 59 * 1259

74289 = 3 * 24763

76029 = 3 * 25343

76363 = 7 * 10909

76623 = 3 * 25541

78361 = 23 * 3407

78493 = 53 * 1481

78619 = 29 * 2711

79683 = 3 * 26561

91501 = 37 * 2473

91687 = 277 * 331

92067 = 3 * 30689

93091 = 127 * 733

98247 = 3 * 32749

99037 = 97 * 1021

That are 34 numbers in the range up to 100,00. Here is the list without factorisation of all the semprimes that satisfy up to ONE MILLION (there are 108 of them):

3099, 9903, 10519, 11707, 13993, 16387, 18247, 19039, 30607, 32667, 36367, 38697, 39487, 39931, 70603, 70711, 72247, 73099, 74227, 74281, 74289, 76029, 76363, 76623, 78361, 78493, 78619, 79683, 91501, 91687, 92067, 93091, 98247, 99037, 100437, 101317, 101899, 104529, 108181, 108789, 120553, 126771, 133243, 134797, 137671, 144523, 147061, 149449, 159427, 160741, 168117, 176731, 176767, 177621, 181801, 184033, 197097, 199879, 312817, 322489, 325441, 328459, 330397, 330481, 331783, 337297, 337897, 338977, 342331, 345493, 350569, 355021, 357393, 365863, 368563, 386197, 387133, 393753, 394543, 711861, 713101, 716779, 717469, 718213, 724951, 734001, 767671, 779833, 790791, 791683, 792733, 793033, 797431, 798733, 925401, 944941, 951679, 954823, 964699, 964717, 965053, 976159, 977617, 978991, 984223, 987801, 996469, 998101

Let's one of these, say 99037, to see that it satisfies. Firstly, we note that 73099 is in the list that we know that it's reversal is an emirpimes. Now its factorisation and concatenations lead to two numbers: 971021 and 102197. Testing confirms that both of these numbers are prime. The emirpimes, 73099 factorises to 13 * 5623 that leads to 135623 and 562313. Again, testing reveals both numbers are prime.

So, out of all the semiprimes in the range up to one million there are only 108 that have the properties listed above. So, we have a very special class of semiprimes indeed. It's interesting to note in the distribution that there are no numbers beginning with 2, 4, 5, 6 or 8 which is to be expected. If a semiprime begins with 2, 4, 5, 6 or 8 then its emirpimes will end in 2, 4, 5, 6 or 8 which means that one of the concatenations of its factors must end in 2 or 5, meaning that it can't be prime. Similarly there are no semiprimes that end in a 0, 2, 4, 5, 6 or 8. In short, all semiprimes that satisfy must start and end with 1, 3, 7 or 9.

Figure 1 shows a table of the frequency of the digital roots of such semiprimes (108 of them in the range up to one million):

Figure 1

ChatGPT and a Tragi-Comedy of Integration

Just when I was getting impressed by ChatGPT's capabilities, I was profoundly shocked by the absolute mess it made when trying to evaluate the following famous integral:$$\int_{-\infty}^{\infty} e^{-x^2} dx=\sqrt{\pi}$$Create a solution to the integral of e^(-x^2) with respect to x from minus infinity to plus infinity. This was therubbish that it came up with:

Create a solution to the integral of e^(-x^2) with respect to x from minus infinity to plus infinity.

To solve the integral of e^(-x^2) with respect to x from minus infinity to plus infinity, we can use the following approach:

First, we can rewrite the integral as follows:

∫e^(-x^2) dx = ∫e^(-(x^2)) dx

Next, we can use the substitution u = x^2 to transform the integral into the following form:

∫e^(-(x^2)) dx = ∫e^(-u) * (du/dx) dx

Note that the substitution u = x^2 implies that du = 2xdx, so we can rewrite the integral as follows:

∫e^(-u) * (du/dx) dx = ∫e^(-u) * (du/(2x) dx)

We can then simplify this expression to get the following:

∫e^(-u) * (du/(2x) dx) = (1/2) ∫e^(-u) du/x

Finally, we can use the formula for the indefinite integral of e^(-u) to find the solution:

(1/2) ∫e^(-u) du/x = (1/2) (-e^(-u)/x + C)

Substituting back in the value of u, we find that the solution is:

(1/2) (-e^(-u)/x + C) = (1/2) (-e^(-x^2)/x + C)

Finally, we can apply the limits of integration to find the definite integral:

∫e^(-x^2) dx from -∞ to ∞ = (1/2) (-e^(-∞)/∞ + e^(-∞)/∞ + C) = 0 + C = C

Therefore, the solution to the integral of e^(-x^2) with respect to x from minus infinity to plus infinity is C, where C is a constant.

Wow. This mathematical howler really surprised me. However, ChatGPT is quick to rectify its mistakes and make amends. The revised solution shown below is not one that I'm familiar with but I know of the error function, even if I don't fully understand it. I need to examine this more closely.

The correct answer is square root of pi. What went wrong with your solution?

I apologize for any confusion caused by my previous solution. You are correct that the correct answer to the integral of e^(-x^2) with respect to x from minus infinity to plus infinity is the square root of pi.

There are a few possible reasons why my previous solution did not yield the correct answer:

One possible reason is that I made an error in my calculations. I apologize if this is the case.

Another possible reason is that I did not take into account the fact that the function e^(-x^2) is an even function. This means that the integral of e^(-x^2) from minus infinity to plus infinity is equal to twice the integral of e^(-x^2) from 0 to plus infinity.

In my solution, I also did not take into account the fact that the integral of e^(-x^2) from 0 to plus infinity can be expressed in terms of the error function, which is defined as:

erf(x) = (2/√π) ∫e^(-t^2) dt from 0 to x

Therefore, the integral of e^(-x^2) from 0 to plus infinity can be expressed as:

∫e^(-x^2) dx from 0 to ∞ = (√π/2) erf(∞) = (√π/2)

Since the integral of e^(-x^2) from minus infinity to plus infinity is equal to twice this value, the correct answer is:

`2 * (√π/2) = √π

I apologize for any confusion caused by my previous solution. Thank you for bringing this to my attention.

I thought that by asking ChatGPT to use a programming language like Python to find the integral, it might have more success. Again, it failed spectacularly:

Write Python code that will find the integral of e^(-x^2) from minus infinity to plus infinity. 

To find the integral of e^(-x^2) from minus infinity to plus infinity, you can use the Gaussian function, also known as the normal distribution function. This function is defined as:

f(x) = 1 / (sqrt(2 * pi) * sigma) * e^(-(x - mu)^2 / (2 * sigma^2))

The integral of this function from minus infinity to plus infinity is equal to 1, because the area under the curve is equal to 1. Therefore, you can use the following code to compute the integral:

import math

def gaussian(x, mu, sigma):

    return (1 / (math.sqrt(2 * math.pi) * sigma)) * math.exp(-(x - mu)**2 / (2 * sigma**2))

# Integrate from -infinity to +infinity

result, error = scipy.integrate.quad(gaussian, -float('inf'), float('inf'), args=(0, 1))

           print(result)  # Output: 1.0

This code uses the quad function from the scipy.integrate module to approximate the integral of the Gaussian function. The args parameter allows you to pass values for the mu and sigma parameters of the function. In this example, we set mu to 0 and sigma to 1.

I hope this helps! Let me know if you have any questions.

What's being said is not wrong but it not what I asked it for. So I tried again.

What you've said is true enough but it's not what I asked you to find. I asked you to integrate the function e^(-x^2) from minus infinity to plus infinity. If done correctly, you should find that the answer is the square root of pi. Can you try again to find the value of this definite integral and show the steps leading to your solution?

I wont' write out in full its response because basically its the same as shown above except that by mentioning the definite integral it finds C to equal to zero.  The solution also comes  up with some bizarre comments:

 Therefore, the value of the integral is 0.

However, it is important to note that the integral of e^(-x^2) from minus infinity to plus infinity is actually undefined, because the function is not integrable over this interval. This is because the function is not defined at x = 0, and it has an infinite number of oscillations as x approaches 0, which means that the area under the curve is not well-defined.

I hope this helps! Let me know if you have any questions.

What? Not integrable over the interval! The function is not defined at \(x=0\)! While the capabilities of ChatGPT may be impressive, it clearly has a lot to learn in the area of integration. However, ChatGPT is a language model and its deficiencies in the area of mathematics are well described by the following Reddit entry:

Why is ChatGPT bad at Math? 

Asking a language model to do math is like asking a chef to perform surgery. While both activities may involve cutting and require a certain level of expertise, they are vastly different fields of knowledge.

A language model, such as ChatGPT, is a type of artificial intelligence that has been trained to generate human-like text. These models are typically trained on large amounts of text data and use complex algorithms to generate coherent and often realistic-sounding sentences.

On the other hand, math is a field of study that deals with numbers, quantities, and equations. It involves logical reasoning, critical thinking, and problem-solving skills. In order to do math well, one must have a deep understanding of mathematical concepts and the ability to apply them to solve complex problems.

In short, while a language model may be able to generate text that sounds like it understands math, it is not capable of actually doing math. Similarly, a chef may be skilled with a knife, but that does not make them qualified to perform surgery.

Asking a language model to do math is not only unrealistic, but it also goes against the intended purpose of these models. A language model's main function is to generate text, not to solve math problems.

In conclusion, asking a language model to do math is like asking a chef to perform surgery. Both require specialized knowledge and skills, but they are fundamentally different fields of expertise.

Formula for Pi

I have to remind myself from time to time what a fertile source of information is the website Cantor's Paradise. Today, I came across a tweet from the site's Twitter feed that mentioned a particularly interesting formula for \(\pi\) from Ramanujan. See Figure 1.

Figure 1: link

I'll rewrite the formula below for clarity:$$\frac{1}{\pi}=\frac{2\sqrt{2}}{9801}  \sum_{n=0}^{\infty} \frac{(4n)! (1103+26390n)}{(n!)^4 396^{4n}}$$Now when \(n=0\), we get$$  \frac{1}{ \pi } \approx \frac{1103 \sqrt{8}}{9801} \text{ and } \pi \approx \frac{9801}{1103 \sqrt{8}} $$Now how accurate is this approximation? Permalink

3.1415926535897932385 ... actual digits of pi to 20 decimal places

3.1415927300133056603 approximated digits of pi when \(n=0\) (first term)

This is a pretty impressive approximation for the first term of an infinite series! As the website says:

This gives the accurate value of \(\pi\) up to 6 decimal places, but this is only the 1st term in another infinite series. This number alone is sufficient to calculate the circumference of the Earth with a maximum error of just 1 meter. It is to be noted that while Ramanujan’s formula takes one formula to calculate up to 6 decimal places, it takes Leibniz about 5 million terms. Ramanujan’s formula could do it in one term though and each successive term adds up another 8 decimal places to the value of π. This formula holds absolutely true for finding the value of π, but there is no clear understanding of how he came up with the numbers in his formula like 9801 and 1103. Mathematicians use this formula today to find the value of π to an insurmountable extent.

Let's set \(n=1\) Permalink:

3.1415926535897932385 actual digits of pi to 20 decimal places

3.1415926535897938780 approximated digits of pi when \(n=1\) (first two terms)

So far we've only displayed \( \pi \) to 20 decimal places but with the addition of another term, there is a need for more accuracy. Here's the result of \(n=2\) Permalink:

3.14159265358979323846264338328 actual digits of pi to 30 decimal places 

3.14159265358979323846264906570 approximated digits of pi when \(n=2\) (first three terms)

There's no need to go further, the improving accuracy is obvious as a result of this amazing formula. Yesterday was Ramanujan's birthday and the twitterverse has been awash with posts about his life and his formulae. He was born on December 22nd 1887 in Tamil Nadu's Erode. 

ADDENDUM: January 16th 2023

Here is another formula for the digits of \( \pi\) that I came across in a tweet:

I won't say anymore about the formula here. I'm just mentioning it and maybe it can form the basis of a future post.

Linear Prime Chains

On Thursday, June 10th of 2019, I posted about Generalised Cunningham Chains while linking to earlier posts that discussed the topic of Cunningham chains. Today I was reminded of these chains because the number associated with my diurnal age has the following property:

A124017

Numbers \(n\) for which \(2n-1\), \(4n-1\), \(8n-1\), \(16n-1\) and \(32n-1\) are primes.

The initial members of the sequence are:

45, 90, 26820, 26925, 30705, 31710, 33375, 63420, 63570, 71805, 83865, 93075, 103185, 127140, 134025, 148050, 170460, 202635, 211035, 223305, 269505, 297225, 303660, 329175, 335625, 362505, 387975, 405270, 405405, 406425, 409755, 463335

In the case of 26925, the sequence of primes generated is 53849, 107699, 215399, 430799 and 861599. At first I thought the sequence was a Cunningham chain of the second kind with a length of 5. This type of chain has the property that  \(p_{i+1} = 2p_{i} − 1\) for all \(1 \leq i \leq 5 \) but, as can be seen, it is the new number that is doubled and not the original one. In other words, a Cunningham chain of the second kind with a length of 6 and a starting prime of \(p\) would give a sequence of \(p, 2p-1, 4p-3, 8p-7, 16p-15, 32p-31\). Not the same sequence at all.

I was interested in pushing the number of primes further and test for \(64n-1\). Up to one million this yields the following sequence of numbers with their prime chains attached:

45 --> 89 179 359 719 1439 2879 

31710 --> 63419 126839 253679 507359 1014719 2029439 

63570 --> 127139 254279 508559 1017119 2034239 4068479 

202635 --> 405269 810539 1621079 3242159 6484319 12968639 

405405 --> 810809 1621619 3243239 6486479 12972959 25945919 

534600 --> 1069199 2138399 4276799 8553599 17107199 34214399 

561330 --> 1122659 2245319 4490639 8981279 17962559 35925119 

589305 --> 1178609 2357219 4714439 9428879 18857759 37715519 

666945 --> 1333889 2667779 5335559 10671119 21342239 42684479 

799350 --> 1598699 3197399 6394799 12789599 25579199 51158399 

903045 --> 1806089 3612179 7224359 14448719 28897439 57794879 

979125 --> 1958249 3916499 7832999 15665999 31331999 62663999 

How many numbers up to one million will generate primes up to \(128n-1\)? Well as it turns out, only one number. Beyond \(128n-1\), there are no chains in the range up to one million.

561330 --> 1122659 2245319 4490639 8981279 17962559 35925119 71850239 

Just as with generalised Cunningham chains, we can also generalise this particular prime chain sequence by changing the value of the subtrahend. Let's change it to 3 and look for prime chains starting with \(2n-3\) and ending with \(128n-3\). Here's the result and there are seven numbers instead of just one in the range up to one million:

3025 --> 6047 12097 24197 48397 96797 193597 387197 

238865 --> 477727 955457 1910917 3821837 7643677 15287357 30574717 

253880 --> 507757 1015517 2031037 4062077 8124157 16248317 32496637 

477730 --> 955457 1910917 3821837 7643677 15287357 30574717 61149437 

507760 --> 1015517 2031037 4062077 8124157 16248317 32496637 64993277 

680185 --> 1360367 2720737 5441477 10882957 21765917 43531837 87063677 

883180 --> 1766357 3532717 7065437 14130877 28261757 56523517 113047037 

We even get 2 numbers that generate a chain that goes up to \(256n-3\). They are:

238865 --> 477727 955457 1910917 3821837 7643677 15287357 30574717 61149437 

253880 --> 507757 1015517 2031037 4062077 8124157 16248317 32496637 64993277  

There are no numbers in the range that generate primes up to \(512n-1\). Let's try a subtrahend of 5 and see what numbers generate primes in the range from \(2n-5\) up to \(128n-5\). The results are:

 273 --> 541 1087 2179 4363 8731 17467 34939 

 933 --> 1861 3727 7459 14923 29851 59707 119419 

 54558 --> 109111 218227 436459 872923 1745851 3491707 6983419 

 59553 --> 119101 238207 476419 952843 1905691 3811387 7622779 

 65811 --> 131617 263239 526483 1052971 2105947 4211899 8423803 

447678 --> 895351 1790707 3581419 7162843 14325691 28651387 57302779 

887226 --> 1774447 3548899 7097803 14195611 28391227 56782459 113564923 

There are no numbers up to one million that generate primes up to \(256n-5\). What about a subtrahend of 7 and a range from \(2n-7\) to \(128n-7\)? The results are:

27360 --> 54713 109433 218873 437753 875513 1751033 3502073 

517875 --> 1035743 2071493 4142993 8285993 16571993 33143993 66287993 

524805 --> 1049603 2099213 4198433 8396873 16793753 33587513 67175033 

549915 --> 1099823 2199653 4399313 8798633 17597273 35194553 70389113 

749580 --> 1499153 2998313 5996633 11993273 23986553 47973113 95946233 

There are no numbers that generate chains up to \(256n-7\). I think that's enough for the subtrahends but what about the coefficients? We could generalise further by looking at even coefficients larger than 2. If we choose odd coefficients, then every odd number will produce an even number when an odd subtrahend is subtracted. Let's try \(4n-11\) in the range up to \(256n-11\). The results are:

187123 --> 748481 2993957 11975861 47903477 191613941 766455797 3065823221 12263292917 

363375 --> 1453489 5813989 23255989 93023989 372095989 1488383989 5953535989 23814143989 

426040 --> 1704149 6816629 27266549 109066229 436264949 1745059829 6980239349 27920957429 

723330 --> 2893309 11573269 46293109 185172469 740689909 2962759669 11851038709 47404154869 

The reason that 187123 is marked in red is that it is the only starting number that is prime. This of course is rather interesting. All the other long chains are generated by composite numbers. Here is a permalink that allows for experimentation with prime chains generated by a number \(n\) using the linear mapping of \(a \cdot n-b\) where \(a\) and \(b\) are integers.

For a post that has relevance to the content of this post, visit Prime Producing Linear Polynomials. After all, what I've been investigating in this post is simply a variety of prime producing linear polynomials of the general from \(y=ax+b\) where \(a,b\) and \(x\) are restricted to integer values.

The Yellowstone Permutation

Before I discovered the "Yellowstone Permutation, I first came across the Enots Wolley sequence where Enots Wolley is Yellowstone with the letters in reverse order. Thus was I lead to the Yellowstone sequence that represents a permutation of the natural numbers and, like them, is infinite. It gets its name from the spiking, geyser like appearance when plotted (see Figure 1). The primes in the sequence appear in their natural order although this is a conjecture for which there is as yet no proof. The sequence is described as follows:

A098550

The Yellowstone permutation: \(a(n) = n\) if \(n \leq 3\), otherwise the smallest number not occurring earlier having at least one common factor with \(a(n-2)\), but none with \(a(n-1)\).

The initial terms are:

1, 2, 3, 4, 9, 8, 15, 14, 5, 6, 25, 12, 35, 16, 7, 10, 21, 20, 27, 22, 39, 11, 13, 33, 26, 45, 28, 51, 32, 17, 18, 85, 24, 55, 34, 65, 36, 91, 30, 49, 38, 63, 19, 42, 95, 44, 57, 40, 69, 50, 23, 48, 115, 52, 75, 46, 81, 56, 87, 62, 29, 31, 58, 93, 64, 99, 68, 77, 54, 119, 60 

The plot of these points is shown in Figure 1:

Figure 1

The only fixed points seem to be 1, 2, 3, 4, 12, 50 and 86. This has been tested up to 100 million terms. To quote from the OEIS comments:

The first 250000 points lie on about 8 roughly straight lines, whose slopes are approximately 0.467, 0.957, 1.15, 1.43, 2.40, 3.38, 5.25 and 6.20. The first six lines seem well-established, but the two lines with highest slope at present are rather sparse. Presumably as the number of points increases, there will be more and more lines of ever-increasing slopes.

However, subsequently in the comments, it's stated that:

The eight roughly straight lines mentioned above are actually curves. A good fit for the "line" with slope \( \approx 1.15\) is:$$a(n) \approx n(1+1.0/\log(n/24.2))$$ and a good fit for the other "lines" is:$$a(n) \approx (c/2) \cdot n(1-0.5/\log(n/3.67))$$for \(c = 1,2,3,5,7,11,13\). The first of these curves consists of most of the odd terms in the sequence. The second family consists of the primes (\(c=1\)), even terms (\(c=2\)), and \(c \cdot \text{prime } (c=3,5,7,11,13, \cdots) \). 

Figure 2 shows the graph for the first 300,000 terms:

 
Figure 2

There is Python code provided at this site but I can't get it to run.  I'll deal with the Enots Wolley sequence in a later post.

Integration Practice

For some reason I was pondering the integration of the function:$$ \frac{x}{\sqrt{1-x^2}}$$the other day and decided to give it a try. I spend so much time on number theory that I know my integration skills are atrophying so this was an opportunity to get in a little practice both with integration and with honing my LaTeX skills. Given that the function is undefined at \(x=1\), I decided that I would integrate between 0 and 1 and thus I was confronted with the integral:$$  \int_0^1 \! \frac{x}{\sqrt{1-x^2}} \, dx$$This of course is a very easy integral once we make the substitution \(x=\sin \theta\) where \(dx=\cos \theta \, d\theta\). The integration limits will change and become 0 to \(\pi/2\). The integral thus becomes:$$ \begin{align} \int_0^{\pi/2} \! \frac{\sin \theta \cdot \cos \theta}{\sqrt{1-\sin^2 \theta}} \, d\theta &= \int_0^{\pi/2} \! \frac{\sin \theta \cdot \cos \theta}{\cos \theta} \, d\theta \\&= \int_0^{\pi/2} \! \sin \theta \, d\theta \\ &= - \biggl [ \cos \theta \biggr ]_0^{\pi/2} \\ &=1 \end{align}$$It's interesting to note that whatever power \(x\) is raised to, the denominator will always disappear after the substitution. That is to say that for an integer \(n>0\) we have:$$\int_0^1 \! \frac{x^n}{\sqrt{1-x^2}} \, dx=\int_0^{\pi/2} \! \sin^n \theta \, d\theta$$Referring to a very useful site - https://www.integral-calculator.com/ - I was reminded of the famous reduction formula for this type of integral, namely: $$ \int \! \sin^n \theta \, d\theta = \frac{n-1}{n} \int \! \sin^{n-2} \, d\theta - \frac{ \cos \theta \cdot \sin^{n-1} \theta}{n}$$Let's apply this formula progressively for \(n\)=2, 3 and 4.

The case of \(n=2\):$$ \begin{align} \int_0^{\pi/2} \! \sin^2 \theta \, d\theta &=\frac{1}{2}\int_0^{\pi/2} \! 1\, d\theta - \biggl [ \frac{ \cos \theta \cdot \sin \theta }{2} \biggr ]_0^{\pi/2}\\&=\frac{\pi}{4} \end{align} $$What we notice is that there will always be a \( \sin \theta \cdot \cos \theta \) term on the far right as the reduction proceeds but this can be ignored because of our limits: the result will always be zero. This simplification means that we now have:$$ \int_0^{\pi/2} \! \sin^n \theta \, d\theta = \frac{n-1}{n} \int_0^{\pi/2} \! \sin^{n-2} \theta \, d\theta$$Let's test this out for the case of \(n=3\) and \(n=4\):

The case of \(n=3\):$$ \begin{align}  \int_0^{\pi/2} \! \sin^3 \theta \, d\theta &= \frac{2}{3} \int_0^{\pi/2} \! \sin \theta \, d\theta \\ &= \frac{2}{3} \end{align}$$The case of \(n=4\):$$ \begin{align}  \int_0^{\pi/2} \! \sin^4 \theta \, d\theta &= \frac{3}{4} \int_0^{\pi/2} \! \sin^2 \theta \, d\theta \\ &= \frac{3 \pi}{16} \end{align}$$The integral calculator site not only provides the steps but also a graph. Figure 1 shows the area under the curve for:$$  \int_0^1 \! \frac{x^4}{\sqrt{1-x^2}} \, dx$$

Figure 1

It's easy to take the integral calculator for granted but it does a rather remarkable job.

Here is a description of how it works, quoted from the website:

For those with a technical background, the following section explains how the Integral Calculator works.

First, a parser analyzes the mathematical function. It transforms it into a form that is better understandable by a computer, namely a tree (see figure below). In doing this, the Integral Calculator has to respect the order of operations. A specialty in mathematical expressions is that the multiplication sign can be left out sometimes, for example we write "5x" instead of "5*x". The Integral Calculator has to detect these cases and insert the multiplication sign.

The parser is implemented in JavaScript, based on the Shunting-yard algorithm, and can run directly in the browser. This allows for quick feedback while typing by transforming the tree into LaTeX code. MathJax takes care of displaying it in the browser.

When the "Go!" button is clicked, the Integral Calculator sends the mathematical function and the settings (variable of integration and integration bounds) to the server, where it is analyzed again. This time, the function gets transformed into a form that can be understood by the computer algebra system Maxima.

Maxima takes care of actually computing the integral of the mathematical function. Maxima's output is transformed to LaTeX again and is then presented to the user. The antiderivative is computed using the Risch algorithm, which is hard to understand for humans. That's why showing the steps of calculation is very challenging for integrals.

In order to show the steps, the calculator applies the same integration techniques that a human would apply. The program that does this has been developed over several years and is written in Maxima's own programming language. It consists of more than 17000 lines of code. When the integrand matches a known form, it applies fixed rules to solve the integral (e. g. partial fraction decomposition for rational functions, trigonometric substitution for integrands involving the square roots of a quadratic polynomial or integration by parts for products of certain functions). Otherwise, it tries different substitutions and transformations until either the integral is solved, time runs out or there is nothing left to try. The calculator lacks the mathematical intuition that is very useful for finding an antiderivative, but on the other hand it can try a large number of possibilities within a short amount of time. The step by step antiderivatives are often much shorter and more elegant than those found by Maxima.

The "Check answer" feature has to solve the difficult task of determining whether two mathematical expressions are equivalent. Their difference is computed and simplified as far as possible using Maxima. For example, this involves writing trigonometric/hyperbolic functions in their exponential forms. If it can be shown that the difference simplifies to zero, the task is solved. Otherwise, a probabilistic algorithm is applied that evaluates and compares both functions at randomly chosen places. In the case of antiderivatives, the entire procedure is repeated with each function's derivative, since antiderivatives are allowed to differ by a constant.

The interactive function graphs are computed in the browser and displayed within a canvas element (HTML5). For each function to be graphed, the calculator creates a JavaScript function, which is then evaluated in small steps in order to draw the graph. While graphing, singularities (e. g. poles) are detected and treated specially. The gesture control is implemented using Hammer.js.

If you have any questions or ideas for improvements to the Integral Calculator, don't hesitate to write me an e-mail.