Divisors and Antidivisors: A Fresh Perspective

I've written about antidivisors before in posts titled Anti-Divisors and More on Anti-Divisors. Here is a fresh take. My diurnal age today is 28134 and the antidivisors of this number are 4, 12, 36, 108, 2084, 6252 and 18756. If we concatenate these numbers from left to right we get a rather large number (412361082084625218756) with the property that:$$ \frac{412361082084625218756}{28134} = 14657037111133334$$Numbers with this property form OEIS A249764:

 

A249764: numbers which divide the concatenation, in ascending order, of their anti-divisors.

Up to 40000, these numbers are relatively rare (permalink): 

15, 30, 105, 120, 150, 222, 375, 585, 1500, 1695, 1755, 1800, 2700, 3449, 3750, 3840, 4891, 6720, 7680, 12000, 13583, 14400, 15000, 18750, 19200, 20940, 28134, 30000, 34800, 35625

The OEIS lists an analog of this, namely OEIS A069872:

 

A069872: numbers \(k\) such that \(k\) divides the concatenation all divisors in ascending order.

Comparatively, members of this sequence are rather more numerous. There are 181 numbers in the range up to 40000 (permalink):

1, 2, 4, 5, 6, 8, 10, 15, 16, 20, 24, 25, 30, 32, 40, 50, 60, 64, 80, 90, 96, 100, 104, 120, 124, 125, 128, 150, 160, 200, 240, 250, 255, 256, 288, 320, 360, 375, 380, 384, 400, 425, 464, 480, 495, 500, 512, 600, 618, 625, 640, 750, 795, 800, 864, 875, 960, 1000, 1024, 1110, 1230, 1250, 1280, 1300, 1390, 1400, 1408, 1440, 1469, 1500, 1525, 1536, 1600, 1632, 1920, 2000, 2048, 2050, 2250, 2400, 2500, 2556, 2560, 2910, 2944, 2952, 3000, 3040, 3125, 3200, 3330, 3360, 3625, 3750, 3825, 3840, 4000, 4096, 4304, 4625, 4650, 4992, 5000, 5120, 5250, 5280, 5300, 5568, 5760, 6000, 6144, 6150, 6250, 6400, 6528, 6750, 7168, 7200, 7560, 7680, 8000, 8192, 9000, 9330, 9375, 9600, 10000, 10240, 10500, 10752, 11875, 12000, 12500, 12800, 13410, 13600, 14000, 15000, 15360, 15625, 15680, 16000, 16384, 16640, 17500, 17920, 18432, 18750, 19710, 20000, 20200, 20250, 20480, 20610, 21600, 21760, 22752, 23040, 23375, 24000, 24120, 24576, 25000, 25600, 25984, 27000, 28125, 30720, 31250, 32000, 32768, 33930, 34480, 34800, 35000, 36000, 37500, 37830, 38400, 39000, 40000

A variation on this is OEIS A240265:

A240265: numbers that divide the concatenation of their aliquot divisors, in ascending order.

These are less numerous. In the range up to 40000, they are:

4, 15, 16, 255, 375, 495, 795, 1469, 3825, 9375, 28125

All these numbers are also in OEIS A069872 and are marked in red in that sequence's list of members. It can be seen that 28125 is rather special and this number marked my diurnal age only 9 days ago. However, at the time, I missed its membership in these two sequences.

Aliquant Parts

I came across a mathematical term today that I hadn't heard of before. The term is "aliquant" defined as follows by contrasting it to similar sounding "aliquot" (source):

Webster defines 'aliquot' as something that contained an exact number of times in something else or to divide into equal parts.

Notice the word "equal". An example being 5 is an aliquot part of 15.

The term 'aliquant', however, is slightly different. Defined as being a part of a number or quantity, but not dividing it without leaving a remainder. An example being 5 is an aliquant part of 16. 

The term occurred in the following context:

 A098743: number of partitions of \(n\) into aliquant parts (i.e., parts that do not divide \(n\)). 

The initial members of the sequence are:

1, 0, 0, 0, 0, 1, 0, 3, 1, 3, 3, 13, 1, 23, 10, 11, 9, 65, 8, 104, 14, 56, 66, 252, 10, 245, 147, 206, 77, 846, 35, 1237, 166, 649, 634, 1078, 60, 3659, 1244, 1850, 236, 7244, 299, 10086, 1228, 1858, 4421, 19195, 243, 17660, 3244, 12268, 4039, 48341, 1819, 27675

The last member shown above, 27675, is my diurnal age today and corresponds to \(n=55\). To find the aliquant parts, simply remove the divisors of the number. For example, 55 has divisors of 1, 5, 11 and 55 and so the number of aliquant parts is 51. The list is as follows:

2, 3, 4, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54

Note that these are different to the numbers that contribute to the total of the totient of a number, where the numbers are coprime. The totient for 55 is 40, made up of the following numbers.

1, 2, 3, 4, 6, 7, 8, 9, 12, 13, 14, 16, 17, 18, 19, 21, 23, 24, 26, 27, 28, 29, 31, 32, 34, 36, 37, 38, 39, 41, 42, 43, 46, 47, 48, 49, 51, 52, 53, 54

A Variation on the Taxi Cab Number

The number 1729 is famous as the so-called "taxi cab number" in memory of the interchange between the mathematicians Hardy and Ramanujan in which the latter observed that the number of the taxi cab in which the former had arrived at the hospital was far from boring (as Hardy had thought). Instead 1729 is the first positive integer that is the sum of two positive cubes in two different ways:$$ \begin{align} 1729 &= 1^3+12^3\\ &= 9^3+10^3 \end{align} $$Today I observed a taxi with the number plate T 3257 and noted that there is a touch of Fibonacci about its digits because:$$ 3 +2 = 5 \text{ and } 2 + 5 = 7$$The digits thus form a Fibonacci-type sequence:$$ 3, 2, 5, 7$$This got me thinking about what numbers with three or more digit have this Fibonacci-like property. Well, up to one million, there are only 82 such numbers so they form a rather exclusive club. 

Here they are (permalink) in a sequence that we'll called the ADDITION SEQUENCE:

101, 112, 123, 134, 145, 156, 167, 178, 189, 202, 213, 224, 235, 246, 257, 268, 279, 303, 314, 325, 336, 347, 358, 369, 404, 415, 426, 437, 448, 459, 505, 516, 527, 538, 549, 606, 617, 628, 639, 707, 718, 729, 808, 819, 909, 1011, 1123, 1235, 1347, 1459, 2022, 2134, 2246, 2358, 3033, 3145, 3257, 3369, 4044, 4156, 4268, 5055, 5167, 5279, 6066, 6178, 7077, 7189, 8088, 9099, 10112, 11235, 12358, 20224, 21347, 30336, 31459, 40448, 101123, 112358, 202246, 303369

I have made a related post titled Additive Fibonacci-like Numbers on the 7th August 2024 but this involved finding the digital roots of numbers unlike what I've done here. So for me 3257 will remain my personal taxi cab number.

Another sequence will emerge if, instead of adding the second number to the first and so on, we SUBTRACT the second from the first and so on. In this scenario, 3211 would satisfy because:$$  3 - 2 = 1 \text{ and } 2 -1 =1$$Up to one million, there are 99 such numbers of three digits or more. Here they are in a sequence we'll call the SUBTRACTION SEQUENCE (permalink):

101, 110, 202, 211, 220, 303, 312, 321, 330, 404, 413, 422, 431, 440, 505, 514, 523, 532, 541, 550, 606, 615, 624, 633, 642, 651, 660, 707, 716, 725, 734, 743, 752, 761, 770, 808, 817, 826, 835, 844, 853, 862, 871, 880, 909, 918, 927, 936, 945, 954, 963, 972, 981, 990, 1101, 2110, 2202, 3211, 3303, 4220, 4312, 4404, 5321, 5413, 5505, 6330, 6422, 6514, 6606, 7431, 7523, 7615, 7707, 8440, 8532, 8624, 8716, 8808, 9541, 9633, 9725, 9817, 9909, 21101, 32110, 42202, 53211, 63303, 64220, 74312, 84404, 85321, 95413, 96330, 321101, 532110, 642202, 853211, 963303

It can be noted that some numbers containing zero feature in both sequences. These numbers are 101, 202, 303, 404, 505, 606, 707, 808 and 909.

While we're at it why not consider multiplication in which the first two digits multiply together to give the third digit and so on. In the range up to one million, there are 78 such numbers and here they are in a sequence we'll call the MULTIPLICATION SEQUENCE (permalink):

100, 111, 122, 133, 144, 155, 166, 177, 188, 199, 200, 212, 224, 236, 248, 300, 313, 326, 339, 400, 414, 428, 500, 515, 600, 616, 700, 717, 800, 818, 900, 919, 1000, 1111, 1224, 1339, 2000, 2122, 2248, 3000, 3133, 4000, 4144, 5000, 5155, 6000, 6166, 7000, 7177, 8000, 8188, 9000, 9199, 10000, 11111, 12248, 20000, 21224, 30000, 31339, 40000, 50000, 60000, 70000, 80000, 90000, 100000, 111111, 200000, 212248, 300000, 400000, 500000, 600000, 700000, 800000, 900000, 1000000

An example is 212248 where we have:$$2 \times 1 = 2, \, 2 \times 1 = 2, \, 2 \times 2 = 4 \text{ and } 4 \times 2 = 8$$The zeros of course make some of these numbers a little trivial and so with the digit 0 excluded we have 41 suitable numbers (permalink) in a sequence we'll call the MULTIPLICATION WITHOUT ZERO SEQUENCE:

111, 122, 133, 144, 155, 166, 177, 188, 199, 212, 224, 236, 248, 313, 326, 339, 414, 428, 515, 616, 717, 818, 919, 1111, 1224, 1339, 2122, 2248, 3133, 4144, 5155, 6166, 7177, 8188, 9199, 11111, 12248, 21224, 31339, 111111, 212248

If we consider dividing the second digit into the first to give the third digit and so on then, excluding numbers with zero, we have the following numbers (permalink) in what we'll call the DIVISION WITHOUT ZERO SEQUENCE:

111, 212, 221, 313, 331, 414, 422, 441, 515, 551, 616, 623, 632, 661, 717, 771, 818, 824, 842, 881, 919, 933, 991, 1111, 2212, 3313, 4221, 4414, 5515, 6616, 7717, 8422, 8818, 9331, 9919, 11111, 42212, 84221, 93313, 111111, 842212

Many of the numbers in the division sequence are not surprisingly the reverse of numbers in the multiplication sequence e.g. 842212 in the division sequence is the reverse of 212248 in the multiplication sequence.

Forming Digit Equations: A Game

Source

It occurred to me how forming digit equations from five digits might make for an interesting game. After all, Wordle consists of guessing the five letters that make up a hidden word. In the game I'm conceiving of, a five digit number between 10000 and 99999 would be generated. Let's say the number generated is 25529. The challenge is to use only the following operators to form a digit equation without altering the order of the digits:

+    addition

-    subtraction

x    multiplication

/    divide by

|    divide into

^    exponentiation

(     left bracket

)    right bracket

Figure 1 shows one possible representation:

Figure 1

The same digit equation would look mathematically as follows:$$ \Big ( 2 + \frac{5}{5} \Big )^2=9$$This is somewhat on the difficult side. Obviously an effective game nowadays involves a catchy user interface so coloured balls that pop up to start the game would be effective. The operators shown above would need to be able to be dragged and dropped between the digits in order to generate the equation. The successful creation of an equation would need to be displayed in mathematically readable way such as was done with the example of 25529.

The game could have different levels starting at the fairly elementary level. For example, 10348 could be rendered as:$$1+0+3+4=8$$This equation involves only the a single operation, addition, and the correct placement of the equal sign. By contrast, 25529 is considerably more challenging and involves the use of brackets, addition, division and exponentiation. Of course, for some numbers it is impossible to form an equation within the imposed constraints. An example is 27497 which is unsolvable as far as I can see. This is a long term project but something that I will keep thinking about or perhaps I'll discover that someone has already created such a game. Who knows?

For earlier posts on this theme see my posts titled The Number Plate Game from the 29th June 2024 and Forming Equations from the Digits of a Number on 14th March 2024. To generate some random five digits numbers follow this permalink.

Forming Equations from the Digits of a Number

There was a post that I made to my Pedagogical Posturing blog in August of 2013 before I created this mathematical blog in the second half of 2015. The title was "Forming Equations from Integer Sequences" and that title was perhaps a little misleading. At the time, I wasn't aware of the OEIS or Online Encyclopedia of Integer Sequences. What I meant was the sequence of digits that define a number. For example, today my diurnal age is 27374 and so the sequence of digits is 2, 7, 3, 7, 4. Concatenation of the digits is not allowed. Thus we cannot have 27, 3, 7, 4 for example. 

In that long ago post, I wrote:

Recently I've been using Twitter to create a daily tweet that records my "day count" (number of days I've been alive) plus its factors (if not prime) and some interesting facts about the number itself or one of its factors. Sometimes there's little to say about the number and in such cases I've found that I can usually form an equation by inserting mathematical operators between one or more of the digits. 

For example, yesterday the count was 23518 and 23 - 5 = 18. Today the count is 23519 and 2 + 3 + 5 - 1 = 9. I was wondering if it's always possible to create an equation from five digits using the standard mathematical operators (addition, subtraction, multiplication, division and exponentiation in combination with brackets). Obviously with just two digits, it's only possible when the digits are repeated e.g. 99 becomes 9=9. With three digits, it's sometimes possible e.g. 819 becomes 8 + 1 = 9 but generally it isn't e.g. 219. With four digits, it's more likely e.g. 2119 becomes -2 + 11 = 9 but I'm doubtful whether this is always so. There must come a point however, where the number of digits is sufficient to ensure that it's always so. Maybe five digits is that point.

From now on, I'll try each day to form an equation to test out this theory. For example, tomorrow the count is 23520 which becomes 2 + 3 - 5 = 2 x 0 and it works for tomorrow but beyond that let's see.

Needless to say I didn't "try each day to form an equation to test out this theory" but it might be time to give it a go. The 27374 of my diurnal age today is an easy one:$$2 \times 7 -3=7+4$$However, yesterday's number, 27373, doesn't prove so easy. It seems that having two 3's and two 7's in the number makes things difficult. The conditions that I imposed in the original blog post were the use of only the standard mathematical operators of addition, subtraction, multiplication, division and exponentiation in combination with brackets. These operators needed to be applied to the digits in the order in which they occurred.

One modification that I will make here is to allow \(x \, | \,y\) meaning \(x\) is divided into \(y\) as opposed to \(x/y\) meaning \(x\) is divided by \(y\). This seems quite reasonable as it still only involves the operation of division but allows more flexibility. Its application doesn't seem to help in the case of 27373. If we allow the operator \(x\) // \(y\) meaning return the whole number part of the dividend, then an equation is possible:$$2|(7-3)=7//3$$If we allow // then we could allow \(x\) % \( y\) meaning return the remainder as a whole number when \(x\) is divided by \(y\). Thus 7 % 3 = 1.  This might be termed modulo division.

I can't see any way to create an equation from 27373 without extending the original conditions. Even concatenation of the digits doesn't seem to help. A common symbol for concatenation is || and thus 2 || 7 = 27. However, I've stipulated that the digits are to be treated as separate so I'll adhere to that condition. The best approach is to stick to the original conditions and if a solution is not possible, then // and % can be resorted to.

In the case of 27374, there is more than one way to create an equation. Here is another way:$$ 2 \times 7 -(3+7)=4 $$So what I'll try to do is to reassert my original goal of trying each day to form an equation and see what patterns emerge.

This activity of forming a "digit equation" is not all that different from one of Quanta's mathematical games called "Hyperjumps". See Figure 1.

Figure 1

Representing Numbers With Digits

My previous post focused on the number 1089 in which I made reference to a newly discovered blogging site at https://math1089.in/ and in particular to a post about the number 1089. In that post it was noted that:$$ \begin{align} 1089 &= 12 \times 3^4 + 5 \times 6 + 78 + 9\\1089 &= 987 + 65 + 4 + 32 + 1 \end{align}$$In another post about the number 108, it was noted that:$$ \begin{align} 108 &= 1 + 2 + 3 + 4 + 5 + 6 + 78 + 9\\108 &= 9 + 8 \times 7 + 6 \times 5 + 4 \times 3 + 2 ‒ 1 \end{align}$$This got me thinking about whether it was possible to represent every number at least once in terms of consecutive single or concatenated digits separated by the basic operations of arithmetic combined with exponentiation and brackets.

For example, my diurnal age today is 27339. Is such a feat possible for this number? The determination is not easy because there are just  so many possible ways to combine the digits from 1 to 9. I spent quite some time playing around with the possibilities and I did come close but not close enough. The exercise is oddly addictive. There's no serious mathematics involved in the exploration. It's more in the nature of a puzzle, like Sudoku. 

There are a variety of strategies, one of which is to establish base points with as few digits as possible. An example of this is:$$35016=1+2+3+4+5+6^7/8+9$$Here the digits 6, 7 and 8 combine to form 34992 and the remaining digits are free to be manipulated. It's true that we can use 34567 to get to a similar number but here five digits are tied up and there's little that can be done with the remaining digits (1, 2 and 8, 9) since they are separated. Here is an example of how we can get close to 27339 using the earlier mentioned base point:$$28084=-(12^3 \times 4+5)+6^7/8+9$$Another approach is to use factors. We know that 27339 = 3 x 13 x 701 and it's easy enough to create the first two factors using 1 + 2 = 3 and 3 x 4 - 5 + 6 = 13. However, we are then left with 789:$$ \begin{align} 27339 &=3 \times 13 \times 701\\30771 &= (1+2) \times (3 \times 4 - 5 +6) \times 789\\ &=3 \times 13 \times 789 \end{align}$$While this works fine for 30771, it's not of much use for 27339. In general, this approach is quite restrictive and the more promising approaches will involve additions and subtractions along with exponentiation, multiplication and division.

At the moment I don't have a solution to the specific problem of representing 27339 in terms of sequential digits and I certainly don't have an answer to the general problem of whether such a representation is always possible or only sometimes possible. Certainly there's an upper limit on the number size and this is imposed by the nine digit restriction but such a limit is huge and my investigation is focusing on numbers in the region of 30000. My suspicion is that it's not always possible within the restrictions imposed. If we relax the requirement that the digits need to be in sequential order or we allows square roots, factorials etc. then maybe it's possible but for the moment I'll keep within the earlier rules and keep revisiting the problem from time to time.

Forming Equations from Integers

Read on!

Once upon a time, before I started this exclusively mathematical blog in 2015, I used to post occasional mathematical content to my Pedagogical Posturing blog at https://voodoo-guru.blogspot.com. I was looking back at some of these posts and noticed this one, from Saturday, 24 August 2013, titled Forming Equations from Integer Sequences. It's only short so I'll quote it in full:

Recently I've been using Twitter to create a daily tweet that records my "day count" (number of days I've been alive) plus its factors (if not prime) and some interesting facts about the number itself or one of its factors. Sometimes there's little to say about the number and in such cases I've found that I can usually form an equation by inserting mathematical operators between one or more of the digits. 

For example, yesterday the count was \(23518\) and \(23 - 5 = 18.\) Today the count is \(23519\) and \(2 + 3 + 5 - 1 = 9\). I was wondering if it's always possible to create an equation from five digits using the standard mathematical operators (addition, subtraction, multiplication, division and exponentiation in combination with brackets). Obviously with just two digits, it's only possible when the digits are repeated e.g. \(99\) becomes \(9 = 9\). With three digits, it's sometimes possible e.g. \(819\) becomes \(8 + 1 = 9\) but generally it isn't e.g. \(219\). With four digits, it's more possible e.g. \(2119\) becomes \(-2 + 11 = 9\) but I'm doubtful whether this is always so. There must come a point however, where the number of digits is sufficient to ensure that it's always so. Maybe five digits is that point.

From now on, I'll try each day to form an equation to test out this theory. For example, tomorrow the count is \(23520\) which becomes \(2+3-5=2 \times 0 \) and it works for tomorrow but beyond that let's see.

Well, I didn't keep my promise of trying to form an equation each day from the digits making up my diurnal age. I have however, written about selfie numbers in a post of Friday, 27th March 2020 to my Mathematical Meanderings blog site. These are somewhat similar in spirit. In that post, I also mention Friedman numbers that can be described as follows:

Consider \(28547 =(8+5)^4−(7 \times 2)\) expressed in base 10, both sides use the same digits. An integer is a Friedman number if it can be put into an equation such that both sides use the same digits but the right hand side has one or more basic arithmetic operators (addition, subtraction, multiplication, division, exponentiation) interspersed. Brackets, as usual, are essential to clarify the order of operations. These numbers are named after Erich Friedman, Assoc. Professor of Mathematics at Stetson University. With the help of his students he has researched Friedman numbers in bases 2 through 10 and even with Roman numerals. When both sides use the digits in the same order, the number is called a ”nice” or ”strong” Friedman number. For example, \( 3125=(3+[1\times2])^5.\)

My approach is similar to this except I'm trying to create an equation from the digits (using them in the same order as they appear in the number). As another example, today I'm \(26121\) days old and this is an easy one because \( (2 \times 6)/12=1 \) if we allow concatenation of digits. If only individual digits are allowed, then \(-(2-6)=1+2+1 \) satisfies. It's probably better to use the individual digits as is done with the Friedman numbers. I'll try to include this as part of my daily number analysis. 

Going back a few days, we have:

• \(26120 \text{ --> } 2 = \frac{6}{1 + 2 +0} \)
• \(26119 \text{ --> } 2 + 6 \times 1 = -1 + 9 \)
• \(26118 \text{ --> } 2 + 6 + 1 = 1 + 8 \)
• \(26117 \text{ --> } 2 + 6 = 1 \times 1 + 7 \)
• \(26116 \text{ --> } 2 + 6 = 1 + 1 + 6 \)
• \(26115 \text{ --> } -2 + 6 + 1 \times 1 = 5 \)
• \(26114 \text{ --> } -2 + 6 = 1 - 1 + 4 \)
• \(26113 \text{ --> } -2 + 6 = 1 \times 1 + 3 \)
• \(26112 \text{ --> } -2 + 6 = 1 + 1 + 2 \)
• \(26111 \text{ --> } 2 = \frac{6}{1+1+1}\)
• \(26110 \text{ --> } 2 \times 6 \times (1-1) = 0 \)

It seems that it's always possible to form an equation using only brackets and the basic arithmetic operators of addition, subtraction, multiplication, division and exponentiation. Perhaps even the exponentiation is not needed, as the examples above show. Let's see if this assumption holds true for future numbers. This is hardly high level mathematics but it's a simple yet oddly satisfying activity.