Mathematical Meanderings: November 2020

Wednesday, 25 November 2020

Polyiamonds

Figure 1

Link to post on Polyominoes from March 1st 2020.

Just recently I turned 26166 days old and one of the properties of this number is that it is a member of OEIS A000577: number of polyiamonds with $n$ cells (turning over allowed, holes allowed, must be connected along edges). In the case of 26166, the value of $n$ is 14. Figure 1 shows a polyiamond with 13 cells that conforms with the sequence's restrictions. It has a hole but if that hole were filled by a triangle then there would be 14 triangles and its shape would account for one of the 26166 possible polyiamonds.

Wolfram Mathworld gives the following definition of a polyiamond:

A generalisation of the polyominoes using a collection of equal-sized equilateral triangles (instead of squares) arranged with coincident sides. Polyiamonds are sometimes simply known as iamonds.

Figure 2 shows the possible polyiamond shapes formed by one up to six polyiamonds. The categories can be identified by a prefix $n$ indicating the number of polyiamonds that comprise them viz. $n$-polyiamonds. So Figure 2 shows the different types of 1-polyiamonds up to 6-polyiamonds. Alternatively, a 6-polyiamond can be referred to as a hexiamond and so forth as shown in the diagram.

Figure 2

Figure 3

Generally, mirror images are not considered distinct. In other words, any polyiamond can be picked up, flipped over and the two mirror images are considered the same. See Figure 3. Once there are nine or more polyiamonds, holes are possible. Some of these are shown in Figure 4. So returning to OEIS A000577 and Figure 2, it can be seen why the initial terms are 1, 1, 1, 3, 4 and 12. The full sequence, up to $n=14$ is 1, 1, 1, 3, 4, 12, 24, 66, 160, 448, 1186, 3334, 9235, 26166 and 73983. The terms getting rapidly larger as $n$ increases.

The hexiamonds are given special names. Again, quoting from Wolfram Mathworld and referring to Figure 2, these are:

Top row: bar, crook, crown, sphinx, snake, and yacht.
Bottom row: chevron, signpost, lobster, hook, hexagon, butterfly.

Polyiamonds with holes are shown in Figure 4. The number of holes associated with the different categories of polyiamonds are given by OEIS A070764: number of polyiamonds with $n$ cells, with holes. The sequence, up to $n$=17, runs: 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 25, 108, 450, 1713, 6267, 21988, 75185.

Figure 4

These shapes can be put together in interesting ways. For example, Figure 5 shows how 9 hexiamonds can be assembled to form a hexagon while Figure 6 shows how 19 hexiamonds can be assembled to form a more complex hexagonal shape:

Figure 5: source

Figure 6: source

Lots of interesting information, puzzles, tables etc. to be found at:

Of course, another approach is to look at the 3-D connections of polyiamonds. Figure 7 shows how a 20-polyiamond folds to form an icosahedron.

Figure 7: source

This leads into the "deltahedron (plural deltahedra), a polyhedron whose faces are all equilateral triangles. The name is taken from the Greek upper case delta (Δ), which has the shape of an equilateral triangle. There are infinitely many deltahedra, but of these only eight are convex, having 4, 6, 8, 10, 12, 14, 16 and 20 faces". Source.

Clearly, there's a lot of room for exploration in this topic but that's enough for this post.

Saturday, 21 November 2020

Twenty Three

With the 23rd November only two days away, I thought that I'd make a post focused simply on this number 23. I know four people who were born on that day. Firstly, I thought I'd look back over my previous posts and see what references to 23 that I could find:

SUM OF CUBES

Most recently I made a post on Sum of Cubes (October 20th 2020) in which I noted that 23 and 239 are the only integers requiring nine positive cubes for their representation. Only 15 integers require eight cubes: 15, 22, 50, 114, 167, 175, 186, 212, 231, 238, 303, 364, 420, 428, and 454. All other numbers require seven cubes or less. In the case of 23 we have:$$23=1^3+1^3+1^3+1^3+1^3+1^3+1^3+2^3+2^3$$CYCLIC NUMBERS

In a post on Cyclic Numbers (November 23rd 2019), I observed that if the digital period of $1/p$ where $p$ is prime is $p$−1, then the digits represent a cyclic number. 23 is such a prime because it has a period of 22:$$\frac{1}{23}=\text{ 0.0434782608695652173913 }$$Multiplying 0434782608695652173913 progressively by 1 to 22 yields all possible cyclic permutations of this number:

0434782608695652173913 multiplication by 1

0869565217391304347826 multiplication by 2

1304347826086956521739 multiplication by 3

1739130434782608695652 multiplication by 4

2173913043478260869565 multiplication by 5

2608695652173913043478 multiplication by 6

3043478260869565217391 multiplication by 7

3478260869565217391304 multiplication by 8

3913043478260869565217 multiplication by 9

4347826086956521739130 multiplication by 10

4782608695652173913043 multiplication by 11

5217391304347826086956 multiplication by 12

5652173913043478260869 multiplication by 13

6086956521739130434782 multiplication by 14

6521739130434782608695 multiplication by 15

6956521739130434782608 multiplication by 16

7391304347826086956521 multiplication by 17

7826086956521739130434 multiplication by 18

8260869565217391304347 multiplication by 19

8695652173913043478260 multiplication by 20

9130434782608695652173 multiplication by 21

9565217391304347826086 multiplication by 22

Those two posts were the only two that I could find that were of interest regarding 23. Now let's look elsewhere.

BIRTHDAY PARADOX

23 pops up in the birthday paradox where, in a group of 23 (or more) randomly chosen people, the probability is more than 50% that some pair of them will have the same birthday. Here is the explanation (source):

With 23 people we have 253 pairs: $\frac{23 \: 22}{2} = 253$. The chance of two people having different birthdays is:$$1 - \frac{1}{365} = \frac{364}{365} = .997260$$Makes sense, right? When comparing one person's birthday to another, in 364 out of 365 scenarios they won't match. Fine. But making 253 comparisons and having them all be different is like getting heads 253 times in a row -- you had to dodge "tails" each time. Let's get an approximate solution by pretending birthday comparisons are like coin flips.We use exponents to find the probability:$$\big( \frac{364}{365} \big )^{253} = 0.4995$$Our chance of getting a single miss is pretty high (99.7260%), but when you take that chance hundreds of times, the odds of keeping up that streak drop. Fast.

SPECIAL PRIMES

Sophie-Germain prime: a prime $p$ is a Sophie-Germain prime if $2 \times p+1$ is prime. In the case of 23, we have 23 x 2 + 1 = 47, a prime.
Safe prime: a prime $p$ is a safe prime if $ \frac{p-1}{2} $ is prime. In the case of 23, we have $ \frac{23-1}{2} =11$ and 11 is prime.
Cunningham chain: 23 is the next to last member of the first Cunningham chain (a sequence of prime numbers) of the first kind to have five terms (2, 5, 11, 23, 47).
Twin prime: 23 is the smallest odd prime that is not a twin prime.
Woodell prime: a Woodell number $W_n$ is any natural number of the form $ W_{n}=n\cdot 2^{n}-1$ for some natural number $n$. A Woodell prime is simply a Woodell number that is prime. 23 is the second such prime after 7. The progression is 7, 23, 383, 32212254719, ... so they are not that common.
Factorial prime: a factorial prime is a prime number that is one less or one more than a factorial (all factorials > 1 are even). 23 = 4!-1 and so it qualifies.
Eisenstein prime: this is a little complicated but here is a Wikipedia link for learning more about them. The initial Eisenstein primes are 2, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, 101, ...
Smarandache–Wellin prime: an integer that in a given base is the concatenation of the first $n$ prime numbers written in that base is called a Smarandache–Wellin number. If the number is prime, it's called a Smarandache–Wellin prime. The initial such primes are 2, 23 and 2357.
Sum of primes: the sum of the first 23 primes is 874, which is divisible by 23, a property shared by few other numbers.
Repunit prime: a number whose digits are all 1 is called a repunit and if that number is prime, then it is called a repunit prime. The 23 digit number 11111111111111111111111 is such a prime

FACTORIALS

The number 23 is the only prime $p$ such that $p$! is $p$ digits long.

23! = 25852016738884976640000

In fact 23 is one of only four numbers $n$ such that $n$! is $n$ digits long. The others are 1, 22, and 24. The number 23! is the smallest pandigital factorial—it contains each digit at least once.

HUMAN GENOME

Human cells (apart from the sex cells) contain 46 chromosomes: 23 from the mother and 23 from the father. The sex cells contain 23 chromosomes. There are$$2^{23}=8,324,608$$ possible combinations of 23 chromosome pairs and thus$$2^{46}=70,368,744,177,664$$ possible combinations when male and female sex cells combine to produce a human.

CHEMISTRY

Atomic Number: the atomic number is the number of protons in the nucleus and this number uniquely identifies the element. The atomic number of Vanadium is 23 meaning that it has 23 protons in its nucleus. To quote from Wikipedia:

Vanadium is a chemical element with the symbol V and atomic number 23. It is a hard, silvery-grey, malleable transition metal. The elemental metal is rarely found in nature, but once isolated artificially, the formation of an oxide layer (passivation) somewhat stabilizes the free metal against further oxidation.

Figure 1: source

Atomic Mass Number: the atomic mass number is the total number of protons and neutrons (together known as nucleons) in an atomic nucleus. The stable isotope of Sodium (Na) has an atomic mass number of 23 (11 protons and 12 neutrons). To quote from Wikipedia again:

Sodium is a chemical element with the symbol Na (from Latin "natrium") and atomic number 11. It is a soft, silvery-white, highly reactive metal. Sodium is an alkali metal, being in group 1 of the periodic table. Its only stable isotope is $^{23}$Na. The free metal does not occur in nature, and must be prepared from compounds. Sodium is the sixth most abundant element in the Earth's crust and exists in numerous minerals such as feldspars, sodalite, and rock salt (NaCl). Many salts of sodium are highly water-soluble: sodium ions have been leached by the action of water from the Earth's minerals over eons, and thus sodium and chlorine are the most common dissolved elements by weight in the oceans.

ASTRONOMY

To quote from this source:

Today, the Earth's axis is tilted 23.5° from the plane of its orbit around the sun. But this tilt changes. During a cycle that averages about 40,000 years, the tilt of the axis varies between 22.1° and 24.5°. The average of 22.1° and 24.5° is of course 23.3°.

COSMOLOGY

Not that I believe in dark matter but for those that do, it's postulated that 23% of the Universe consists of it. Here is a graph to convince you it's true!

*********************************************

There's a lot more that could be said about the number 23 but that will do for now.

Personal Investigation: Part Four

This post builds on my three previous posts so it's best to look at those first in order to properly understand what I'm doing. These posts are:

It occurred to me that there's probably no reason to maintain $k$ at a constant value and that the exponents could be variable. For example:$$a_n= \text{ sum of digits } \big (a_{n-1}^2+a_{n-2}^3+a_{n-3}^4\big ) \text { with }a_0=0, a_1=1,a_2=2$$Do we eventually end up in a loop? This is what I set out to investigate.

Let's start with the following SageMath code (permalink):

a, b, c= 0, 1, 2
L=[a, b, c]
for x in [1..31]:
   d=a^2+b^3+c^4
   d=sum(d.digits())
   L.append(d)
   a, b, c=b, c, d
print(L)

Here, instead of a single value for $k$, the values are 2,3 and 4. The output of the algorithm reveals that, very quickly, a loop arises:

1, 2, 8, 10, 13, 24, 20, 32, 21, 30, 25, 25, 17, 33, 30, 28, 28, 29, 18, 18, 22, 13, 23, 21, 24, 25, 25, 26, 24, 39, 28, 28, 29

It would seem that, regardless of the starting values or the exponents, the series always ends up looping. For example, take starting values of $a, b, c$= 100, 191, 227 and exponents of 112, 223 and 454. Thus:$$a_n= \text{ sum of digits } \big (a_{n-1}^{112}+a_{n-2}^{223}+a_{n-3}^{454}\big ) \text { with }a_0=100, a_1=191,a_2=227$$Here the series reaches a maximum value 8328 and contains 275 distinct terms.

I've only explored the triad here but there's no reason to suppose that similar results hold true for the tetrad, pentad, hexad and beyond. It would seem that for constant powers, no matter how high, the series eventually repeats.

Thursday, 19 November 2020

Personal Investigation: Part Three

This post builds on my two previous posts so it's best to look at those first in order to properly understand what I'm doing. These posts are:

Looking at the maximal values associated with progressively summing the squares of dyads, triads, tetrads etc. and then finding the sum of their digits, the need for a consistent notational system arose. I wanted to be able to describe that specific term in the sequence with the maximum value for a particular $k$. Thus I came up with $ a_{k_{max, \, N}} $ for this maximum value where $k$ is the exponent and $N$ is the number of terms that are being added together. Let's look at some specific examples:

DYAD

$a_{k_{max, \, N=2}}$ can be used to represent the maximum value attained by $ a_n $ for a particular $k$ when summing two terms progressively, where:$$a_n= \text{ sum of digits } \big (a_{n-1}^k+a_{n-2}^k \big ) \text { with }a_0=0, a_1=1$$The sequence for $2 \leq k \leq 50$ is as follows:

[14, 28, 35, 45, 56, 91, 83, 110, 98, 135, 128, 151, 155, 181, 197, 218, 200, 241, 275, 271, 296, 286, 308, 319, 341, 351, 350, 376, 353, 410, 443, 433, 395, 495, 443, 501, 551, 521, 548, 565, 614, 620, 614, 604, 611, 646, 641, 716, 701]

So we can write $a_{2_{max, \, N=2}}=14$, $a_{3_{max, \, N=2}}=28$ etc.

TRIAD

$a_{k_{max, \, N=3}}$ can be used to represent the maximum value attained by $ a_n $ for a particular $k$ when summing three terms progressively, where:$$a_n= \text{ sum of digits } \big (a_{n-1}^k+a_{n-2}^k +a_{n-3}^k\big ) \text { with }a_0=0, a_1=1,a_2=2$$The sequence for $2 \leq k \leq 50$ is:

[24, 34, 35, 58, 66, 93, 98, 99, 117, 140, 147, 165, 191, 181, 197, 218, 236, 248, 266, 297, 297, 311, 309, 347, 344, 362, 360, 408, 416, 436, 425, 449, 479, 482, 479, 544, 546, 546, 566, 570, 597, 624, 632, 702, 665, 664, 696, 710, 725]

So we can write $a_{2_{max, \, N=3}}=24$, $a_{3_{max, \, N=3}}=34$ etc.

TETRAD

$a_{k_{max, \, N=4}}$ can be used to represent the maximum value attained by $ a_n $ for a particular $k$ when summing four terms progressively, where:$$a_n= \text{ sum of digits } \big (a_{n-1}^k+a_{n-2}^k +a_{n-3}^k+a_{n-4}^k\big )$$ $$ \text { with }a_0=0, a_1=1, a_2=2, a_3=3$$The sequence for $2 \leq k \leq 50$ is:

[26, 28, 47, 56, 57, 82, 99, 118, 119, 147, 173, 183, 188, 206, 209, 214, 236, 282, 263, 271, 297, 311, 335, 338, 371, 357, 389, 409, 399, 442, 459, 476, 485, 457, 525, 541, 575, 567, 566, 633, 579, 651, 659, 666, 666, 699, 722, 762, 764]

So we can write $a_{2_{max, \, N=4}}=26$, $a_{3_{max, \, N=4}}=28$ etc.

PENTAD

$a_{k_{max, \, N=5}}$ can be used to represent the maximum value attained by $ a_n $ for a particular $k$ when summing five terms progressively, where:$$a_n= \text{ sum of digits } \big (a_{n-1}^k+a_{n-2}^k +a_{n-3}^k+a_{n-4}^k+a_{n-5}^k\big )$$ $$ \text { with }a_0=0, a_1=1, a_2=2, a_3=3, a_4=4$$The sequence for $2 \leq k \leq 50$ is:

[26, 37, 38, 63, 65, 98, 92, 118, 122, 147, 156, 186, 197, 199, 198, 234, 255, 266, 281, 306, 324, 324, 335, 362, 357, 380, 405, 408, 426, 458, 474, 496, 504, 510, 516, 559, 557, 577, 563, 604, 588, 628, 657, 667, 695, 709, 686, 754, 764]

So we can write $a_{2_{max, \, N=5}}=26$, $a_{3_{max, \, N=5}}=37$ etc.

HEXAD

$a_{k_{max, \, N=6}}$ can be used to represent the maximum value attained by $ a_n $ for a particular $k$ when summing six terms progressively, where:$$a_n= \text{ sum of digits } \big (a_{n-1}^k+a_{n-2}^k +a_{n-3}^k+a_{n-4}^k+a_{n-5}^k+a_{n-6}^k\big )$$ $$ \text { with }a_0=0, a_1=1, a_2=2, a_3=3, a_4=4,a_5=4$$The sequence for $2 \leq k \leq 50$ is:

[26, 39, 47, 66, 78, 91, 98, 118, 131, 147, 167, 175, 191, 207, 234, 232, 230, 269, 281, 288, 311, 335, 329, 355, 380, 393, 452, 412, 437, 440, 464, 486, 509, 510, 528, 544, 585, 575, 576, 608, 617, 630, 654, 656, 684, 705, 725, 734, 764]

So we can write $a_{2_{max, \, N=6}}=26$, $a_{3_{max, \, N=6}}=39$ etc.

To be continued:

Wednesday, 18 November 2020

Personal Investigation: Part Two

This post follows on from a previous one so it's best to read that first in order to understand what I'm on about. Here is the link:

Personal Investigation: Part 1

After my previous post on Triads, it occurred to me that the same behaviour might occur with dyads. Starting with two numbers instead of three, I investigated the behaviour of the sequence:$$a_n=\text{ sum of digits }\big (a_{n-1}^k+a_{n-2}^k \big ),a_0=0,a_1=1, k \geq 2$$Not surprisingly, the behaviour was the same, with loops developing for $2 \leq k \leq 50$.

For example, consider the case of$$a_n= \text{ sum of digits } \big ( a_{n-1}^2+a_{n-2}^2 \big ),a_0=0, a_1=1$$Here we the sequence begins 0, 1, 1, 2, 5, 11, 11, 8, 14, 8, 8, 11, 14, 11, 11, 8, ... but we see that we have the loop shown in bold. The total number of terms generated is seven and these are 0, 1, 2, 5, 8, 11 and 14. The number of terms generated by the different values of $k$ is as follows (the first element in the ordered pair is the $k$ value and the second element is the number of terms):

[(2, 7), (3, 13), (4, 8), (5, 18), (6, 7), (7, 30), (8, 13), (9, 29), (10, 12), (11, 27), (12, 11), (13, 37), (14, 19), (15, 22), (16, 15), (17, 37), (18, 13), (19, 41), (20, 22), (21, 37), (22, 13), (23, 28), (24, 12), (25, 34), (26, 27), (27, 48), (28, 14), (29, 23), (30, 11), (31, 39), (32, 35), (33, 19), (34, 10), (35, 48), (36, 12), (37, 44), (38, 26), (39, 40), (40, 28), (41, 35), (42, 22), (43, 59), (44, 35), (45, 32), (46, 24), (47, 47), (48, 15), (49, 61), (50, 46)

It can be seen that the largest number of terms (61) occurs when $k=49$. As for the maximum values, the sequence that arises is as follows:

14, 28, 35, 45, 56, 91, 83, 110, 98, 135, 128, 151, 155, 181, 197, 218, 200, 241, 275, 271, 296, 286, 308, 319, 341, 351, 350, 376, 353, 410, 443, 433, 395, 495, 443, 501, 551, 521, 548, 565, 614, 620, 614, 604, 611, 646, 641, 716, 701

Figure 1 shows the graph of this sequence and highlights its basically linear behaviour:

Figure 1

Figure 2 shows the code (permalink) to generate the necessary details:

Figure 2

A logical continuation of this investigation would be to look at tetrads, pentads, hexads, heptads, octads, enneads, decads and so on. Let's take the case of the tetrad with $k=2$:$$a_n=\text{ sum of digits } \big (a_{n-1}^2+a_{n-2}^2 +a_{n-3}^2 +a_{n-4}^2 \big ),a_0=0,a_1=1,a_2=2,a_3=3$$Here the sequence of terms generated is:

0, 1, 2, 3, 5, 12, 11, 20, 15, 17, 9, 23, 8, 18, 26, 18, 20, 14, 21, 11, 15, 20, 17, 9, 23, 21, 8, 8, 18, 20, 15, 5, 20, 6, 20, 15, 8, 14, 21, 17, 18, 8, 11, 24, 14, 21, 11, 11, 24, 17, 9, 14, 8, 9, 8, 9, 11, 14, 12, 11, 15, 20, 17, ...

Once again, a loop arises, shown in bold typeface above.

The total number of terms generated is 19:

0, 1, 2, 3, 5, 6, 8, 9, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24 and 26.

For $2 \leq k \leq 50 $, the sequence of terms for maximum values associated with each $k$ is:

26, 28, 47, 56, 57, 82, 99, 118, 119, 147, 173, 183, 188, 206, 209, 214, 236, 282, 263, 271, 297, 311, 335, 338, 371, 357, 389, 409, 399, 442, 459, 468, 485, 457, 525, 541, 558, 567, 566, 633, 579, 651, 659, 666, 666, 699, 722, 714, 735, ...

Here is the permalink for the tetrad calculations. Figure 3 shows a graph of this sequence (up to the 50th term) and highlights once again the fact that the progression is basically linear.

Figure 3

Tuesday, 17 November 2020

Personal Investigation: Part One

It is difficult to find a title for this post but as it certainly involves triads, or groups of three, I've settled on that. Let's start with the triad 0, 1, 2 and square each term so that we have:$$0^2+1^2+2^2=0+1+4 =5$$The result is 5 and, because it only contains one digit, the sum of its digits is 5. Now let's make 1, 2, 5 our new triad and carry out the same procedure:$$1^2+2^2+5^2=1+4+25=29 \text{ with sum of digits }=11$$If we continue this process, we get the following sequence of terms:$$ 0, 1, 2, 5, 3, 11, 11, 8, 9, 14, 8, 8, 9, 11, \\14, 20, 15, 11, 17, 14, 12, 17, 17, 11, 24, 23, 11, 11,\\ 15, 17,14, 8, 18, 17, 20, 5, 12, 20, 20, 17, 18, 5, \\17, 17, 9, 20, 14, 20, 24, 11,17, 23, 21, 17, 17, 11, ... $$ With the last three terms listed here, we have entered a loop because 17, 17, 11 appeared earlier. This is the loop:$$ 17, 17, 11, 24, 23, 11, 11, 15, 17, 14, 8, 18, 17, 20, 5, \\12, 20, 20, 17, 18, 5, 17, 17, 9, 20, 14, 20, 24, 11, 17, 23, 21 $$We find that there are only 17 numbers that appear, including our starting values of 0, 1 and 2. Here are the numbers expressed in order:$$0, 1, 2, 3, 5, 8, 9, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24$$What happens if we raise our triad of starting numbers to the third power. Here we have:$$0^3+1^3+2^3=0+1+8=9$$Once again, because 9 has only one digit, its sum of digits in also 9. Let's use 1, 2 and 9 as our new triad:$$1^3+2^3+9^3=1+8+729=738 \text{ with sum of digits }=18$$Continuing this process, we get the following sequence of terms:$$0, 1, 2, 9, 18, 26, 17, 16, 26, 26, 26, 24, \\ 34, 18, 28, 29, 18, 18, 17, 26, 16, 26, 26, ...$$With the last three terms listed, we have again entered a loop because 16, 16 and 26 occurred earlier. Here is our loop:$$16, 26, 26, 26, 24, 34, 18, 28, 29, 18, 18, 17, 26$$We find that only twelve numbers appearing, including our starting values of 0, 1 and 2:$$0, 1, 2, 9, 16, 17, 18, 24, 26, 28, 29, 34$$If we continue raising our initial triads to higher and higher powers, it seems that we always get a finite set of numbers. In other words, sooner or later, a loop develops but must ensure to take enough terms. When raising to the 50th power for example, we must take at least 561 iterations and there are total of 106 terms. So what are we doing in general terms:$$a_n=\text{ sum of digits } \big (a_{n-1}^k+a_{n-2}^k+a_{n-3}^k \big )$$

where $a_0=0, a_1=1 \text{ and }a_3=2, 2 \leq k \leq 50$

For every value of $k$, I've taken the largest value in the set of numbers generated and formed a sequence of these largest values for $2 \leq k \leq 50$. Let's denote this term by $a_{max, \, k}$ where $k$ ranges from 2 to 50. The result is the following sequence:$$[24, 34, 35, 58, 66, 93, 98, 99, 117, 140, \\147, 165, 191, 181, 197, 218, 236, 248, 266, 297,\\ 297, 311, 309, 347, 344, 362, 360, 408, 416, 436, \\425, 449, 479, 482, 479, 544, 546, 546, 566, 570,\\ 597, 624, 632, 702, 665, 664, 696, 710, 725]$$We see that $a_{max,\,4}=35$ and $a_{max,\,49}=710$. The sequence more or less increases but not strictly so. I supposing that these loops (and consequent maximum values) continue for $k >50$ but I haven't tested so far. The sequence is probably worthy of submission to the OEIS.

Sunday, 15 November 2020

Another OEIS Submission

After tinkering with this sequence for a couple of weeks, I decided to submit it to the OEIS in its original form. I'd tried various ways to make it more interesting but, in the end, it was its inherent simplicity that won out. Here it is:$$a_n=\text{ sum of digits of }\big (a_{n-1} \big )^{a_{n-2}} \text{ where } a_0=1 \text{ and } a_1=2$$I was able to use SageMathCell to generate the terms up to and including $a_{17}$ but after that the algorithm times out. Here is a permalink with the SageMath code (in blue) and the output (in red) as follows:

a, b=1,2
L=[a, b]
for n in [1..17]:
c=b^a
c=sum(c.digits())
L.append(c)
a, b=b, c
print(L)

[1, 2, 2, 4, 7, 7, 25, 34, 151, 331, 1690, 3265, 26449, 64528, 574513, 1671208, 16090657, 54199564, 559922497]

Without the sum of digits, this exponentiation would become too large too quickly and the resultant sequence would be fairly meaningless. The sum of digits brings the exponent back to a much lower number but, even so, by $n=17$ the term is more than half a billion. In my submission, I was able to provide a link to GeeksforGeeks that shows the code to calculate the sum of digits of a given number to a given power in Python3, C++, Java, C# and PHP. Maybe this will give my submission a little more weight. See Figure 1 for a screenshot.

Figure 1

I'm well aware that this is quite a simple sequence but it will have to do until I come up with something like the Collatz or PrimeLatz trajectory sequences. I'll keep trying. Meanwhile, it will be interesting to see how my submission is greeted by the guardians of OEIS who guard their sequences jealously and are suspicious of newcomers. I'll add to this post as news arrives. Here is a link to my previous contribution: OEIS A335789 under the name Sean Lestrange (August 14th 2020). I have an earlier entry, OEIS A301938, under the name Sean Reeves (March 28th 2018). As to the name change, well that's another story.

UPDATE: well, that was quick. When I checked today (16th November 2020), the sequence had already been approved: OEIS A338917. See Figure 2:

Figure 2

Thursday, 5 November 2020

The Perrin Sequence

Many years ago, in the bookstore at the bottom of SOGO in Pondok Indah Mall (Jakarta), I acquired a book by David Wells titled Prime Numbers: The Most Mysterious Figures in Math. I still have the physical copy but I also now have an electronic copy. On page 103, there is the information shown in Figure 1:

Figure 1

It so happens that the number representing my diurnal age today, 26149, is a member of this "very generalized Fibonacci sequence" and is in fact OEIS A050443:

A050443

a(0)=4, a(1)=0, a(2)=0, a(3)=3; thereafter a(n) = a(n-3) + a(n-4).

4, 0, 0, 3, 4, 0, 3, 7, 4, 3, 10, 11, 7, 13, 21, 18, 20, 34, 39, 38, 54, 73, 77, 92, 127, 150, 169, 219, 277, 319, 388, 496, 596, 707, 884, 1092, 1303, 1591, 1976, 2395, 2894, 3567, 4371, 5289, 6461, 7938, 9660, 11750, 14399, 17598, 21410, 26149

I was alerted to this connection to the book thanks to an OEIS reference that also remarks that the sequence is related to the Perrin sequence that also has the property that a($p$) is divisible by $p$ for primes $p$. So what is the Perrin sequence? WolframMathWorld defines it as:$$P(n)=P(n-2)+P(n-3) \text{ where }P(0)=3,P(1)=0,P(2)=2$$Unusually, the Wolfram article also displays a cartoon (see Figure 2):

Figure 2: The above cartoon (Amend 2005) shows an unconventional sports
application of the Perrin sequence (right panel).
(The left two panels instead apply the Fibonacci numbers).

$P(n)$ is also the solution of a third-order linear homogeneous recurrence equation having characteristic equation $x^3-x-1=0$. The solutions to this equation are given by:$$\begin{align} x& = -\frac{1}{2} \, {\left(\frac{1}{18} \, \sqrt{23} \sqrt{3} + \frac{1}{2}\right)}^{\frac{1}{3}} {\left(i \, \sqrt{3} + 1\right)} - \frac{-i \, \sqrt{3} + 1}{6 \, {\left(\frac{1}{18} \, \sqrt{23} \sqrt{3} + \frac{1}{2}\right)}^{\frac{1}{3}}}\\x &= -\frac{1}{2} \, {\left(\frac{1}{18} \, \sqrt{23} \sqrt{3} + \frac{1}{2}\right)}^{\frac{1}{3}} {\left(-i \, \sqrt{3} + 1\right)} - \frac{i \, \sqrt{3} + 1}{6 \, {\left(\frac{1}{18} \, \sqrt{23} \sqrt{3} + \frac{1}{2}\right)}^{\frac{1}{3}}}\\ x &= {\left(\frac{1}{18} \, \sqrt{23} \sqrt{3} + \frac{1}{2}\right)}^{\frac{1}{3}} + \frac{1}{3 \, {\left(\frac{1}{18} \, \sqrt{23} \sqrt{3} + \frac{1}{2}\right)}^{\frac{1}{3}}}\end{align} $$The third and only real-valued solution above is the so-called plastic constant $ \approx 1.32471795... $, also called the:

le nombre radiant
minimal Pisot number
plastic number
plastic ratio
platin number
Siegel's number
silver number
silver constant

It is the limiting ratio of the successive terms of the Padovan sequence or Perrin sequence. Thus we have:$$\lim_{n \to \infty} \frac{P(n)}{P(n-1)}=P \approx 1.32471795$$The Padovan sequence mentioned is similar to the Perrin except that it has different starting values:$$P(n)=P(n-2)+P(n-3) \text{ where }P(0)=0,P(1)=1,P(2)=1$$It should be noted that for OEIS A050443, the ratio of successive terms does not approach the plastic constant but instead 1.22074... which as far as I can determine doesn't have a particular name.

Figure 3

It was only after composing this post that I realised I had already made a post about the plastic number on Friday, 5th June 2020. This is well worth the read because it illustrates some applications of the number (as shown in Figure 3). The Greek letter $ \rho $ can be used to represent the plastic ratio. In which case, to quote from the earlier post and referencing Figure 3:

There are precisely three ways of partitioning a square into three similar rectangles:
The trivial solution given by three congruent rectangles with aspect ratio 3:1.
The solution in which two of the three rectangles are congruent with the third one of twice the linear dimension of the congruent pair and where the rectangles have aspect ratio 3:2.
The solution in which the three rectangles are mutually non-congruent (all of different sizes) and where they have aspect ratio $ \rho^2 $. The ratios of the linear sizes of the three rectangles are: $ \rho $ (large : medium); $ \rho^2$(medium : small); and $ \rho^3 $ (large : small). The internal, long edge of the largest rectangle (the square's fault line) divides two of the square's four edges into two segments each that stand to one another in the ratio ρ. The internal, coincident short edge of the medium rectangle and long edge of the small rectangle divides one of the square's other, two edges into two segments that stand to one another in the ratio $\rho^4$.

Figure 4 shows the approximate measurements for a unit square in which the three rectangles are mutually non-congruent. We have:$$ \frac{b}{1-b}=\rho$$

Figure 4

Tuesday, 3 November 2020

Osculators

In this post I'm describing a very interesting method of determining the divisors of any given number that I came across in a post by Sohel Sahoo:

All credit is given to Sohel Sahoo for his method and all I've done in this post is to rephrase his method so that it is easier for me to understand.

The assertion is made that there exist two specific numbers (called osculators: one positive, the other negative) for any divisor. The divisor's positive or the negative osculator can be used to determine if a given number has that particular divisor. The sum of the absolute values of the osculators is equal to the divisor.

Procedures to find the osculators of a given divisor

(7 and 13 will be used as examples):

If the divisor is a single digit, work with the smallest multiple that has two digits.

In the case of 7, the smallest two digit multiple is 14 so we work with that.

In the case of 13, there is no need to modify it.
Let the osculator be $x$ and multiply the unit digit of the divisor by $x$.

In the case of 14, this gives $4x$.

In the case of 13 this gives $3x$.
Add the result obtained in step 2 to the remaining digits to obtain an expression in $x$.

In the case of 14 this gives the expression $1+4x$.

In the case of 13 this gives the expression $1+3x$.
Find the smallest positive and negative values of $x$ that make the expression divisible by the divisor. These are the positive and negative osculators for the divisor.

In the case of 7, $1+4 \times 5=21$ and $1+4 \times -2 =-7$ and so the osculators are 5 and -2. Note that |5|+|-2|=7.

In the case of 13, $1 + 3 \times 4=13$ and $1+ 3 \times -9 = -26$ and so the osculators are 4 and -9. Note that |4|+|-9|=13.

Doing this for other divisors generates the table shown below:

Number	Positive Osculator	Negative Osculator
3	1	-2
7	5	-2
9	1	-8
11	10	-1
13	4	-9
17	12	-5
19	2	-17
21	19	-2
23	7	-16
27	19	-8
29	3	-26
31	28	-3
33	10	-23
37	26	-11
39	4	-35
41	37	-4
43	13	-30
47	33	-14
49	5	-44
51	46	-5
53	16	-37
57	40	-17
59	6	-53
61	55	-6
63	19	-44
67	47	-20
69	7	-62
71	64	-7
73	22	-51
77	54	-23
79	8	-71
81	73	-8
83	25	-58
87	61	-26
89	9	-80

Some tips for remembering the osculators:

Any divisor ending in 9 can generally be written as $a9$.

Let the osculator be $x$.

According to rules the expression will be $9x+a$.

In decimal representation we have:

$a9=10a+9=9a+a+9=9a+9+a=9(a+1)+a$

Obviously, the positive osculator is ($a+1$). Henceforth, owing to this, the positive osculator for divisor numbers ending in 9 is just one more than its previous digits.

For 9, 19, 29, 39 etc.(all ending in 9), the positive osculators are 1,2,3,4 etc.
For 3, 13, 23, 33 etc. (all ending in 3) multiply them by 3 in order to get 9 in the unit place as 9, 39, 69, 99 etc. Thus you get 1, 4, 7, 10 etc. as positive osculators.
For 7, 17, 27, 37 etc.(all ending in 7) multiply them by 7 so as to attain 9 in unit place like 49, 119, 189, 259 etc. Hence you get 5, 12, 19, 26 etc. as positive osculators.
For 1, 11, 21, 31 etc. (all ending in 1) multiply them by 9 thereof attain 9 as last digit as 9, 99, 189, 279 etc. So, you get positive osculators viz. 1, 10, 19, 28 etc.

Special Divisibility Rule:

If any number is made by repeating a digit 6 times then the no. will be divisible by 3, 7, 11, 13, 21, 37, 77, 91, 143 and 1001 e.g. 111111, 222222 and 333333 are divisible by these numbers.

Formula for using the osculators to test divisibility:

We'll use 69125 and divisibility by 7 as an example to illustrate the use of the formula. We'll use both the positive and negative osculators (although in practice, only one is needed):

Firstly, multiply the osculator of the divisor by the unit digit of the number that is being tested for divisibility by that divisor.

In the case of 69125 being test for divisibility by 7, this means 5 x 5 = 35 using the positive osculator of 5 and 5 x -2 = 10 using the negative osculator of -2.
Add the result so obtained on multiplication to the remaining of digits if using the positive osculator, subtract the result if using the negative osculator.

Thus 69125 --> 6912 + 25 = 6937 or 6912 - 10 = 6902.
Repeat processes 1 and 2 until the number is small enough to be recognised as a multiple of the divisor or not.

Thus 6937 --> 693 + 35 = 728 --> 72 + 40 = 112 --> 11 + 10 --> 21 which is 7 x 3.
6902 --> 690 - 4 = 686 --> 68 -12 = 56 which is 7 x 8.
If the result is a multiple of the divisor then the divisor is confirmed.

69125 does reduce to a multiple of 7 and so 7 is a divisor. It can also be seen that using the smaller of the two osculators (ignoring signs) makes for easier calculations.

Another example:

What happens if 69125 were tested for divisibility by 13? We'll use the 4 as the osculator as it is the smaller of the two.

69125 --> 6912 - 20 = 6892 --> 689 - 8 = 681 --> 68 - 4 = 64 which is clearly not a multiple of 13 and so we conclude that 69125 is not divisible by 13.

Usefulness of Divisibility Tests

One can argue that learning divisibility tests like this is pointless in this technological age but I think such mental activities combat mental decline (I am a septuagenarian so that's important) and reduce our reliance on technology by making us more confident and self-sufficient (as a counterbalance to the encroachments of AI).

Note on Osculators and Osculation

It should be noted that the terms osculator and osculation have special meanings in Vedic Mathematics. An osculator is an algorithm for performing osculation while osculation means the determination of whether a number is divisible by another by means of certain operations on its digits. The meaning of osculation is thus different from that of mainstream mathematics where the term means a contact between curves or surfaces, at which point they have a common tangent. Thus we can speak of osculating circles, meaning two circles that touch at a point through which a common tangent passes. Here is a link to more information on Vedic Mathematics.