No two points on an infinite line are infinitely distant

I mentioned this in passing in "What if there was no beginning?" in connection with the Kalām Paradox. but I think it deserves a post of its own.

The Kalām Paradox, for those who came in late, asserts that nothing temporal can have always existed. If it had, it would be infinitely old, meaning that an infinite amount of time must already have elapsed. However, it is impossible for an infinite amount of time to elapse, because time elapses one finite step at a time, and adding up finite quantities can never yield an infinite quantity.

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First, I should make it explicit what sort of infinity we're talking about here. We're not talking about Zeno's paradoxes which find an uncountable infinity within countably finite quantities. In the racecourse paradox, for example, it is supposed to be impossible to run a mile, because first you have to run half a mile; and before you can run half a mile, you have to run a quarter of a mile; and so on -- an infinite number of tasks. A number-theoretical way of expressing this is that between any two integers there are (uncountably) infinitely many real numbers. To apply Zeno's spatial paradox to time, we could say it is impossible for anyone or anything to be a year old -- because before a year can elapse, 6 months have to elapse; before 6 months, 3 months; and so on -- a sum of infinitely many finite quantities, which must therefore add up to infinity. This is not the Kalām argument, which takes it for granted that things can have a finite age.

I will not bother to address Zeno-type paradoxes because no one is in any danger of believing them. The Johnsonian "I refute it thus!" is sufficient. It's very obvious that fleet-footed warriors can catch tortoises, and that it is in fact possible to run a mile. These are matters of everyday observation, as the subject matter of Kalām -- the question of whether the universe had a beginning -- cannot be.

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When the Kalām Paradox denies that the past can be infinite, it is referring to countable infinity, of which the paradigm case is not the real numbers but the integers. (To be sure, time may also be infinitely divisible and thus uncountably infinite, but that is a separate question.) The image we want to keep in mind, then, is the integer number line.

I propose that linear time is like this number line. The origin (zero) corresponds to the present, the negative integers to points in the past, and the positive integers to points in the future. Let's say that -1 on the number line means "one year ago," 1 means "one year hence," and so on. (There are of course many -- possibly infinitely many -- intermediate points in time between each of these points, just as there are infinitely many real numbers between one integer and the next, but that is not germane to Kalām. We take it for granted that Zeno is wrong and that countably-finite intervals of time can and do elapse.)

The integer number line is countably infinite -- meaning not that it is infinitely divisible but that it is infinitely long. There is no "first" or "last" integer. This does not mean that there is a first integer, -∞, which is infinitely distant from the origin; there is no such integer as -∞; there is no first (or last) integer. Every integer is finite (meaning finitely distant from the origin), and therefore the difference between any two integers is also finite.

Applied to time, this means that there was no first moment of time, no absolute beginning -- not that the beginning was "infinity years ago" (-∞), but that there was no beginning.

Now according to the Kalām Paradox, this means that an infinite amount of time must already have elapsed to get to the present. This is impossible, and yet we manifestly have reached the present; therefore, there was a beginning.

The question to ask is: An infinite amount of time must have elapsed from what point to the present? If the answer is "from the beginning," the argument fails, because the claim it is trying to disprove is that there was no beginning, just as there is no first integer. If the answer is "not from the beginning, since there is none, but from any arbitrary point infinitely distant from the present," the argument also fails, because there is no such point, just as there is no such thing as an infinite integer (infinitely distant from the origin). The only answer left, then, is "a point finitely distant from the present," and then there is no more paradox.

If there was no beginning, then we can say that, for any finite number n, n years have already elapsed. This is true no matter how arbitrarily big n is, which is the sense in which the past is infinite. But it does not mean that an infinite number of years have already elapsed. Therefore, this presupposition of the Kalām Paradox conclusively fails.

I am open to corrections from readers more mathematically gifted than myself, but I have to say I'm pretty darn sure I'm right about this.

Visualizing gnomon series for the figurate numbers modulo 10

In modular arithmetic, the integers modulo k form a closed figure -- a polygon with k vertices -- rather than a line. The decagon below represents the integers modulo 10. To count, start at +1 and follow the black lines clockwise. The numbers on the left side of the figure are negative because n ≡ n - 10 (mod 10); thus, 5 ≡ -5, 6 ≡ -4, 7 ≡ -3, and so on.

The number that may be added to the nth figurate number to yield the (n + 1)th is called a gnomon. To generate the series of triangular numbers, you start with 0, then add 1, then add 2, then add 3, and so on through the natural numbers. In other words, the gnomon series for triangular numbers is (1, 2, 3, 4, 5, 6, 7, ...) -- which is congruent (mod 10) to (+1, +2, +3, +4, ±5, -4, -3, -2, -1, 0) endlessly repeated. To get the gnomon series for the triangular numbers, start at +1 on the decagon and follow the black lines clockwise.

For the triangular numbers, the difference between the nth gnomon is the (n - 1)th gnomon is 1. I shall express this by saying that the gnomon interval for the triangular numbers is 1. For the squares, the gnomon interval is 2; for the pentagonal numbers, it is 3; and so on. The gnomon interval for the n-gonal numbers is always equal to n - 2.

The gnomon series for the n-gonal numbers modulo 10 may be read off our decagon by starting at +1 and going clockwise, reading every (n - 2)th vertex. It is readily apparent that there are only 10 possible gnomon series, since reading every (n + 10)th vertex is the same as reading every nth vertex. The gnomon series are:

• 3-gonal: black lines clockwise (+1, +2, +3, +4, ±5, -4, -3, -2, -1, 0)
• 4-gonal: red lines clockwise (+1, +3, ±5, -3, -1)
• 5-gonal: green lines clockwise (+1, +4, -3, 0, +3, -4, -1, +2, ±5, -2)
• 6-gonal: purple lines clockwise (+1, ±5, -1, +3, -3)
• 7-gonal: orange line (+1, -4)
• 8-gonal: purple lines counterclockwise (+1, -3, +3, -1, ±5)
• 9-gonal: green lines counterclockwise (+1, -2, ±5, +2, -1, -4, +3, 0, -3, +4, +1)
• 10-gonal: red lines couterclockwise (+1, -1, -3, ±5, +3)
• 11-gonal: black lines counterclockwise (+1, 0, -1, -2, -3, -4, ±5, +4, +3, +2)
• 12-gonal: only one vertex (+1)

After the 12-gonal numbers, the gnomon series repeat; the (n + 10)-gonal numbers are congruent to the n-gonal numbers (mod 10).

Which of these gnomon series will generate a repeating palindromic series? All those, and only those, whose representation on the decagon exhibits left-right symmetry -- that is, all figurate numbers except the 7-gonal (the orange line), the 12-gonal (a single non-centered point), and those congruent to them (the 17-gonal, 22-gonal, etc.).

When first looking for RPSs in figurate numbers mod 10, I only got as far as the 10-gonal numbers, so the 7-gonal numbers seemed to be the only exceptions to the RPS rule. This new postulate predicts that the 12-gonal numbers will also be an exception -- and indeed they are. The 12-gonal numbers mod 10 are (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) endlessly repeated, which is not a palindrome.

Two tasks remain: (1) proving what I have just asserted, and (2) devising a way to predict, for any modulus, which gnomon intervals will yield a left-right symmetrical pattern. 

Applying Kevin McCall's logic to squares and other non-centered figurate numbers

Note: this post uses special terminology and notation introduced in the last post. You should read that first in order to understand what follows.

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Kevin McCall's proof of the RPS theorem for reduced triangular numbers is based on the following observation:

The series of triangular numbers is generated by starting with 0, then adding 1, then adding 2, then 3, and so on through the succession of natural numbers. Due to the fact that k - n ≡ -n (mod k), one you've added numbers up to a certain point, you start doing the modular equivalent of subtracting those same numbers in reverse order, creating a palindrome.

Thus, if we consider the sequence of triangular numbers reduced modulo 10:

• 0
• +1
• +2
• +3
• +4
• +5
• +6 ≡ -4 (mod 10)
• +7 ≡ -3 (mod 10)
• +8 ≡ -2 (mod 10)
• +9 ≡ -1 (mod 10)
• +10 ≡ 0 (mod 10)
• +11 ≡ +1 (mod 10)
• +12 ≡ +2 (mod 10)
• +13 ≡ +3 (mod 10)
• +14 ≡ +4 (mod 10)
• +15 ≡ -5 (mod 10)
• +16 ≡ -4 (mod 10)
• +17 ≡ -3 (mod 10)
• +18 ≡ -2 (mod 10)
• +19 ≡ -1 (mod 10)
• +20 ≡ +0 (mod 10)
• etc.

At the end of this 20-step cycle, we are back where we started, with a number that is congruent to 0 (mod 10), and the cycle starts again.

Note that we have to go through two cycles of adding and subtracting the numbers because +5 ≡ -5 (mod 10). We count it as +5 the first time around and -5 the second time, so that it cancels out. If the modulus is odd, only one cycle is necessary.

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Now let's consider the sequence of square numbers. We generate this series by starting with 0, then adding 1, then adding 3, then 5, and so on through the succession of odd natural numbers. This results in an RPS for much the same reason that the triangular series does: adding successive numbers is the modular equivalent of adding up to a certain point and then subtracting the same numbers in reverse order. Here's how it works modulo 10.

• 0
• +1
• +3
• +5
• +7 ≡ -3 (mod 10)
• +9 ≡ -1 (mod 10)
• +11 ≡ +1 (mod 10)
• +13 ≡ +3 (mod 10)
• +15 ≡ -5 (mod 10)
• +17 ≡ -3 (mod 10)
• +19 ≡ -1 (mod 10)
• etc.

As with the triangular numbers, we have to go through two cycles so that the two 5s cancel each other out. Notice that, unlike the triangular numbers, this sequence never returns to adding 0 (mod 10). That is why the triangular numbers reduced mod 10 = RPS (0136051865), while the squares are RPS (0)1496(5) -- the extra parentheses indicating that there are not two 5s in a row in the middle of the cycle, nor two 0s in a row at the end of one cycle and the beginning of the next.

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Moving on to the pentagonal numbers, they are generated by starting with 0, then adding 1, then 4, then 7, then 10, and so on -- every third natural number. The pattern should be obvious by now: The sequence of n-gonal numbers is generated by starting with 0 and adding, successively, every (n - 2)th natural number, beginning with 1. Here's the generation of the pentagonal sequence modulo 10.

• 0
• +1
• +4
• +7
• +10 ≡ +0 (mod 10)
• +13 ≡ -7 (mod 10)
• +16 ≡ -4 (mod 10)
• +19 ≡ -1 (mod 10)
• +22 ≡ +2 (mod 10)
• +25 ≡ +5 (mod 10)
• +28 ≡ +8 (mod 10)
• +31 ≡ +1 (mod 10)
• +34 ≡ +4 (mod 10)
• +37 ≡ +7 (mod 10)
• +40 ≡ +0 (mod 10)
• +43 ≡ -7 (mod 10)
• +46 ≡ -4 (mod 10)
• +49 ≡ -1 (mod 10)
• +52 ≡ -8 (mod 10)
• +55 ≡ -5 (mod 10)
• +58 ≡ -2 (mod 10)
• +61 ≡ +1 (mod 10)
• etc.

The cycle here is more involved because we are adding every third natural number, which means that after we reach the -1 which cancels out the original +1, we do not go on to either - or +1 and the cycle does not yet begin anew.

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Skipping hexagonal numbers for the time being, let's jump straight to what we're really interested in: the heptagonal numbers -- the only figurate numbers yet examined which do not yield an RPS when reduced modulo 10. In keeping with the pattern, the heptagonal numbers are generated by adding, successively, every 5th natural number -- yielding, modulo 10:

• 0
• +1
• +6
• +11 ≡ +1 (mod 10)
• +16 ≡ +6 (mod 10)
• +21 ≡ +1 (mod 10)
• +26 ≡ +6 (mod 10)
• +31 ≡ +1 (mod 10)
• +36 ≡ +6 (mod 10)
• etc.

As can be seen, we just continue adding 1 and 6 (or subtracting 9 and 4) forever. This gives us a repeating cycle with a period of 20 -- because 10(1 + 6) ≡ 0 (mod 10) -- but no palindrome is created because we never reach -1/+9 or -6/+4.

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My tentative conclusion is that the sequence of (non-centered) n-gonal numbers reduced modulo k will always be an RPS if n - 2 and k are relatively prime. When that condition holds, adding every (n - 2)th natural number in succession will (I think) mean in hitting all possible modular values, resulting in an RPS. The triangular numbers are a special case because for that sequence n - 2 = 1, which is coprime to every integer.

Where n - 2 and k are not coprime, an RPS may result, but not necessarily. I need to think a little more about what exactly determines which such sequences are RPSs and which are not.

Are all reduced sequences of figurate numbers repeating palindromes?

Time for another mathematical interlude.

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Preliminaries: Terminology and notation

A palindrome is any series of elements that is the same forwards and backwards. Each palindrome thus consists of two parts, which we shall call the head and the tail. The tail consists of the same series of elements as the head, but in reverse order. For example, in the palindromic word “noon,” the string “no” is the head, and “on” is the tail. In “noon,” the head and tail are entirely separate, but in a palindrome with an odd number of elements, the end of the head will overlap with the beginning of the tail. For example, the head of the palindromic word “level” is “lev,” and the tail is “vel”; a single “v” does double duty as the last element of the head and the first element of the tail.

No special notation is required to write a simple palindrome such as “noon” or “level,” but we are concerned in this post with repeating palindromic series (RPSs). If a palindrome consists of a head followed by a tail (possibly overlapping), an RPS is a head followed by a tail, then the head again, then the tail again, and so on to infinity. In the notation we will be using here, an RPS is represented by "RPS" followed by its head enclosed in parens (round brackets), thus:

• RPS (no) = noonnoonnoonnoonnoon...

If the end of the head overlaps with the beginning of the tail, an additional open-paren is placed before the first element of the tail, thus:

• RPS (le(v) = levellevellevellevellevel...

In an RPS, unlike a simple palindrome, overlap at the other end is also possible. For example, the word “grammar” is not itself a palindrome, but “grammar” endlessly repeated is an RPS. The letter “g” is both the beginning of the head and the end of the tail. This sort of overlap is indicated by placing a close-paren after the last element of the tail, thus:

• RPS (g)ram) = grammargrammargrammar...

It is possible to have overlap at both ends, as in this example.

• RPS (v)oo(d) = voodoovoodoovoodoo...

Overlap need not be limited to a single element; it can be a series of elements, provided that series is itself a palindrome. For example, consider the RPS created by endlessly repeating the word “sestet.”

• RPS (ses)t(e) = sestetsestetsestetsestet...

I should also mention that, much like a repeating decimal, an RPS has a beginning but not an end. As with a decimal like 0.16666666..., there may be a non-repeating segment at the beginning, before the RPS proper starts. Our notation can deal with this by putting this non-repeating segment before the first paren, as in this example.

• RPS n(eve)(r) = neverevereverevereverever...

One final note about the notation: If each element in the RPS can be represented by a single character (a letter, a one-digit number, etc.), it can be written as in the examples above, without commas or spaces. If the elements of an RPS are words, multidigit numbers, etc., the elements should be separated by commas and spaces.

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RPSs in the sequence of triangular numbers

Several posts on this blog have dealt with the fact that, for any modulus k, the sequence of triangular numbers reduced modulo k will be an RPS. Two different proofs of this have been given (here and here). The pattern is easiest to see when k = 10, since (in the decimal system) any number reduced modulo 10 is equal to the final digit of that number.

The first several triangular numbers are: 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666...

If we reduce that sequence modulo 10 (by replacing each number in the sequence with its final digit), we get: 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6...

This reduced sequence is RPS (0136051865). Any other modulus will also yield an RPS.

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Square numbers

Only some three months after proving the RPS theorem for triangular numbers did I notice that the sequence of square numbers shows a similar pattern. Below are the square numbers from 0² to 50². Start at the upper left, follow the zigzag down to the bottom, and then come back up the zigzag on the right.

Numbers in the same column have the same last digit (i.e., are congruent modulo 10). Numbers in the same row have the same last two digits (i.e., are congruent modulo 100).

Square numbers reduced modulo 10 = RPS (0)1496(5).

Square numbers reduced modulo 100 = RPS (0), 1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 21, 44, 69, 96, 25, 56, 89, 24, 61, 0, 41, 84, 29, 76, (25).

I haven't checked if other moduli also yield RPSs, but, based on my experience with triangular numbers, and on the general principle that there is nothing mathematically special about powers of 10, I feel quite certain that they do.

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Other non-centered figurate numbers

Pentagonal numbers: 0, 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, 210, 247, 287, 330, 376, 425, 477, 532, 590, 651, 715, 782, 852, 925, 1001, 1080, 1162, 1247, 1335, 1426, 1520, 1617, 1717, 1820, 1926, 2035, 2147, 2262, 2380, 2501, 2625, 2752, 2882, 3015, 3151, 3290, 3432, 3577, 3725, 3876, 4030, 4187...

Pentagonal numbers reduced modulo 10 = RPS (01522510)275607)

Hexagonal numbers: 0, 1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, 276, 325, 378, 435, 496, 561, 630, 703, 780, 861, 946, 1035, 1128, 1225, 1326, 1431, 1540, 1653, 1770, 1891, 2016, 2145, 2278, 2415, 2556, 2701, 2850, 3003, 3160, 3321, 3486, 3655, 3828, 4005, 4186, 4371, 4560

Hexagonal numbers reduced modulo 10 = RPS (016585610(3)

Heptagonal numbers: 0, 1, 7, 18, 34, 55, 81, 112, 148, 189, 235, 286, 342, 403, 469, 540, 616, 697, 783, 874, 970, 1071, 1177, 1288, 1404, 1525, 1651, 1782, 1918, 2059, 2205, 2356, 2512, 2673, 2839, 3010, 3186, 3367, 3553, 3744, 3940, 4141, 4347, 4558, 4774, 4995, 5221, 5452, 5688

Heptagonal numbers reduced modulo 10 = 01784512895623906734 endlessly repeated -- not a palindrome!

Octagonal numbers: 0, 1, 8, 21, 40, 65, 96, 133, 176, 225, 280, 341, 408, 481, 560, 645, 736, 833, 936, 1045, 1160, 1281, 1408, 1541, 1680, 1825, 1976, 2133, 2296, 2465, 2640, 2821, 3008, 3201, 3400, 3605, 3816, 4033, 4256, 4485, 4720, 4961, 5208, 5461

Octagonal numbers reduced modulo 10 = RPS (01810)56(3)

Enneagonal numbers: 0, 1, 9, 24, 46, 75, 111, 154, 204, 261, 325, 396, 474, 559, 651, 750, 856, 969, 1089, 1216, 1350, 1491, 1639, 1794, 1956, 2125, 2301, 2484, 2674, 2871, 3075, 3286, 3504, 3729, 3961, 4200, 4446, 4699, 4959, 5226, 5500, 5781, 6069, 6364

Enneagonal numbers reduced modulo 10 = RPS (0194651441564910)69)

Decagonal numbers: 0, 1, 10, 27, 52, 85, 126, 175, 232, 297, 370, 451, 540, 637, 742, 855, 976, 1105, 1242, 1387, 1540, 1701, 1870, 2047, 2232, 2425, 2626, 2835, 3052, 3277, 3510, 3751, 4000, 4257, 4522, 4795, 5076, 5365, 5662, 5967, 6280, 6601, 6930, 7267, 7612, 7965, 8326

Decagonal numbers reduced modulo 10 = RPS (010)725(6)

This is so bizarre that I almost think I must have made some mistake, but I'm pretty sure I haven't. Inductively, it looks like virtually all non-centered figurate numbers reduce to RPSs, but the heptagonal numbers are an exception! Why? Are there other exceptions?

⁂

Centered figurate numbers

Centered triangular numbers: 1, 4, 10, 19, 31, 46, 64, 85, 109, 136, 166, 199, 235, 274, 316, 361, 409, 460, 514, 571, 631, 694, 760, 829, 901, 976, 1054, 1135, 1219, 1306, 1396, 1489, 1585, 1684, 1786, 1891, 1999, 2110, 2224, 2341, 2461, 2584, 2710, 2839, 2971, 3106, 3244, 3385, 3529

Centered triangular numbers reduced modulo 10 = RPS (1409164596)

Centered square numbers: 1, 5, 13, 25, 41, 61, 85, 113, 145, 181, 221, 265, 313, 365, 421, 481, 545, 613, 685, 761, 841, 925, 1013, 1105, 1201, 1301, 1405, 1513, 1625, 1741, 1861, 1985, 2113, 2245, 2381, 2521, 2665, 2813, 2965, 3121, 3281, 3445, 3613, 3785, 3961, 4141, 4325, 4513

Centered square numbers reduced modulo 10 = RPS (15(3)

Centered pentagonal numbers: 1, 6, 16, 31, 51, 76, 106, 141, 181, 226, 276, 331, 391, 456, 526, 601, 681, 766, 856, 951, 1051, 1156, 1266, 1381, 1501, 1626, 1756, 1891, 2031, 2176, 2326, 2481, 2641, 2806, 2976, 3151, 3331, 3516, 3706, 3901, 4101, 4306, 4516, 4731, 4951, 5176, 5406

Centered pentagonal numbers reduced modulo 10 = RPS (16)

Hex numbers: 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141, 1261, 1387, 1519, 1657, 1801, 1951, 2107, 2269, 2437, 2611, 2791, 2977, 3169, 3367, 3571, 3781, 3997, 4219, 4447, 4681, 4921, 5167, 5419, 5677, 5941, 6211, 6487

Hex numbers reduced modulo 10 = RPS (17(9)

Centered heptagonal numbers: 1, 8, 22, 43, 71, 106, 148, 197, 253, 316, 386, 463, 547, 638, 736, 841, 953, 1072, 1198, 1331, 1471, 1618, 1772, 1933, 2101, 2276, 2458, 2647, 2843, 3046, 3256, 3473, 3697, 3928, 4166, 4411, 4663, 4922, 5188, 5461, 5741, 6028, 6322, 6623, 6931, 7246

Centered heptagonal numbers reduced modulo 10 = RPS (1823168736)

Centered octagonal numbers (i.e., odd squares): 1, 9, 25, 49, 81, 121, 169, 225, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1225, 1369, 1521, 1681, 1849, 2025, 2209, 2401, 2601, 2809, 3025, 3249, 3481, 3721, 3969, 4225, 4489, 4761, 5041, 5329, 5625, 5929, 6241, 6561, 6889, 7225, 7569

Centered octagonal numbers reduced modulo 10 = RPS (19(5)

Centered enneagonal numbers (i.e., every third triangular number): 1, 10, 28, 55, 91, 136, 190, 253, 325, 406, 496, 595, 703, 820, 946, 1081, 1225, 1378, 1540, 1711, 1891, 2080, 2278, 2485, 2701, 2926, 3160, 3403, 3655, 3916, 4186, 4465, 4753, 5050, 5356, 5671, 5995, 6328, 6670, 7021, 7381, 7750, 8128, 8515, 8911, 9316

Centered enneagonal numbers reduced modulo 10 = RPS (1085160356)

Centered decagonal numbers: 1, 11, 31, 61, 101, 151, 211, 281, 361, 451, 551, 661, 781, 911, 1051, 1201, 1361, 1531, 1711, 1901, 2101, 2311, 2531, 2761, 3001, 3251, 3511, 3781, 4061, 4351, 4651, 4961, 5281, 5611, 5951, 6301, 6661, 7031, 7411, 7801, 8201, 8611, 9031, 9461, 9901

Centered decagonal numbers reduced modulo 10 = RPS (1)

Centered decagonal numbers reduced modulo 100 = RPS (1, 11, 31, 61, 1, 51, 11, 81, 61, 51)

Star numbers (i.e., centered dodecagonal numbers): 1, 13, 37, 73, 121, 181, 253, 337, 433, 541, 661, 793, 937, 1093, 1261, 1441, 1633, 1837, 2053, 2281, 2521, 2773, 3037, 3313, 3601, 3901, 4213, 4537, 4873, 5221, 5581, 5953, 6337, 6733, 7141, 7561, 7993, 8437, 8893, 9361, 9841, 10333, 10837

Star numbers reduced modulo 10 = RPS (13(7)

It certainly looks as if all such sequences reduce to RPSs, but the unexpected exception of the (non-centered) heptagonal numbers makes me hesitant to jump to that conclusion.

⁂

So the new mission (paging Kevin McCall!) is to come up with a general proof that almost all figurate number sequences reduce to RPSs -- a proof that makes it clear what the exceptions are and why. There's obviously a pattern here that goes beyond the triangular numbers, and it should be possible to express that pattern mathematically.

The difference between proof and understanding

The mathematical proofs laid out in my previous post (which I am sure very few of you have bothered to read) left me both satisfied and disappointed. Having had only the patchiest of mathematical educations (basic algebra and statistics, plus such rudiments of set theory and symbolic logic as linguists require; no trigonometry or calculus), I took a certain satisfaction in having been able to do it at all -- but it was disappointing to realize that I didn't seem much closer to understanding the patterns than I had been before. Why are they always palindromic, for instance? Saying that their palindromicity can be expressed algebraically as n(n + 1) ÷ 2 ≡ (2k - (n + 1))(2k - n) ÷ 2 (mod k), and that that equation turns out to be true for all natural number values of n and k, just doesn't count as an answer to that question. I can follow each step of the algebra, but in the end I do not feel enlightened; I do not think, "Oh, now I get it!" It is possible to prove something without really understanding it.

Then I read Kevin McCall's much better proofs of the same postulates. What a difference! Where I had hammered out my proofs by algebraic brute force, McCall had understood. -- and left me thinking, in T. H. Huxley's much-quoted words, "How extremely stupid not to have thought of that!"

⁂

The heart of McCall's proof is the observation that, in modular arithmetic with modulus k, adding k - n is equivalent to subtracting n. You can easily see this in the most familiar everyday use of modular arithmetic, which is our 12-hour clock, with modulus 12. If you want to get from 11:00 to 7:00, for example, you can either subtract 4 hours or add 12 - 4 = 8 hours.

The series of triangular numbers is generated by starting with 0, then adding 1, then adding 2, then 3, and so on through the succession of natural numbers. Due to the fact that k - n ≡ -n (mod k), one you've added numbers up to a certain point, you start doing the modular equivalent of subtracting those same numbers in reverse order, creating a palindrome. For example, if the modulus is 7:

• 0
• +1
• +2
• +3
• +4 ≡ -3 (mod 7)
• +5 ≡ -2 (mod 7)
• +6 ≡ -1 (mod 7) 
• etc.

Obviously, this will create a repeating palindromic pattern with a period of 7.

Why is the period twice as long for even moduli? Consider the case when the modulus is 6.

• 0
• +1
• +2
• +3
• +4 ≡ -2 (mod 6)
• +5 ≡ -1 (mod 6)
• +6 ≡ 0 (mod 6)
• +7 ≡ +1 (mod 6)
• +8 ≡ +2 (mod 6)
• +9 ≡ -3 (mod 6)
• +10 ≡ -2 (mod 6)
• +11 ≡ -1 (mod 6)
• etc.

Because the modulus k is even, k ÷ 2 ≡ -k ÷ 2 (mod k). In this case 3 ≡ -3 (mod 6). That means that, when we go through the first cycle of 6 integers, we add 1, add 2, add/subtract 3, subtract 2, and subtract 1. We can think of this as adding or subtracting 3, but whichever it is, we only perform the operation once in a single cycle, so at the end of the cycle we still have something congruent to 3 (mod 6); we have not yet returned to the 0 with which we began. However, if we go through two cycles, we can think of ourselves as adding 3 the first time around and then subtracting it the second time, bringing us back to our starting point. Thus, the period is twice as long for even moduli, and the palindrome is centered on k ÷ 2, a number which is special because it alone is congruent to its own negation.

⁂

Now I understand -- and just about anyone else can understand, too, without any need to pore over complicated algebraic operations. Congratulations, Kevin McCall; you've really solved this, whereas I was just crunching numbers.

Proving triangular number congruence patterns

For the patterns proven in this post, see here and here.

1. The formula for triangular numbers is T_n = n(n + 1) ÷ 2

This is common knowledge, but I include it for the benefit of any readers who may be even less mathematically inclined than myself.

The n^th triangular number (written T_n) is the sum of the natural numbers from 0 to n. Such numbers are called "triangular" because that number of points can be arranged in a triangular configuration as shown below.

Deriving the formula for triangular numbers is fairly straightforward. Take a given number, n, and write out 1 + 2 + ... + n. The sum of those numbers will be T_n. Now, below that, write the same sum in the opposite direction, n + n-1 + ... + 1. The example below shows what this looks like for n = 5.

Each of the two rows adds up to T_n, so the total of the two rows is 2T_n. If we look at the vertical columns, each of them adds up to n + 1. The first column is 1 + n; the second is 2 + (n - 1); the third is 3 + (n - 2); and so on. There are n columns in all, so 2T_n = n(n + 1). The formula for T_n, therefore, is n(n + 1) ÷ 2.

(Sorry if using an obelus instead of fractional notation makes these equations a little hard to read. It allows me to type them inline instead of inserting an image file for each equation.)

⁂

2. For any modulus k, the series of triangular numbers reduced modulo k repeats, with a period no greater than 2k.

This part was discovered by Kevin McCall, though I've reformulated it somewhat.

It is postulated that, for any modulus k, T_2k ≡ 0 (mod k). This means that T_2k is evenly divisible by k; in other words, that T_2k ÷ k is an integer.

Plugging 2k into our triangular number formula, we get T_2k = 2k(2k + 1) ÷ 2. Dividing by k gives us T_2k ÷ k = 2k(2k + 1) ÷ 2k = 2k + 1. Since k is itself an integer, 2k + 1 is an integer as well. Therefore, T_2k ≡ 0 (mod k).

Now consider the next triangular number in the series, T_{2k + 1}. For any given triangular number,  T_n , T_{n + 1} = T_n  + n + 1. (For example, for n = 10, the 10th triangular number is 55, and 55 + 10 + 1 = 66, which is the 11th triangular number.) Since 2k + 1 ≡ 1 (mod k), this amounts (in the modular arithmetic we are using) to adding 1, and T_{2k + 1} will be congruent to 1 (mod k).

Now the whole series of triangular numbers begins with 0. Then we add 1, then 2, then 3, and so on to infinity to generate the whole series.

• T₀ = 0
• T₁ = T₀ + 1 = 1
• T₂ = T₁ + 2 = 3
• T₃ = T₂ + 3 = 6
• T₄ = T₃ + 4 = 10
• etc.

It should be clear that, when we reach T_2k , the whole process begins again.

• T_2k ≡ 0 (mod k)
• T_{2k + 1} = T_2k + 2k + 1 ≡ 1 (mod k)
• T_{2k + 2} = T_{2k + 1} + 2k + 2 ≡ 1 + 2 ≡ 3 (mod k)
• T_{2k + 3} = T_{2k + 2} + 2k + 3 ≡ 3 + 3 ≡ 6 (mod k)
• T_{2k + 4} = T_{2k + 3} + 2k + 4 ≡ 6 + 4 ≡ 10 (mod k)
• etc.

We can see that any time T_n ≡ 0 and T_{n + 1} ≡ 1 (mod k), the series will repeat. We know that this always happens where n = 2k. Therefore, the period of the repetition must be either 2k or else a number by which 2k is evenly divisible.

⁂

3. The period of repetition is always either k (if k is odd) or 2k (if k is even).

If k is and odd number, the series of triangular numbers reduced modulo k will begin to repeat at k rather than 2k because T_k ≡ 0 (mod k), meaning that T_k ÷ k is an integer. T_k ÷ k = k(k + 1) ÷ 2k = (k + 1) ÷ 2. Since k is odd, k + 1 is even, so (k + 1) ÷ 2 is an integer. T_{k + 1} will then be equal to T_k + k + 1 ≡ 1 (mod k), and so on.

At this point my direct reliance on Kevin McCall's proof stops. Everything below is my own. (Kevin has proved it all, too, but I have not yet read his proof and will not do so until I have first proved it all myself.)

How do we know that the period is never smaller than k? Might it sometimes be k ÷ 2, for instance? No. While it sometimes happens that T_j ≡ 0 (mod k), where j < k, it is never also true that T_{j + 1} ≡ 1 (mod k), which is what is necessary for the series to repeat. Recall that T_{n + 1} = T_n  + n + 1. Therefore, where T_j ≡ 0 (mod k) and j < k, it follows that T_{j + 1} ≡ j + 1 (mod k). The only way j + 1 can be congruent to 1 (mod k) is if j ≡ 0 (mod k), which we have specified that it is not. (With the exception of 0, no number less than k can be congruent to 0 modulo k.)

Therefore, the period of repetition is always precisely k (if k is odd) or 2k (if k is even).

⁂

4. The repeating series is always palindromic.

It is postulated that for any modulus k, the series for triangular numbers from T₀ to T_{2k - 1} reduced modulo k (which series goes on to repeat itself forever, as proven above) is palindromic. This means that

• T₀ ≡ T_{2k - 1} (mod k)
• T₁ ≡ T_{2k - 2} (mod k)
• T₂ ≡ T_{2k - 3} (mod k)
• etc.

Stating this generally, we can say that T_n ≡ T_{2k - (n +1)} (mod k). Replacing T_n with the formula for triangular numbers, we arrive at the following equation: n(n + 1) ÷ 2 ≡ (2k - (n + 1))(2k - n) ÷ 2 (mod k).

Since multiplying both sides by the same number preserves congruence, we can simplify this to: n(n + 1) ≡ (2k - (n + 1))(2k - n) (mod k).

Doing the math, we find that this is equivalent to: n(n + 1) ≡ n(n + 1) + k(4k - 2(2n + 1)) (mod k).

Subtracting n(n + 1) from both sides, we get: k(4k - 2(2n + 1)) ≡ 0 (mod k). Trivially, any multiple of k is congruent to 0 modulo k, so this is true. Therefore, the series is palindromic.

⁂

5. Where k is even, the number at the center of the palindrome is k ÷ 2

The period of the repeating palindrome is 2k where k is even, so the number at the center of the palindrome is T_k = k(k + 1) ÷ 2. This number should be congruent to k ÷ 2 (mod k).

Multiplying both sides by 2, we get k(k + 1) ≡ k (mod k). Trivially, nk ≡ k ≡ 0 (mod k), for any integer value of n, so this is true.

⁂

Now that everything has been proven that I set out to prove, I will look at Kevin McCall's full proof and, if it is different from my own (as I suspect it will be), post it as well.

Triangular number congruence patterns proven

Kevin McCall, whose name you may recognize because his thoughts on dice and the Tarot (qv) have appeared on this blog in the past, has worked out a proof of the congruence patterns in the series of triangular numbers which I postulated here. He has proven that the series of triangular numbers reduced modulo k repeats itself, that the repetition has a period of k (if k is odd) or 2k (if k is even), and that the repeating series is always palindromic.

I have only seen his proof of the first of those statements; I am not going to look at the remainder of his proof until after I have proven it myself independently -- at which point I will post both his proof and mine.

Patterns in the digits of triangular numbers

Here's another way of saying what I said in my post on congruence patterns in the series of triangular numbers.

List the triangular numbers, starting with 0. The final digits of the numbers in the series will exhibit a repeating pattern with a period of 20, and the pattern will be palindromic. The penultimate (tens-place) digits will also exhibit a repeating palindromic pattern, with a period of 200. The repeating palindrome for the hundreds-place digits will have a period of 2000, and so on for any "place" you care to choose.

This is assuming you write the numbers in the familiar decimal system -- but any number base will yield corresponding results. If the triangular numbers are written in base b, the pth-to-last digit will exhibit a repeating palindromic pattern with a period of b^p (if b is odd) or 2b^p (if b is even).

(Oswald Wirth, whose writings first alerted me to the existence of such patterns, was doing what he called "Theosophic reduction" -- i.e., adding up the digits of a number until a single digit is arrived at. "Theosophically reducing" a number in base b corresponds to taking its final digit in base b - 1.)

The patterns are easiest to see in very small bases such as binary and ternary, shown below with color-coding to highlight the repeating palindromes.

I have still not figured out why these patterns exist. Even stating the pattern algebraically is proving somewhat difficult for someone whose mathematical training is as limited as mine.

Congruence patterns in the series of triangular numbers

The Swiss occultist Oswald Wirth writes of what he calls "Theosophic addition and reduction." I don't know what it's got to do with Theosophy, but I found it mathematically interesting.

By "Theosophic addition," Wirth simply means the operation which yields the series of triangular numbers, where the nth triangular number is the sum of all integers from 0 to n, inclusive; in other words, the nth triangular number is equal to (n² + n) ÷ 2. The first several triangular numbers are: 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666.

"Theosophic reduction" means adding up all the digits of a number, and then repeating the process as necessary until a one-digit number is arrived at. Mathematically, this amounts to finding the smallest positive integer to which it is congruent modulo 9.

Wirth took the first 21 triangular numbers (beginning with 1) and "Theosophically reduced" them, yielding this series: 1, 3, 6, 1, 6, 3, 1, 9, 9, 1, 3, 6, 1, 6, 3, 1, 9, 9, 1, 3, 6. As you can see, the same series of 9 numbers (1, 3, 6, 1, 6, 3, 1, 9, 9) repeats itself; and if you keep going beyond 21 (which is where Wirth stopped because he was considering the numerology of the 21 Tarot trumps), it becomes apparent that it keeps repeating itself forever.

Wirth took this pattern as confirmation of the traditional (ultimately Pythagorean) numerological idea that the first 9 natural numbers are the building blocks of all the rest, and that that "the decad is a new monad." However, it seemed pretty obvious to me that there is nothing special about the number 9, and that the 9-based pattern Wirth found was almost certainly an artifact of the use of the decimal system in the "Theosophical reduction" -- such that "reduction" meant finding congruence modulo 9 -- and that other moduli would yield other patterns.

I also saw that Wirth had missed an interesting pattern in his integer series because he had started with 1 rather than 0, and because he was thinking in terms of "reduction" rather than congruence. (Nine is congruent to 0 modulo 9, but you can't arrive at that 0 by adding up digits in the "Theosophic" fashion.) If we start with the 0th triangular number (which is 0), and if we "reduce" multiples of 9 to 0 rather than to 9, the repeating series becomes (0, 1, 3, 6, 1, 6, 3, 1, 0) -- which is a palindrome!

I decided to check other moduli, starting with the easiest, which is 10. Any decimal number is congruent modulo 10 to its final digit, so look back at that list of triangular numbers and look at the final digits. The first thing you will notice is that certain final digits (2, 4, 7, 9) never occur at all. Look a little further, and you will see that there is a repeating pattern. It has a period of 20 (not 10, as we might have expected) -- and, sure enough, it is a palindrome: (0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0).

It is natural to jump from this to the induction that every modulus will yield a palindromic repeating pattern, and this turns out to be true for all the moduli I have looked at, as the table below shows. I have used centered alignment to highlight the palindromic nature of the repeating series.

From this sample, it appears that:

1. Any modulus, m, yields a repeating palindromic series of congruences for the series of triangular numbers.
2. If m is odd, the period of the repeating series is equal to m.
3. If m is even, the period is equal to 2m, and each of the two numbers at the center of the palindrome is equal to m/2.

Now, I'm roughly 100% sure that I'm not the first person to have noticed these patterns, and that someone else has already mathematically proven what I have only induced. So, to those of my readers who have had a proper mathematical education -- no spoilers, please! I want to try to figure this out for myself.