Spiral Artefact #3

Tuesday, 19th Mar, 2024 by Krilling for Company in Arithmetic, Integer Sequences, Mathematics, Ulam Spirals and tagged A059270, adding and subtracting consecutive integers, diagonal of square numbers, grid of square blocks, OEIS, recreational math, recreational mathematics, recreational maths, spiral of integers, Square Numbers, sums of consecutive integers, sums of integers, Ulam Spirals, Ulam-like spirals | Leave a comment

What’s the next number in this sequence?

1, 3, 0, 4, 9, 15, 8, 0, 9, 19, 30, 42, 29, 15, 0, 16, 33, 51, 70, 90, 69, 47, 24, 0, 25, ?

Even if you can’t work out the full rule generating the sequence, you may be able to deduce that the next number is… 51. There’s a pattern involving 0:

1, 3, 0, 4, 9, 15, 8, 0, 9, 19, 30, 42, 29, 15, 0, 16, 33, 51, 70, 90, 69, 47, 24, 0, 25... → 3, 0, 4, 9, [...] 8, 0, 9, 19, [...] 15, 0, 16, 33, [...] 24, 0, 25...

The first number after each 0 is 1 more than the first number before the 0. The second number after the 0 is equal to 2 * (first-number-after 0) + 1. So:

1, 3, 0, 4, 2*4+1 = 9, [...] 8, 0, 9, 2*9+1 = 19, [...] 15, 0, 16, 2*16+1 = 33, [...] 24, 0, 25, 2*25+1 = 51...

But what is the full rule for generating the sequence? It’s based on this pattern of sums I noticed:

1+2 = 3 4+5+6 = 7+8 = 15 9+10+11+12 = 13+14+15 = 42 16+17+18+19+20 = 21+22+23+24 = 90 25+26+27+28+29+30 = 31+32+33+34+35 = 165 36+37+38+39+40+41+42 = 43+44+45+46+47+48 = 273 49+50+51+52+53+54+55+56 = 57+58+59+60+61+62+63 = 420 64+65+66+67+68+69+70+71+72 = 73+74+75+76+77+78+79+80 = 612 — See A059270 at the OEIS

The sum of the first two integers (1+2) equals the next integer (3). The sum of the next three integers (4+5+6) equals the sum of the next two integers (7+8). The sum of the next four integers (9+10+11+12) equals the sum of the next three integers (13+14+15). And so on. The sequence is based on an adaptation of that pattern:

1 + 2 - 3 = 0 4 + 5 + 6 - 7 - 8 = 0 9 + 10 + 11 + 12 - 13 - 14 - 15 = 0 16 + 17 + 18 + 19 + 20 - 21 - 22 - 23 - 24 = 0 25 + 26 + 27 + 28 + 29 + 30 - 31 - 32 - 33 - 34 - 35 = 0
↓
1 + 2 - 3 + 4 + 5 + 6 - 7 - 8 + 9 + 10 + 11 + 12 - 13 - 14 - 15 + 16 + 17 + 18 + 19 + 20 - 21 - 22 - 23 - 24 + 25 + 26 + 27 + 28 + 29 + 30 - 31 - 32 - 33 - 34 - 35...

If you work out the partial sums of the additions and subtractions, you get the sequence I started with, which regularly rises to a new high, then falls back to 0:

1, 3, 0, 4, 9, 15, 8, 0, 9, 19, 30, 42, 29, 15, 0, 16, 33, 51, 70, 90, 69, 47, 24, 0, 25, 51, 78, 106, 135, 165, 134, 102, 69, 35, 0, 36, 73, 111, 150, 190, 231, 273, 230, 186, 141, 95, 48, 0, 49, 99, 150, 202, 255, 309, 364, 420, 363, 305, 246, 186, 125, 63, 0, 64, 129, 195, 262, 330, 399, 469, 540, 612, 539, 465, 390, 314, 237, 159, 80, 0, 8 1, 163, 246, 330, 415, 501, 588, 676, 765, 855, 764, 672, 579, 485, 390, 294, 197, 99, 0, 100...

When you represent the numbers of the sequence on an Ulam-like spiral, you get this pattern of lines (and zigzags) against a haze of less regular points:

Spiral for pos2neg1 = 1, 3, 0, 4, 9, 15, 8, 0, 9, 19, 30, 42, 29, 15, 0, 16, 33…

I’ll call the lines spiral artefacts. I don’t know what generates all of them, but the zigzag diagonal from top left to bottom right is partly created by the square numbers. Here’s the spiral at higher resolutions:

Spiral for pos2neg1 (x2)

Spiral for pos2neg1 (x4)

You’ll find more of the lines if you look at Ulam-like spirals for adaptations of the original sequence. Suppose you add the first three integers, then take away the next two, then add the next four integers, then take away the next three, and so on: 1 + 2 + 3 – 4 – 5 + 6 + 7 + 8 + 9 – 10 – 11 – 12 + 13 + 14… Here are the partials sums of these additions and subtractions:

1, 3, 6, 2, -3, 3, 10, 18, 27, 17, 6, -6, 7, 21, 36, 52, 69, 51, 32, 12, -9, 13, 36, 60, 85, 111, 138, 110, 81, 51, 20, -12, 21, 55, 90, 126, 163, 201, 240, 200, 159, 117, 74, 30, -15, 31, 78, 126, 175, 225, 276, 328, 381, 327, 272, 216, 159, 101, 42, -18, 43, 105, 168, 232, 297, 363, 430, 498, 567, 497, 426, 354, 281, 207, 132, 56, -21, 57, 136, 216, 297, 379, 462, 546, 631, 717, 804, 716, 627, 537, 446, 354, 261, 167, 72, -24, 73, 171, 270, 370...

If the original sequence is pos2neg1 (add first two integers, take away next one integer, etc), this adapted sequence is pos3neg2 (add first three integers, take away next two, etc). Here’s the spiral for pos3neg2 (with negative numbers represented as positive):

Spiral for pos3neg2 = 1, 3, 6, 2, -3, 3, 10, 18, 27, 17, 6, -6, 7, 21, 36, 52,
69, 51, 32, 12…

Note that the spiral is incomplete and some of the lines not fully extended, because the lines are easier to see when the sequence doesn’t carry on too long and clutter the screen. Here are more adapted sequences shown on Ulam-like spirals (again, some of the spirals are incomplete):

Spiral for pos4neg3 = 1, 3, 6, 10, 5, -1, -8, 0, 9, 19, 30, 42, 29, 15, 0, -16, 1, 19, 38, 58…

Spiral for pos5neg4 = 1, 3, 6, 10, 15, 9, 2, -6, -15, -5, 6, 18, 31, 45, 60, 44, 27, 9, -10, -30…

Spiral for pos6neg5 = 1, 3, 6, 10, 15, 21, 14, 6, -3, -13, -24, -12, 1, 15, 30, 46, 63, 81, 62, 42…

Spiral for pos7neg6 = 1, 3, 6, 10, 15, 21, 28, 20, 11, 1, -10, -22, -35, -21, -6, 10, 27, 45, 64, 84…

Spiral for pos8neg7 = 1, 3, 6, 10, 15, 21, 28, 36, 27, 17, 6, -6, -19, -33, -48, -32, -15, 3, 22, 42…

Spiral for pos9neg8 = 1, 3, 6, 10, 15, 21, 28, 36, 45, 35, 24, 12, -1, -15, -30, -46, -63, -45, -26, -6…

Spiral for pos10neg9 = 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 44, 32, 19, 5, -10, -26, -43, -61, -80, -60…

Spiral for pos11neg10 = 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 54, 41, 27, 12, -4, -21, -39, -58, -78…

Elsewhere Other-Engageable

• Spiral Artefact #1 — different patterns on an Ulam-like spiral
• Spiral Artefact #2 — more different patterns

Spiral Artefact #2

Thursday, 29th Feb, 2024 by Krilling for Company in Arithmetic, Integer Sequences, Mathematics, Ulam Spirals and tagged diagonal of square numbers, grid of square blocks, recreational math, recreational mathematics, recreational maths, spiral of integers, Square Numbers, sums of consecutive integers, sums of integers, Ulam Spirals | Leave a comment

Why stop at primes? Those are the numbers the Ulam spiral is usually used for. You get a grid of square blocks, then move outward from the middle of the grid in a spiral, counting as you go. If the count matches a prime, you fill the block in. The first block is 1. Not filled. The second block is 2, which is prime. So the block is filled. The third block is 3, which is prime. Filled again. And so on. In the end, the Ulam spiral for primes looks like this:

The Ulam spiral of prime numbers

But why stop at primes? If you change the fill-test, you get different patterns. I’ve recently tried a test based on how many ways a number can be represented as the sum of consecutive integers. For example, 5, 208 and 536 can be represented in only one way:

5 = 2+3 208 = 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 536 = sum(26..41) = 26 + 27 + 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41

Let’s use “runsum” to mean a sum of consecutive integers. If the function runsum(n) returns the count of runsums for n, then runsum(5) = runsum(208) = runsum(536) = 1. Here are spirals for runsum(n) = 1:

A spiral for runsum(n) = 1, i.e. numbers that are the sum of consecutive integers in only one way

runsum(n) = 1 (higher resolution)

runsum(n) = 1 (higher resolution still)

Now try runsum(n) = 2, i.e. numbers that are the sum of consecutive integers in exactly two ways:

A spiral for runsum(n) = 2

runsum(n) = 2 (hi-res #1)

runsum(n) = 2 (hi-res #2)

runsum(n) = 2 (hi-res #3)

Why do most of the numbers fall on a diagonal? I don’t know, but I know that the diagonal represents square numbers:

9 = sum(4..5) = sum(2..4) 25 = sum(12..13) = sum(3..7) 36 = sum(11..13) = sum(1..8) 49 = sum(24..25) = sum(4..10)

Now try runsum(n) = 3:

A spiral for runsum(n) = 3

runsum(n) = 3 (hi-res)

It’s a densely packed spiral, unlike the spiral for runsum(n) = 4:

A spiral for runsum(n) = 4

runsum(n) = 4 (hi-res)

Like the spiral for runsum(n) = 2, the numbers are disproportionately falling on the diagonal of square numbers:

81 = 9^2 = sum(40..41) = sum(26..28) = sum(11..16) = sum(5..13) 324 = 18^2 = sum(107..109) = sum(37..44) = sum(32..40) = sum(2..25) 2500 = 50^2 = sum(498..502) = sum(309..316) = sum(88..112) = sum(43..82)

Here are spirals for runsum(n) = 5:

A spiral for runsum(n) = 5 (note patterns in green)

runsum(n) = 5 (hi-res #1)

runsum(n) = 5 (hi-res #2)

There are two interesting patterns in the spiral, marked in green above and enlarged below:

Pattern #1 in spiral for runsum(n) = 5

Pattern #2 in spiral for runsum(n) = 5

Are the patterns merely artefacts or does one or both represent something mathematically significant? I don’t know.

More spirals:

A spiral for runsum(n) = 6

A spiral for runsum(n) = 7

runsum(n) = 7 (hi-res)

A spiral for runsum(n) = 8

runsum(n) = 8 (hi-res #1)

runsum(n) = 8 (hi-res #2)

Numbers in the spiral for runsum(n) = 8 are again falling disproportionately on the diagonal of square numbers. Here’s one of those squares:

441 = 21^2 = sum(220..221) = sum(146..148) = sum(71..76) = sum(60..66) = sum(45..53) = sum(25..38) = sum(16..33) = sum(11..31)

Previously Pre-Posted…

• Spiral Artefact #1 — a look at patterns in spirals with different tests

RevNumSum

Monday, 23rd Aug, 2021 by Krilling for Company in Base Behavior, Integer Sequences, Mathematics and tagged base 2, base 3, bases, integer reversal, Integer Sequences, integers, math, mathematics, maths, recreational math, recreational mathematics, recreational maths, reversed numbers, revnumsum, sums of integers, Triangular Numbers | Leave a comment

If you take an integer, n, and reverse its digits to get the integer r, there are three possibilities:

n > r (e.g. 85236 > 63258) n < r (e.g. 17783 < 38771) n = r (e.g. 45154 = 45154)

If n = r, n is a palindrome. If n > r, I call n a major number. If n < r, I call n a minor number. And here are the minor and major numbers represented as white squares on an Ulam-like spiral (the negative of a minor spiral is a major spiral, and vice versa — sometimes one looks better than the other):

b=2 (minor numbers)

b=3

b=4

b=5

b=6

b=7 (major numbers)

b=8 (minor numbers)

b=9 (mjn)

b=10 (mjn)

b=11 (mjn)

b=12 (mjn)

b=13 (mjn)

b=14 (mjn)

b=15 (mjn)

b=16 (mjn)

b=17 (mjn)

b=18 (mjn)

b=19 (mjn)

b=20 (mjn)

Minor numbers, b=2..20 (animated)

Now let’s look at a sequence formed by summing the reversed numbers, minor ones, major ones and palindromes. Here are the standard integers:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17...

If you sum the integers, you get what are called the triangular numbers:

1 = 1 3 = 1 + 2 6 = 1 + 2 + 3 10 = 1 + 2 + 3 + 4 15 = 1 + 2 + 3 + 4 + 5 21 = 1 + 2 + 3 + 4 + 5 + 6 28 = 1 + 2 + 3 + 4 + 5 + 6 + 7 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 45 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 55 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 66 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 78 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 91 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 105 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 120 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 136 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 153 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 171 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 190 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 210 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20

But what happens if you reverse the integers before summing them? Here side-by-side are the triangular numbers and the underlined revnumsums (as they might be called):

45 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 45 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 55 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 46 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 66 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 57 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 78 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 78 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 91 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 109 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 105 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 150 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 120 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 201 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 + 51 136 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 262 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 + 51 + 61 153 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 333 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 + 51 + 61 + 71 171 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 414 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 + 51 + 61 + 71 + 81 190 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 505 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 + 51 + 61 + 71 + 81 + 91 210 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 507 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 + 51 + 61 + 71 + 81 + 91 + 2

Unlike triangular numbers, revnumsums are dependent on the base they’re calculated in. In base 2, the revnumsum is always smaller than the triangular number, except at step 1. In base 3, the revnumsum is equal to the triangular number at steps 1, 2 and 15 (= 120 in base 3). Otherwise it’s smaller than the triangular number.

And in higher bases? In bases > 3, the revnumsum rises and falls above the equivalent triangular number. When it’s higher, it tends towards a maximum height of (base+1)/4 * triangular number.

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Tag Archives: sums of integers

Spiral Artefact #3

Spiral Artefact #2

RevNumSum