“2n2 + 29 is prime for all values of n for 1 to 28.” — The Penguin Dictionary of Curious and Interesting Numbers, David Wells (1986).
• 31, 37, 47, 61, 79, 101, 127, 157, 191, 229, 271, 317, 367, 421, 479, 541, 607, 677, 751, 829, 911, 997, 1087, 1181, 1279, 1381, 1487, 1597.
Elsewhere other-posted:
• Prime Time #1
• Poulet’s Propellor — Musings on Math and Mathculinity
• La Spirale è Mobile
Papyrocentric Performativity Presents:
• Plates from the Great – Shots from the Front: The British Soldier 1914-18, Richard Holmes (HarperPress 2008; paperback 2010)
• Math for the Mistress – A Mathematician’s Apology, G.H. Hardy (1940)
• Sinister Sinema – Scalarama: A Celebration of Subterranean Cinema at Its Sleazy, Slimy and Sinister Best, ed. Norman Foreman, B.A. (TransVisceral Books 2015)
• Rick Pickings – Lost, Stolen or Shredded: Stories of Missing Works of Art and Literature, Rick Gekoski (Profile Books 2013/2014)
• Slug is a Drug – Collins Complete Guide to British Coastal Wildlife, Paul Sterry and Andrew Cleave (HarperCollins 2012) (posted @ Overlord of the Über-Feral)
Or Read a Review at Random: RaRaR
“The study of mathematics is the indispensable basis for all intellectual and spiritual progress.” — F.M. Cornford (1874-1943) quoted in The Sacred in Music (see also Pythagoreanism).
Challenger chopped and changed. That is to say, in one important respect, Arthur Conan Doyle’s character Professor Challenger lacked continuity. His philosophical views weren’t consistent. At one time he espoused materialism, at another he opposed it. He espoused it in “The Land of Mist” (1927):
“Don’t tell me, Daddy, that you with all your complex brain and wonderful self are a thing with no more life hereafter than a broken clock!”
“Four buckets of water and a bagful of salts,” said Challenger as he smilingly detached his daughter’s grip. “That’s your daddy, my lass, and you may as well reconcile your mind to it.”
But earlier, in “The Poison Belt” (1913), he had opposed it:
“No, Summerlee, I will have none of your materialism, for I, at least, am too great a thing to end in mere physical constituents, a packet of salts and three bucketfuls of water. Here ― here” ― and he beat his great head with his huge, hairy fist ― “there is something which uses matter, but is not of it ― something which might destroy death, but which death can never destroy.”
That story was published just over a century ago, but Challenger’s boast has not been vindicated in the meantime. So far as science can see, matter rules mind, not vice versa. Conan Doyle thought the same as the earlier Challenger, but Conan Doyle’s rich and teeming brain seems to have ended in “mere physical constituents”. To all appearances, when the organization of his brain broke down, so did his consciousness. And that concluded the cycle described by A.E. Housman in “Poem XXXII” of A Shropshire Lad (1896):
From far, from eve and morning
And yon twelve-winded sky,
The stuff of life to knit me
Blew hither: here am I.Now – for a breath I tarry
Nor yet disperse apart –
Take my hand quick and tell me,
What have you in your heart.Speak now, and I will answer;
How shall I help you, say;
Ere to the wind’s twelve quarters
I take my endless way. (ASL, XXXII)
Continue reading This Mortal Doyle…
Interesting patterns emerge when primes are represented as white blocks in a series of n-width left-right lines laid vertically, one atop the other. When the line is five blocks wide, the patterns look like this (the first green block is 1, followed by primes 2, 3 and 5, then 7 in the next line):
(Click for larger version)
Right at the bottom of the first column is an isolated prime diamond, or priamond (marked with a green block). It consists of the four primes 307-311-313-317, where the three latter primes equal 307 + 4 and 6 and 10, or 307 + 5-1, 5+1 and 5×2 (the last prime in the first column is 331 and the first prime in the second is 337). About a third of the way down the first column is a double priamond, consisting of 97, 101, 103, 107, 109 and 113. For a given n, then, a priamond is a set of primes, p1, p2, p3 and p4, such that p2 = p1 + n-1, p3 = p + n+1 and p4 = p1 + 2n.
There are also fragments of pearl-necklace in the columns. One is above the isolated priamond. It consists of four prime-blocks slanting from left to right: 251-257-263-269, or 251 + 6, 12 and 18. A prearl-necklace, then, is a set of primes, p1, p2, p3…, such that p2 = p1 + n+i, p3 = p + 2(n+i)…, where i = +/-1. Now here are the 7-line and 9-line:
Above: 7-line for primes
Above: 9-line for primes
In the 9-line, you can see a prime-ladder marked with a red block. It consists of the primes 43-53-61-71-79-89-97-107, in alternate increments of 10 and 8, or 9+1 and 9-1. A prime-ladder, then, is a set of primes, p1, p2, p3, p4…, such that p2 = p1 + n+1, p3 = p + 2n, p3 = p + 3n+1…
And here is an animated gif of lines 5 through 51:
(Click or open in new window for larger version or if file fails to animate)
“Exchange rate behaves like particles in a molecular fluid” — ScienceDaily, 13/iii/2014.
1,729,404 is seven digits long. If you drop one digit at a time, you can create seven more numbers from it, each six digits long. If you add these numbers, something special happens:
1,729,404 → 729404 (missing 1) + 129404 (missing 7) + 179404 (missing 2) + 172404 + 172904 + 172944 + 172940 = 1,729,404
So 1,729,404 is narcissistic, or equal to some manipulation of its own digits. Searching for numbers like this might seem like a big task, but you can cut the search-time considerably by noting that the final two digits determine whether a number is a suitable candidate for testing. For example, what if a seven-digit number ends in …38? Then the final digit of the missing-digit sum will equal (3 x 1 + 8 x 6) modulo 10 = (3 + 48) mod 10 = 51 mod 10 = 1. This means that you don’t need to check any seven-digit number ending in …38.
But what about seven-digit numbers ending in …57? Now the final digit of the sum will equal (5 x 1 + 7 x 6) modulo 10 = (5 + 42) mod 10 = 47 mod 10 = 7. So seven-digit numbers ending in …57 are possible missing-digit narcissistic sums. Then you can test numbers ending …157, …257, …357 and so on, to determine the last-but-one digit of the sum. Using this method, one quickly finds the only two seven-digit numbers of this form in base-10:
1,729,404 → 729404 + 129404 + 179404 + 172404 + 172904 + 172944 + 172940 = 1,729,404
1,800,000 → 800000 + 100000 + 180000 + 180000 + 180000 + 180000 + 180000 = 1,800,000
What about eight-digit numbers? Only those ending in these two digits need to be checked: …00, …23, …28, …41, …46, …64, …69, …82, …87. Here are the results:
• 13,758,846 → 3758846 + 1758846 + 1358846 + 1378846 + 1375846 + 1375846 + 1375886 + 1375884 = 13,758,846
• 13,800,000 → 3800000 + 1800000 + 1300000 + 1380000 + 1380000 + 1380000 + 1380000 + 1380000 = 13,800,000
• 14,358,846 → 4358846 + 1358846 + 1458846 + 1438846 + 1435846 + 1435846 + 1435886 + 1435884 = 14,358,846
• 14,400,000 → 4400000 + 1400000 + 1400000 + 1440000 + 1440000 + 1440000 + 1440000 + 1440000 = 14,400,000
• 15,000,000 → 5000000 + 1000000 + 1500000 + 1500000 + 1500000 + 1500000 + 1500000 + 1500000 = 15,000,000
• 28,758,846 → 8758846 + 2758846 + 2858846 + 2878846 + 2875846 + 2875846 + 2875886 + 2875884 = 28,758,846
• 28,800,000 → 8800000 + 2800000 + 2800000 + 2880000 + 2880000 + 2880000 + 2880000 + 2880000 = 28,800,000
• 29,358,846 → 9358846 + 2358846 + 2958846 + 2938846 + 2935846 + 2935846 + 2935886 + 2935884 = 29,358,846
• 29,400,000 → 9400000 + 2400000 + 2900000 + 2940000 + 2940000 + 2940000 + 2940000 + 2940000 = 29,400,000
But there are no nine-digit sumbers, or nine-digit numbers that supply missing-digit narcissistic sums. What about ten-digit sumbers? There are twenty-one:
1,107,488,889; 1,107,489,042; 1,111,088,889; 1,111,089,042; 3,277,800,000; 3,281,400,000; 4,388,888,889; 4,388,889,042; 4,392,488,889; 4,392,489,042; 4,500,000,000; 5,607,488,889; 5,607,489,042; 5,611,088,889; 5,611,089,042; 7,777,800,000; 7,781,400,000; 8,888,888,889; 8,888,889,042; 8,892,488,889; 8,892,489,042 (21 numbers)
Finally, the nine eleven-digit sumbers all take this form:
30,000,000,000 → 0000000000 + 3000000000 + 3000000000 + 3000000000 + 3000000000 + 3000000000 + 3000000000 + 3000000000 + 3000000000 + 3000000000 + 3000000000 = 30,000,000,000
So that’s forty-one narcissistic sumbers in base-10. Not all of them are listed in Sequence A131639 at the Encyclopedia of Integer Sequences, but I think I’ve got my program working right. Other bases show similar patterns. Here are some missing-digit narcissistic sumbers in base-5:
• 1,243 → 243 + 143 + 123 + 124 = 1,243 (b=5) = 198 (b=10)
• 1,324 → 324 + 124 + 134 + 132 = 1,324 (b=5) = 214 (b=10)
• 1,331 → 331 + 131 + 131 + 133 = 1,331 (b=5) = 216 (b=10)
• 1,412 → 412 + 112 + 142 + 141 = 1,412 (b=5) = 232 (b=10)
• 100,000 → 00000 + 10000 + 10000 + 10000 + 10000 + 10000 = 100,000 (b=5) = 3,125 (b=10)
• 200,000 → 00000 + 20000 + 20000 + 20000 + 20000 + 20000 = 200,000 (b=5) = 6,250 (b=10)
• 300,000 → 00000 + 30000 + 30000 + 30000 + 30000 + 30000 = 300,000 (b=5) = 9,375 (b=10)
• 400,000 → 00000 + 40000 + 40000 + 40000 + 40000 + 40000 = 400,000 (b=5) = 12,500 (b=10)
And here are some sumbers in base-16:
5,4CD,111,0EE,EF0,542 = 4CD1110EEEF0542 + 5CD1110EEEF0542 + 54D1110EEEF0542 + 54C1110EEEF0542 + 54CD110EEEF0542 + 54CD110EEEF0542 + 54CD110EEEF0542 + 54CD111EEEF0542 + 54CD1110EEF0542 + 54CD1110EEF0542 + 54CD1110EEF0542 + 54CD1110EEE0542 + 54CD1110EEEF542 + 54CD1110EEEF042 + 54CD1110EEEF052 + 54CD1110EEEF054 (b=16) = 6,110,559,033,837,421,890 (b=10)
6,5DD,E13,CEE,EF0,542 = 5DDE13CEEEF0542 + 6DDE13CEEEF0542 + 65DE13CEEEF0542 + 65DE13CEEEF0542 + 65DD13CEEEF0542 + 65DDE3CEEEF0542 + 65DDE1CEEEF0542 + 65DDE13EEEF0542 + 65DDE13CEEF0542 + 65DDE13CEEF0542 + 65DDE13CEEF0542 + 65DDE13CEEE0542 + 65DDE13CEEEF542 + 65DDE13CEEEF042 + 65DDE13CEEEF052 + 65DDE13CEEEF054 (b=16) = 7,340,270,619,506,705,730 (b=10)
10,000,000,000,000,000 → 0000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 + 1000000000000000 = 10,000,000,000,000,000 (b=16) = 18,446,744,073,709,551,616 (b=10)
F0,000,000,000,000,000 → 0000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 + F000000000000000 = F0,000,000,000,000,000 (b=16) = 276,701,161,105,643,274,240 (b=10)
Next I’d like to investigate sumbers created by missing two, three and more digits at a time. Here’s a taster:
1,043,101 → 43101 (missing 1 and 0) + 03101 (missing 1 and 4) + 04101 (missing 1 and 3) + 04301 + 04311 + 04310 + 13101 + 14101 + 14301 + 14311 + 14310 + 10101 + 10301 + 10311 + 10310 + 10401 + 10411 + 10410 + 10431 + 10430 + 10431 = 1,043,101 (b=5) = 18,526 (b=10)
Here’s a simple sequence. What’s the next number?
1, 2, 4, 8, 16, 68, 100, ?
The rule I’m using is this: Reverse the number, then add the sum of the digits. So 1 doubles till it becomes 16. Then 16 becomes 61 + 6 + 1 = 68. Then 68 becomes 86 + 8 + 6 = 100. Then 100 becomes 001 + 1 = 2. And the sequence falls into a loop.
Reversing the number means that small numbers can get big and big numbers can get small, but the second tendency is stronger for the first few seeds:
• 1 → 2 → 4 → 8 → 16 → 68 → 100 → 2
• 2 → 4 → 8 → 16 → 68 → 100 → 2
• 3 → 6 → 12 → 24 → 48 → 96 → 84 → 60 → 12
• 4 → 8 → 16 → 68 → 100 → 2 → 4
• 5 → 10 → 2 → 4 → 8 → 16 → 68 → 100 → 2
• 6 → 12 → 24 → 48 → 96 → 84 → 60 → 12
• 7 → 14 → 46 → 74 → 58 → 98 → 106 → 608 → 820 → 38 → 94 → 62 → 34 → 50 → 10 → 2 → 4 → 8 → 16 → 68 → 100 → 2
• 8 → 16 → 68 → 100 → 2 → 4 → 8
• 9 → 18 → 90 → 18
• 10 → 2 → 4 → 8 → 16 → 68 → 100 → 2
An 11-seed is a little more interesting:
11 → 13 → 35 → 61 → 23 → 37 → 83 → 49 → 107 → 709 → 923 → 343 → 353 → 364 → 476 → 691 → 212 → 217 → 722 → 238 → 845 → 565 → 581 → 199 → 1010 → 103 → 305 → 511 → 122 → 226 → 632 → 247 → 755 → 574 → 491 → 208 → 812 → 229 → 935 → 556 → 671 → 190 → 101 → 103 (11 leads to an 18-loop from 103 at step 26; total steps = 44)
Now try some higher bases:
• 1 → 2 → 4 → 8 → 15 → 57 → 86 → 80 → 15 (base=11)
• 1 → 2 → 4 → 8 → 14 → 46 → 72 → 34 → 4A → B6 → 84 → 58 → 96 → 80 → 14 (base=12)
• 1 → 2 → 4 → 8 → 13 → 35 → 5B → C8 → A6 → 80 → 13 (base=13)
• 1 → 2 → 4 → 8 → 12 → 24 → 48 → 92 → 36 → 6C → DA → C8 → A4 → 5A → B6 → 80 → 12 (base=14)
• 1 → 2 → 4 → 8 → 11 → 13 → 35 → 5B → C6 → 80 → 11 (base=15)
• 1 → 2 → 4 → 8 → 10 → 2 (base=16)
Does the 1-seed always create a short sequence? No, it gets pretty long in base-19 and base-20:
• 1 → 2 → 4 → 8 → [16] → 1D → DF → [17]3 → 4[18] → 107 → 709 → 914 → 424 → 42E → E35 → 54[17] → [17]5C → C7D → D96 → 6B3 → 3C7 → 7D6 → 6EE → E[16]2 → 2[18]8 → 90B → B1A → A2E → E3[17] → [17]5A → A7B → B90 → AC→ DD → F1 → 2C → C[16] → [18]2 → 40 → 8 (base=19)
• 1 → 2 → 4 → 8 → [16] → 1C → CE → F[18] → 108 → 80A → A16 → 627 → 731 → 13[18] → [18]43 → 363 → 36F → F77 → 794 → 4A7 → 7B5 → 5CA → ADC → CF5 → 5[17]4 → 4[18]B → B[19][17] → [18]1[18] → [18]3F → F5E → E79 → 994 → 4AB → BB9 → 9D2 → 2ED → DFB → B[17]C → C[19]B → C1E → E2[19] → [19]49 → 96B → B7F → F94 → 4B3 → 3C2 → 2D0 → D[17] → [19]3 → 51 → 1B → BD → EF → [17]3 → 4[17] → [18]5 → 71 → 1F → F[17] → [19]7 → 95 → 63 → 3F → [16]1 → 2D → D[17] (base=20)
Then it settles down again:
• 1 → 2 → 4 → 8 → [16] → 1B → BD → EE → [16]0 → 1B (base=21)
• 1 → 2 → 4 → 8 → [16] → 1A → AC → DA → BE → FE → [16]0 → 1A (base=22)
• 1 → 2 → 4 → 8 → [16] → 19 → 9B → C6 → 77 → 7[21] → [22]C → EA → BF → [16]E → [16]0 → 19 (base=23)
Base-33 is also short:
1 → 2 → 4 → 8 → [16] → [32] → 1[31] → [32]0 → 1[31] (base=33)
And so is base-35:
1 → 2 → 4 → 8 → [16] → [32] → 1[29] → [29][31] → [33][19] → [21]F → [16][22] → [23][19] → [20][30] → [32]0 → 1[29] (base=35)
So what about base-34?
1 → 2 → 4 → 8 → [16] → [32] → 1[30] → [30][32] → 10[24] → [24]0[26] → [26]26 → 63[26] → [26]47 → 75[29] → [29]6E → E8A → A9C → CA7 → 7B7 → 7B[32] → [32]C[23] → [23]E[31] → [31][16][23] → [23][18][33] → [33][20][29] → [29][23]D → D[25][26] → [26][27]9 → 9[29][20] → [20][30][33] → [33][33]1 → 21[32] → [32]23 → 341 → 14B → B4[17] → [17]59 → 96E → E74 → 485 → 58[21] → [21]95 → 5A[22] → [22]B8 → 8C[29] → [29]D[23] → [23]F[26] → [26][17][19] → [19][19][20] → [20][21]9 → 9[23]2 → 2[24]9 → 9[25]3 → 3[26]C → C[27]A → A[28][27] → [27][30]7 → 7[32][23] → [24]01 → 11F → F1[18] → [18]2F → F3[19] → [19]4[18] → [18]5[26] → [26]6[33] → [33]8[23] → [23]A[29] → [29]C[17] → [17]E[19] → [19]F[33] → [33][17][18] → [18][19][33] → [33][21][20] → [20][24]5 → 5[26]1 → 1[27]3 → 3[27][32] → [32][28][31] → [31][31][21] → [22]0C → C1[22] → [22]2D → D3[25] → [25]4[20] → [20]66 → 67[18] → [18]83 → 39D → D9[28] → [28]A[29] → [29]C[27] → [27]E[29] → [29][16][29] → [29][19]1 → 1[21]A → A[21][33] → [33][23]6 → 6[25][27] → [27][26][30] → [30][29]8 → 8[31][29] → [29][33]8 → 91[31] → [31]2[16] → [16]4C → C5E → E69 → 979 → 980 → 8[26] → [27]8 → 9[28] → [29]C → E2 → 2[30] → [31]0 → 1[28] → [28][30] → [32][18] → [20]E → F[20] → [21][16] → [17][24] → [25][24] → [26]6 → 7[24] → [25]4 → 5[20] → [20][30] → [32]2 → 3[32] → [33]4 → 62 → 2E → E[18] → [19]C → D[16] → [17]8 → 98 → 8[26] (1 leads to a 30-loop from 8[26] / 298 in base-34 at step 111; total steps = 141)
An alternative rule is to add the digit-sum first and then reverse the result. Now 8 becomes 8 + 8 = 16 and 16 becomes 61. Then 61 becomes 61 + 6 + 1 = 68 and 68 becomes 86. Then 86 becomes 86 + 8 + 6 = 100 and 100 becomes 001 = 1:
• 1 → 2 → 4 → 8 → 61 → 86 → 1
• 2 → 4 → 8 → 61 → 86 → 1 → 2
• 3 → 6 → 21 → 42 → 84 → 69 → 48 → 6
• 4 → 8 → 61 → 86 → 1 → 2 → 4
• 5 → 1 → 2 → 4 → 8 → 62 → 7 → 48 → 6 → 27 → 63 → 27
• 6 → 21 → 42 → 84 → 69 → 48 → 6
• 7 → 41 → 64 → 47 → 85 → 89 → 601 → 806 → 28 → 83 → 49 → 26 → 43 → 5 → 6 → 27 → 63 → 27
• 8 → 61 → 86 → 1 → 2 → 4 → 8
• 9 → 81 → 9
• 10 → 11 → 31 → 53 → 16 → 32 → 73 → 38 → 94 → 701 → 907 → 329 → 343 → 353 → 463 → 674 → 196 → 212 → 712 → 227 → 832 → 548 → 565 → 185 → 991 → 101 → 301 → 503 → 115 → 221 → 622 → 236 → 742 → 557 → 475 → 194→ 802 → 218 → 922 → 539 → 655 → 176 → 91 → 102 → 501 → 705 → 717 → 237 → 942 → 759 → 87 → 208 → 812 → 328 → 143 → 151 → 851 → 568 → 785 → 508 → 125 → 331 → 833 → 748 → 767 → 787 → 908 → 529 → 545 → 955 → 479 → 994 → 6102 → 1116 → 5211 → 225 → 432 → 144 → 351 → 63 → 27 → 63
Papyrocentric Performativity Presents:
• Book in Black – Black Sabbath: Symptom of the Universe, Mick Wall (Orion Books 2013)
• Critical Math – A Mathematician Reads the Newspaper, John Allen Paulos (Penguin 1996)
• Rude Boys – Ruthless: The Global Rise of the Yardies, Geoff Small (Warner 1995)
• K-9 Konundrum – Dog, Peter Sotos (TransVisceral Books 2014)
• Ghosts in the Cathedral – The Neutrino Hunters: The Chase for the Ghost Particle and the Secrets of the Universe, Ray Jayawardhana (Oneworld 2013) (posted @ Overlord of the Über-Feral)
Or Read a Review at Random: RaRaR
123456789. How many ways are there to insert + and – between the numbers and create a formula for 100? With pen and ink it takes a long time to answer. With programming, the answer will flash up in an instant:
01. 1 + 2 + 3 - 4 + 5 + 6 + 78 + 9 = 100 02. 1 + 2 + 34 - 5 + 67 - 8 + 9 = 100 03. 1 + 23 - 4 + 5 + 6 + 78 - 9 = 100 04. 1 + 23 - 4 + 56 + 7 + 8 + 9 = 100 05. 12 - 3 - 4 + 5 - 6 + 7 + 89 = 100 06. 12 + 3 + 4 + 5 - 6 - 7 + 89 = 100 07. 12 + 3 - 4 + 5 + 67 + 8 + 9 = 100 08. 123 - 4 - 5 - 6 - 7 + 8 - 9 = 100 09. 123 + 4 - 5 + 67 - 89 = 100 10. 123 + 45 - 67 + 8 - 9 = 100 11. 123 - 45 - 67 + 89 = 100
And the beauty of programming is that you can easily generalize the problem to other bases. In base b, how many ways are there to insert + and – in the block [12345…b-1] to create a formula for b^2? When b = 10, the answer is 11. When b = 11, it’s 42. Here are two of those formulae in base-11:
123 - 45 + 6 + 7 - 8 + 9 + A = 100[b=11] 146 - 49 + 6 + 7 - 8 + 9 + 10 = 121 123 + 45 + 6 + 7 - 89 + A = 100[b=11] 146 + 49 + 6 + 7 - 97 + 10 = 121
When b = 12, it’s 51. Here are two of the formulae:
123 + 4 + 5 + 67 - 8 - 9A + B = 100[b=12] 171 + 4 + 5 + 79 - 8 - 118 + 11 = 144 123 + 4 + 56 + 7 - 89 - A + B = 100[b=12] 171 + 4 + 66 + 7 - 105 - 10 + 11 = 144
So that’s 11 formulae in base-10, 42 in base-11 and 51 in base-12. So what about base-13? The answer may be surprising: in base-13, there are no +/- formulae for 13^2 = 169 using the numbers 1 to 12. Nor are there any formulae in base-9 for 9^2 = 81 using the numbers 1 to 8. If you reverse the block, 987654321, the same thing happens. Base-10 has 15 formulae, base-11 has 54 and base-12 has 42. Here are some examples:
9 - 8 + 7 + 65 - 4 + 32 - 1 = 100 98 - 76 + 54 + 3 + 21 = 100 A9 + 87 - 65 + 4 - 3 - 21 = 100[b=11] 119 + 95 - 71 + 4 - 3 - 23 = 121 BA - 98 + 76 - 5 - 4 + 32 - 1 = 100[b=12] 142 - 116 + 90 - 5 - 4 + 38 - 1 = 144
But base-9 and base-13 again have no formulae. What’s going on? Is it a coincidence that 9 and 13 are each one more than a multiple of 4? No. Base-17 also has no formulae for b^2 = 13^2 = 169. Here is the list of formulae for bases-7 thru 17:
1, 2, 0, 11, 42, 51, 0, 292, 1344, 1571, 0 (block = 12345...) 3, 2, 0, 15, 54, 42, 0, 317, 1430, 1499, 0 (block = ...54321)
To understand what’s going on, consider any sequence of consecutive integers starting at 1. The number of odd integers in the sequence must always be greater than or equal to the number of even integers:
1, 2 (1 odd : 1 even) 1, 2, 3 (2 odds : 1 even) 1, 2, 3, 4 (2 : 2) 1, 2, 3, 4, 5 (3 : 2) 1, 2, 3, 4, 5, 6 (3 : 3) 1, 2, 3, 4, 5, 6, 7 (4 : 3) 1, 2, 3, 4, 5, 6, 7, 8 (4 : 4)
The odd numbers in a sequence determine the parity of the sum, that is, whether it is odd or even. For example:
1 + 2 = 3 (1 odd number) 1 + 2 + 3 = 6 (2 odd numbers) 1 + 2 + 3 + 4 = 10 (2 odd numbers) 1 + 2 + 3 + 4 + 5 = 15 (3 odd numbers) 1 + 2 + 3 + 4 + 5 + 6 = 21 (3 odd numbers) 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 (4 odd numbers)
If there is an even number of odd numbers, the sum will be even; if there is an odd number, the sum will be odd. Consider sequences that end in a multiple of 4:
1, 2, 3, 4 → 2 odds : 2 evens 1, 2, 3, 4, 5, 6, 7, 8 → 4 : 4 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 → 6 : 6
Such sequences always contain an even number of odd numbers. Now, consider these formulae in base-10:
1. 12 + 3 + 4 + 56 + 7 + 8 + 9 = 99 2. 123 - 45 - 67 + 89 = 100 3. 123 + 4 + 56 + 7 - 89 = 101
They can be re-written like this:
1. 1×10^1 + 2×10^0 + 3×10^0 + 4×10^0 + 5×10^1 + 6×10^0 + 7×10^0 + 8×10^0 + 9×10^0 = 99
2. 1×10^2 + 2×10^1 + 3×10^0 – 4×10^1 – 5×10^0 – 6×10^1 – 7×10^0 + 8×10^1 + 9×10^0 = 100
3. 1×10^2 + 2×10^1 + 3×10^0 + 4×10^0 + 5×10^1 + 6×10^1 + 7×10^0 – 8×10^1 – 9×10^0 = 101
In general, the base-10 formulae will take this form:
1×10^a +/- 2×10^b +/- 3×10^c +/– 4×10^d +/– 5×10^e +/– 6×10^f +/– 7×10^g +/– 8×10^h +/– 9×10^i = 100
It’s important to note that the exponent of 10, or the power to which it is raised, determines whether an odd number remains odd or becomes even. For example, 3×10^0 = 3×1 = 3, whereas 3×10^1 = 3×10 = 30 and 3×10^2 = 3×100 = 300. Therefore the number of odd numbers in a base-10 formula can vary and so can the parity of the sum. Now consider base-9. When you’re trying to find a block-formula for 9^2 = 81, the formula will have to take this form:
1×9^a +/- 2×9^b +/- 3×9^c +/- 4×9^d +/- 5×9^e +/- 6×9^f +/- 7×9^g +/- 8×9^h = 81
But no such formula exists for 81 (with standard exponents). It’s now possible to see why this is so. Unlike base-10, the odd numbers in the formula will remain odd what the power of 9. For example, 3×9^0 = 3×1 = 3, 3×9^1 = 3×9 = 27 and 3×9^2 = 3×81 = 243. Therefore base-9 formulae will always contain four odd numbers and will always produce an even number. Odd numbers in base-2 always end in 1, even numbers always end in 0. Therefore, to determine the parity of a sum of integers, convert the integers to base-2, discard all but the final digit of each integer, then sum the 1s. In a base-9 formula, these are the four possible results:
1 + 1 + 1 + 1 = 4 1 + 1 + 1 - 1 = 2 1 + 1 - 1 - 1 = 0 1 - 1 - 1 - 1 = -2
The sum represents the parity of the answer, which is always even. Similar reasoning applies to base-13, base-17 and all other base-[b=4n+1].
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