Magistra Rules the Waves

One of my favourite integer sequences has the simple formula n(i) = n(i-1) + digitsum(n(i-1)). If it’s seeded with 1, its first few terms go like this:

n(1) = 1
n(2) = n(1) + digitsum(n(1)) = 1 + digitsum(1) = 2
n(3) = 2 + digitsum(2) = 4
n(4) = 4 + digitsum(4) = 8
n(5) = 8 + digitsum(8) = 16
n(6) = 16 + digitsum(16) = 16 + 1+6 = 16 + 7 = 23
n(7) = 23 + digitsum(23) = 23 + 2+3 = 23 + 5 = 28
n(8) = 28 + digitsum(28) = 28 + 2+8 = 28 + 10 = 38

As a sequence, it looks like this:

1, 2, 4, 8, 16, 23, 28, 38, 49, 62, 70, 77, 91, 101, 103, 107, 115, 122, 127, 137, 148, 161, 169, 185, 199, 218, 229, 242, 250, 257, 271, 281, 292, 305, 313, 320, 325, 335, 346, 359, 376, 392, 406, 416, 427, 440, 448, 464, 478, 497, 517, 530, 538, 554, 568, 587, 607, 620, 628, 644, 658, 677, 697, 719, 736, 752, 766, 785, 805, 818, 835, 851, 865, 884, 904, 917, 934, 950, 964, 983, 1003…

Given a number at random, is there a quick way to say whether it appears in the sequence seeded with 1? Not that I know, with one exception. If the number is divisible by 3, it doesn’t appear, at least in base 10. In base 2, that rule doesn’t apply:

n(1) = 1
n(2) = 1 + digitsum(1) = 10 = 1 + 1 = 2
n(3) = 10 + digitsum(10) = 10 + 1 = 11 = 2 + 1 = 3
n(4) = 11 + digitsum(11) = 11 + 1+1 = 101 = 3 + 2 = 5
n(5) = 101 + digitsum(101) = 101 + 1+0+1 = 111 = 5 + 2 = 7
n(6) = 111 + digitsum(111) = 111 + 11 = 1010 = 7 + 3 = 10
n(7) = 1010 + digitsum(1010) = 1010 + 10 = 1100 = 10 + 2 = 12
n(8) = 1100 + digitsum(1100) = 1100 + 10 = 1110 = 12 + 2 = 14

1, 2, 3, 5, 7, 10, 12, 14, 17, 19, 22, 25, 28, 31, 36, 38, 41, 44, 47, 52, 55, 60, 64, 65, 67, 70, 73, 76, 79, 84, 87, 92, 96, 98, 101, 105, 109, 114, 118, 123, 129, 131, 134, 137, 140, 143, 148, 151, 156, 160, 162, 165, 169, 173, 178, 182, 187, 193, 196, 199, 204, 208, 211, 216, 220, 225, 229, 234, 239, 246, 252, 258, 260, 262, 265, 268, 271, 276, 279, 284, 288, 290, 293, 297, 301, 306, 310, 315, 321, 324, 327, 332, 336, 339, 344, 348, 353, 357, 362, 367, 374…

What patterns are there in these sequences? It’s easier to check when they’re represented graphically, so I converted them into patterns à la the Ulam spiral, where n is represented as a dot on a spiral of integers. This is the spiral for base 10:

ulambase10Base 10


And these are the spirals for bases 2 and 3:

ulambase2

Base 2


ulambase3

Base 3


These sequences look fairly random to me: there are no obvious patterns in the jumps from n(i) to n(i+1), i.e. in the values for digitsum(n(i)). Now try the spirals for bases 9 and 33:

ulambase9

Base 9


ulambase33

Base 33


Patterns have appeared: there is some regularity in the jumps. You can see these regularities more clearly if you represent digitsum(n(i)) as a graph, with n(i) on the x axis and digitsum(n(i)) on the y axis. If the graph starts with n(i) = 1 on the lower left and proceeds left-right, left-right up the screen, it looks like this in base 10:

base10

Base 10 (click to enlarge)


Here are bases 2 and 3:

base2

Base 2


base3

Base 3


The jumps seem fairly random. Now try bases 9, 13, 16, 17, 25, 33 and 49:

base9

Base 9


base13

Base 13


base16

Base 16


base17

Base 17


base25

Base 25


base33

Base 33


base49

Base 49


In some bases, the formula n(i) = n(i-1) + digitsum(n(i-1)) generates mild randomness. In others, it generates strong regularity, like waves rolling ashore under a steady wind. I don’t understand why, but regularity seems to occur in bases that are one more than a power of 2 and also in some bases that are primes or squares.


Elsewhere other-posted:

Mathematica Magistra Mundi
8200_idf_insignia

Reverssum

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

Six Six Nix

4 x 3 = 13. A mistake? Not in base-9, where 13 = 1×9^1 + 3 = 12 in base-10. This means that 13 is a sum-product number in base-9: first add its digits, then multiply them, then multiply the digit-sum by the digit-product: (1+3) x (1×3) = 13[9]. There are four more sum-product numbers in this base:

2086[9] = 17 x 116 = (2 + 8 + 6) x (2 x 8 x 6) = 1536[10] = 16 x 96
281876[9] = 35 x 7333 = (2 + 8 + 1 + 8 + 7 + 6) x (2 x 8 x 1 x 8 x 7 x 6) = 172032[10] = 32 x 5376
724856[9] = 35 x 20383 = (7 + 2 + 4 + 8 + 5 + 6) x (7 x 2 x 4 x 8 x 5 x 6) = 430080[10] = 32 x 13440
7487248[9] = 44 x 162582 = (7 + 4 + 8 + 7 + 2 + 4 + 8) x (7 x 4 x 8 x 7 x 2 x 4 x 8) = 4014080[10] = 40 x 100352

And that’s the lot, apart from the trivial 0 = (0) x (0) and 1 = (1) x (1), which are true in all bases.

What about base-10?

135 = 9 x 15 = (1 + 3 + 5) x (1 x 3 x 5)
144 = 9 x 16 = (1 + 4 + 4) x (1 x 4 x 4)
1088 = 17 x 64 = (1 + 8 + 8) x (1 x 8 x 8)

1088 is missing from the list at Wikipedia and the Encyclopedia of Integer Sequences, but I like the look of it, so I’m including it here. Base-11 has five sum-product numbers:

419[11] = 13 x 33 = (4 + 1 + 9) x (4 x 1 x 9) = 504[10] = 14 x 36
253[11] = [10] x 28 = (2 + 5 + 3) x (2 x 5 x 3) = 300[10] = 10 x 30
2189[11] = 19 x 121 = (2 + 1 + 8 + 9) x (2 x 1 x 8 x 9) = 2880[10] = 20 x 144
7634[11] = 19 x 419 = (7 + 6 + 3 + 4) x (7 x 6 x 3 x 4) = 10080[10] = 20 x 504
82974[11] = 28 x 3036 = (8 + 2 + 9 + 7 + 4) x (8 x 2 x 9 x 7 x 4) = 120960[10] = 30 x 4032

But the record for bases below 50 is set by 7:

22[7] = 4 x 4 = (2 + 2) x (2 x 2) = 16[10] = 4 x 4
505[7] = 13 x 34 = (5 + 5) x (5 x 5) = 250[10] = 10 x 25
242[7] = 11 x 22 = (2 + 4 + 2) x (2 x 4 x 2) = 128[10] = 8 x 16
1254[7] = 15 x 55 = (1 + 2 + 5 + 4) x (1 x 2 x 5 x 4) = 480[10] = 12 x 40
2343[7] = 15 x 132 = (2 + 3 + 4 + 3) x (2 x 3 x 4 x 3) = 864[10] = 12 x 72
116655[7] = 33 x 2424 = (1 + 1 + 6 + 6 + 5 + 5) x (1 x 1 x 6 x 6 x 5 x 5) = 21600[10] = 24 x 900
346236[7] = 33 x 10362 = (3 + 4 + 6 + 2 + 3 + 6) x (3 x 4 x 6 x 2 x 3 x 6) = 62208[10] = 24 x 2592
424644[7] = 33 x 11646 = (4 + 2 + 4 + 6 + 4 + 4) x (4 x 2 x 4 x 6 x 4 x 4) = 73728[10] = 24 x 3072

And base-6? Six Nix. There are no sum-product numbers unique to that base (to the best of my far-from-infallible knowledge). Here is the full list for base-3 to base-50 (not counting 0 and 1 as sum-product numbers):

5 in base-11 4 in base-21 3 in base-31 2 in base-41
4 in base-12 5 in base-22 1 in base-32 3 in base-42
0 in base-3 3 in base-13 4 in base-23 3 in base-33 4 in base-43
2 in base-4 3 in base-14 2 in base-24 4 in base-34 5 in base-44
1 in base-5 2 in base-15 3 in base-25 2 in base-35 6 in base-45
0 in base-6 2 in base-16 6 in base-26 2 in base-36 7 in base-46
8 in base-7 6 in base-17 0 in base-27 3 in base-37 3 in base-47
1 in base-8 5 in base-18 1 in base-28 3 in base-38 7 in base-48
5 in base-9 7 in base-19 0 in base-29 1 in base-39 5 in base-49
3 in base-10 3 in base-20 2 in base-30 2 in base-40 3 in base-50

DeVil to Power

666 is the Number of the Beast described in the Book of Revelation:

13:18 Here is wisdom. Let him that hath understanding count the number of the beast: for it is the number of a man; and his number is Six hundred threescore and six.

But 666 is not just diabolic: it’s narcissistic too. That is, it mirrors itself using arithmetic, like this:

666^47 =

5,049,969,684,420,796,753,173,148,798,405,
  564,772,941,516,295,265,408,188,117,632,
  668,936,540,446,616,033,068,653,028,889,
  892,718,859,670,297,563,286,219,594,665,
  904,733,945,856 → 5 + 0 + 4 + 9 + 9 + 6 + 9 + 6 + 8 + 4 + 4 + 2 + 0 + 7 + 9 + 6 + 7 + 5 + 3 + 1 + 7 + 3 + 1 + 4 + 8 + 7 + 9 + 8 + 4 + 0 + 5 + 5 + 6 + 4 + 7 + 7 + 2 + 9 + 4 + 1 + 5 + 1 + 6 + 2 + 9 + 5 + 2 + 6 + 5 + 4 + 0 + 8 + 1 + 8 + 8 + 1 + 1 + 7 + 6 + 3 + 2 + 6 + 6 + 8 + 9 + 3 + 6 + 5 + 4 + 0 + 4 + 4 + 6 + 6 + 1 + 6 + 0 + 3 + 3 + 0 + 6 + 8 + 6 + 5 + 3 + 0 + 2 + 8 + 8 + 8 + 9 + 8 + 9 + 2 + 7 + 1 + 8 + 8 + 5 + 9 + 6 + 7 + 0 + 2 + 9 + 7 + 5 + 6 + 3 + 2 + 8 + 6 + 2 + 1 + 9 + 5 + 9 + 4 + 6 + 6 + 5 + 9 + 0 + 4 + 7 + 3 + 3 + 9 + 4 + 5 + 8 + 5 + 6 = 666

666^51 =

993,540,757,591,385,940,334,263,511,341,
295,980,723,858,637,469,431,008,997,120,
691,313,460,713,282,967,582,530,234,558,
214,918,480,960,748,972,838,900,637,634,
215,694,097,683,599,029,436,416 → 9 + 9 + 3 + 5 + 4 + 0 + 7 + 5 + 7 + 5 + 9 + 1 + 3 + 8 + 5 + 9 + 4 + 0 + 3 + 3 + 4 + 2 + 6 + 3 + 5 + 1 + 1 + 3 + 4 + 1 + 2 + 9 + 5 + 9 + 8 + 0 + 7 + 2 + 3 + 8 + 5 + 8 + 6 + 3 + 7 + 4 + 6 + 9 + 4 + 3 + 1 + 0 + 0 + 8 + 9 + 9 + 7 + 1 + 2 + 0 + 6 + 9 + 1 + 3 + 1 + 3 + 4 + 6 + 0 + 7 + 1 + 3 + 2 + 8 + 2 + 9 + 6 + 7 + 5 + 8 + 2 + 5 + 3 + 0 + 2 + 3 + 4 + 5 + 5 + 8 + 2 + 1 + 4 + 9 + 1 + 8 + 4 + 8 + 0 + 9 + 6 + 0 + 7 + 4 + 8 + 9 + 7 + 2 + 8 + 3 + 8 + 9 + 0 + 0 + 6 + 3 + 7 + 6 + 3 + 4 + 2 + 1 + 5 + 6 + 9 + 4 + 0 + 9 + 7 + 6 + 8 + 3 + 5 + 9 + 9 + 0 + 2 + 9 + 4 + 3 + 6 + 4 + 1 + 6 = 666

But those are tiny numbers compared to 6^(6^6). That means 6^46,656 and equals roughly 2·6591… x 10^36,305. It’s 36,306 digits long and its full digit-sum is 162,828. However, 666 lies concealed in those digits too. To see how, consider the function Σ(x1,xn), which returns the sum of digits 1 to n of x. For example, π = 3·14159265…, so Σ(π14) = 3 + 1 + 4 + 1 = 9. The first 150 digits of 6^(6^6) are these:

26591197721532267796824894043879185949053422002699
24300660432789497073559873882909121342292906175583
03244068282650672342560163577559027938964261261109
… (150 digits)

If x = 6^(6^6), then Σ(x1,x146) = 666, Σ(x2,x148) = 666, and Σ(x2,x149) = 666.

There’s nothing special about these patterns: infinitely many numbers are narcissistic in similar ways. However, 666 has a special cultural significance, so people pay it more attention and look for patterns related to it more carefully. Who cares, for example, that 667 = digit-sum(667^48) = digit-sum(667^54) = digit-sum(667^58)? Fans of recreational maths will, but not very much. The Number of the Beast is much more fun, narcissistically and otherwise:

666 = digit-sum(6^194)
666 = digit-sum(6^197)

666 = digit-sum(111^73)
666 = digit-sum(111^80)

666 = digit-sum(222^63)
666 = digit-sum(222^66)

666 = digit-sum(333^58)
666 = digit-sum(444^53)
666 = digit-sum(777^49)
666 = digit-sum(999^49)


Previously pre-posted (please peruse):

More Narcissisum
Digital Disfunction
The Hill to Power
Narcissarithmetic #1
Narcissarithmetic #2

More Narcissisum

The number 23 is special, inter alia, because it’s prime, divisible by only itself and 1. It’s also special because its reciprocal has maximum period. That is, the digits of 1/23 come in repeated blocks of 22, like this:

1/23 = 0·0434782608695652173913  0434782608695652173913  0434782608695652173913…

But 1/23 fails to be special in another way: you can’t sum its digits and get 23:

0 + 4 + 3 + 4 + 7 = 18
0 + 4 + 3 + 4 + 7 + 8 = 26
0 + 4 + 3 + 4 + 7 + 8 + 2 + 6 + 0 + 8 + 6 + 9 + 5 + 6 + 5 + 2 + 1 + 7 + 3 + 9 + 1 + 3 = 99

1/7 is different:

1/7 = 0·142857… → 1 + 4 + 2 = 7

This means that 7 is narcissistic: it reflects itself by manipulation of the digits of 1/7. But that’s in base ten. If you try base eight, 23 becomes narcissistic too (note that 23 = 2 x 8 + 7, so 23 in base eight is 27):

1/27 = 0·02620544131… → 0 + 2 + 6 + 2 + 0 + 5 + 4 + 4 = 27 (base=8)

Here are more narcissistic reciprocals in base ten:

1/3 = 0·3… → 3 = 3
1/7 = 0·142857… → 1 + 4 + 2 = 7
1/8 = 0·125 → 1 + 2 + 5 = 8
1/13 = 0·076923… → 0 + 7 + 6 = 13
1/14 = 0·0714285… → 0 + 7 + 1 + 4 + 2 = 14
1/34 = 0·02941176470588235… → 0 + 2 + 9 + 4 + 1 + 1 + 7 + 6 + 4 = 34
1/43 = 0·023255813953488372093… → 0 + 2 + 3 + 2 + 5 + 5 + 8 + 1 + 3 + 9 + 5 = 43
1/49 = 0·020408163265306122448979591836734693877551… → 0 + 2 + 0 + 4 + 0 + 8 + 1 + 6 + 3 + 2 + 6 + 5 + 3 + 0 + 6 + 1 + 2 = 49
1/51 = 0·0196078431372549… → 0 + 1 + 9 + 6 + 0 + 7 + 8 + 4 + 3 + 1 + 3 + 7 + 2 = 51
1/76 = 0·01315789473684210526… → 0 + 1 + 3 + 1 + 5 + 7 + 8 + 9 + 4 + 7 + 3 + 6 + 8 + 4 + 2 + 1 + 0 + 5 + 2 = 76
1/83 = 0·01204819277108433734939759036144578313253… → 0 + 1 + 2 + 0 + 4 + 8 + 1 + 9 + 2 + 7 + 7 + 1 + 0 + 8 + 4 + 3 + 3 + 7 + 3 + 4 + 9 = 83
1/92 = 0·010869565217391304347826… → 0 + 1 + 0 + 8 + 6 + 9 + 5 + 6 + 5 + 2 + 1 + 7 + 3 + 9 + 1 + 3 + 0 + 4 + 3 + 4 + 7 + 8 = 92
1/94 = 0·01063829787234042553191489361702127659574468085… → 0 + 1 + 0 + 6 + 3 + 8 + 2 + 9 + 7 + 8 + 7 + 2 + 3 + 4 + 0 + 4 + 2 + 5 + 5 + 3 + 1 + 9 + 1 + 4 = 94
1/98 = 0·0102040816326530612244897959183673469387755… → 0 + 1 + 0 + 2 + 0 + 4 + 0 + 8 + 1 + 6 + 3 + 2 + 6 + 5 + 3 + 0 + 6 + 1 + 2 + 2 + 4 + 4 + 8 + 9 + 7 + 9 + 5 = 98


Previously pre-posted (please peruse):

Digital Disfunction
The Hill to Power
Narcissarithmetic #1
Narcissarithmetic #2

Digital Disfunction

It’s fun when functions disfunc. The function digit-sum(n^p) takes a number, raises it to the power of p and sums its digits. If p = 1, n is unchanged. So digit-sum(1^1) = 1, digit-sum(11^1) = 2, digit-sum(2013^1) = 6. The following numbers set records for the digit-sum(n^1) from 1 to 1,000,000:

digit-sum(n^1): 1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 29, 39, 49, 59, 69, 79, 89, 99, 199, 299, 399, 499, 599, 699, 799, 899, 999, 1999, 2999, 3999, 4999, 5999, 6999, 7999, 8999, 9999, 19999, 29999, 39999, 49999, 59999, 69999, 79999, 89999, 99999, 199999, 299999, 399999, 499999, 599999, 699999, 799999, 899999, 999999.

The pattern is easy to predict. But the function disfuncs when p = 2. Digit-sum(3^2) = 9, which is more than digit-sum(4^2) = 1 + 6 = 7 and digit-sum(5^2) = 2 + 5 = 7. These are the records from 1 to 1,000,000:

digit-sum(n^2): 1, 2, 3, 7, 13, 17, 43, 63, 83, 167, 264, 313, 707, 836, 1667, 2236, 3114, 4472, 6833, 8167, 8937, 16667, 21886, 29614, 32617, 37387, 39417, 42391, 44417, 60663, 63228, 89437, 141063, 221333, 659386, 791833, 976063, 987917.

Higher powers are similarly disfunctional:

digit-sum(n^3): 1, 2, 3, 4, 9, 13, 19, 53, 66, 76, 92, 132, 157, 353, 423, 559, 842, 927, 1192, 1966, 4289, 5826, 8782, 10092, 10192, 10275, 10285, 10593, 11548, 11595, 12383, 15599, 22893, 31679, 31862, 32129, 63927, 306842, 308113.

digit-sum(n^4): 1, 2, 3, 4, 6, 8, 13, 16, 18, 23, 26, 47, 66, 74, 118, 256, 268, 292, 308, 518, 659, 1434, 1558, 1768, 2104, 2868, 5396, 5722, 5759, 6381, 10106, 12406, 14482, 18792, 32536, 32776, 37781, 37842, 47042, 51376, 52536, 84632, 255948, 341156, 362358, 540518, 582477.

digit-sum(n^5): 1, 2, 3, 5, 6, 14, 15, 18, 37, 58, 78, 93, 118, 131, 139, 156, 179, 345, 368, 549, 756, 1355, 1379, 2139, 2759, 2779, 3965, 4119, 4189, 4476, 4956, 7348, 7989, 8769, 9746, 10566, 19199, 19799, 24748, 31696, 33208, 51856, 207198, 235846, 252699, 266989, 549248, 602555, 809097, 814308, 897778.

You can also look for narcissistic numbers with this function, like digit-sum(9^2) = 8 + 1 = 9 and digit-sum(8^3) = 5 + 1 + 2 = 8. 9^2 is the only narcissistic square in base ten, but 8^3 has these companions:

17^3 = 4913 → 4 + 9 + 1 + 3 = 17
18^3 = 5832 → 5 + 8 + 3 + 2 = 18
26^3 = 17576 → 1 + 7 + 5 + 7 + 6 = 26
27^3 = 19683 → 1 + 9 + 6 + 8 + 3 = 27

Twelfth powers are as unproductive as squares:

108^12 = 2518170116818978404827136 → 2 + 5 + 1 + 8 + 1 + 7 + 0 + 1 + 1 + 6 + 8 + 1 + 8 + 9 + 7 + 8 + 4 + 0 + 4 + 8 + 2 + 7 + 1 + 3 + 6 = 108

But thirteenth powers are fertile:

20 = digit-sum(20^13)
40 = digit-sum(40^13)
86 = digit-sum(86^13)
103 = digit-sum(103^13)
104 = digit-sum(104^13)
106 = digit-sum(106^13)
107 = digit-sum(107^13)
126 = digit-sum(126^13)
134 = digit-sum(134^13)
135 = digit-sum(135^13)
146 = digit-sum(146^13)

There are also numbers that are narcissistic with different powers, like 90:

90^19 = 1·350851717672992089 x 10^37 → 1 + 3 + 5 + 0 + 8 + 5 + 1 + 7 + 1 + 7 + 6 + 7 + 2 + 9 + 9 + 2 + 0 + 8 + 9 = 90
90^20 = 1·2157665459056928801 x 10^39 → 1 + 2 + 1 + 5 + 7 + 6 + 6 + 5 + 4 + 5 + 9 + 0 + 5 + 6 + 9 + 2 + 8 + 8 + 0 + 1 = 90
90^21 = 1·09418989131512359209 x 10^41 → 1 + 0 + 9 + 4 + 1 + 8 + 9 + 8 + 9 + 1 + 3 + 1 + 5 + 1 + 2 + 3 + 5 + 9 + 2 + 0 + 9 = 90
90^22 = 9·84770902183611232881 x 10^42 → 9 + 8 + 4 + 7 + 7 + 0 + 9 + 0 + 2 + 1 + 8 + 3 + 6 + 1 + 1 + 2 + 3 + 2 + 8 + 8 + 1 = 90
90^28 = 5·23347633027360537213511521 x 10^54 → 5 + 2 + 3 + 3 + 4 + 7 + 6 + 3 + 3 + 0 + 2 + 7 + 3 + 6 + 0 + 5 + 3 + 7 + 2 + 1 + 3 + 5 + 1 + 1 + 5 + 2 + 1 = 90

One of the world’s most famous numbers is also multi-narcissistic:

666 = digit-sum(666^47)
666 = digit-sum(666^51)

1423 isn’t multi-narcissistic, but I like the way it’s a prime that’s equal to the sum of the digits of its power to 101, which is also a prime:

1423^101 = 2,
976,424,759,070,864,888,448,625,568,610,774,713,351,233,339,
006,775,775,271,720,934,730,013,444,193,709,672,452,482,197,
898,160,621,507,330,824,007,863,598,230,100,270,989,373,401,
979,514,790,363,102,835,678,646,537,123,754,219,728,748,171,
764,802,617,086,504,534,229,621,770,717,299,909,463,416,760,
781,260,028,964,295,036,668,773,707,186,491,056,375,768,526,
306,341,717,666,810,190,220,650,285,746,057,099,312,179,689,
423 →

2 + 9 + 7 + 6 + 4 + 2 + 4 + 7 + 5 + 9 + 0 + 7 + 0 + 8 + 6 + 4 + 8 + 8 + 8 + 4 + 4 + 8 + 6 + 2 + 5 + 5 + 6 + 8 + 6 + 1 + 0 + 7 + 7 + 4 + 7 + 1 + 3 + 3 + 5 + 1 + 2 + 3 + 3 + 3 + 3 + 9 + 0 + 0 + 6 + 7 + 7 + 5 + 7 + 7 + 5 + 2 + 7 + 1 + 7 + 2 + 0 + 9 + 3 + 4 + 7 + 3 + 0 + 0 + 1 + 3 + 4 + 4 + 4 + 1 + 9 + 3 + 7 + 0 + 9 + 6 + 7 + 2 + 4 + 5 + 2 + 4 + 8 + 2 + 1 + 9 + 7 + 8 + 9 + 8 + 1 + 6 + 0 + 6 + 2 + 1 + 5 + 0 + 7 + 3 + 3 + 0 + 8 + 2 + 4 + 0 + 0 + 7 + 8 + 6 + 3 + 5 + 9 + 8 + 2 + 3 + 0 + 1 + 0 + 0 + 2 + 7 + 0 + 9 + 8 + 9 + 3 + 7 + 3 + 4 + 0 + 1 + 9 + 7 + 9 + 5 + 1 + 4 + 7 + 9 + 0 + 3 + 6 + 3 + 1 + 0 + 2 + 8 + 3 + 5 + 6 + 7 + 8 + 6 + 4 + 6 + 5 + 3 + 7 + 1 + 2 + 3 + 7 + 5 + 4 + 2 + 1 + 9 + 7 + 2 + 8 + 7 + 4 + 8 + 1 + 7 + 1 + 7 + 6 + 4 + 8 + 0 + 2 + 6 + 1 + 7 + 0 + 8 + 6 + 5 + 0 + 4 + 5 + 3 + 4 + 2 + 2 + 9 + 6 + 2 + 1 + 7 + 7 + 0 + 7 + 1 + 7 + 2 + 9 + 9 + 9 + 0 + 9 + 4 + 6 + 3 + 4 + 1 + 6 + 7 + 6 + 0 + 7 + 8 + 1 + 2 + 6 + 0 + 0 + 2 + 8 + 9 + 6 + 4 + 2 + 9 + 5 + 0 + 3 + 6 + 6 + 6 + 8 + 7 + 7 + 3 + 7 + 0 + 7 + 1 + 8 + 6 + 4 + 9 + 1 + 0 + 5 + 6 + 3 + 7 + 5 + 7 + 6 + 8 + 5 + 2 + 6 + 3 + 0 + 6 + 3 + 4 + 1 + 7 + 1 + 7 + 6 + 6 + 6 + 8 + 1 + 0 + 1 + 9 + 0 + 2 + 2 + 0 + 6 + 5 + 0 + 2 + 8 + 5 + 7 + 4 + 6 + 0 + 5 + 7 + 0 + 9 + 9 + 3 + 1 + 2 + 1 + 7 + 9 + 6 + 8 + 9 + 4 + 2 + 3 = 1423


Previously pre-posted (please peruse):

The Hill to Power
Narcissarithmetic #1
Narcissarithmetic #2

The Hill to Power

89 is special because it’s a prime number, divisible by only itself and 1. It’s also a sum of powers in a special way: 89 = 8^1 + 9^2. In base ten, no other two-digit number is equal to its own ascending power-sum like that. But the same pattern appears in these three-digit numbers, as the powers climb with the digits:

135 = 1^1 + 3^2 + 5^3 = 1 + 9 + 125 = 135
175 = 1^1 + 7^2 + 5^3 = 1 + 49 + 125 = 175
518 = 5^1 + 1^2 + 8^3 = 5 + 1 + 512 = 518
598 = 5^1 + 9^2 + 8^3 = 5 + 81 + 512 = 598

And in these four-digit numbers:

1306 = 1^1 + 3^2 + 0^3 + 6^4 = 1 + 9 + 0 + 1296 = 1306
1676 = 1^1 + 6^2 + 7^3 + 6^4 = 1 + 36 + 343 + 1296 = 1676
2427 = 2^1 + 4^2 + 2^3 + 7^4 = 2 + 16 + 8 + 2401 = 2427

The pattern doesn’t apply to any five-digit number in base-10 and six-digit numbers supply only this near miss:

263248 + 1 = 2^1 + 6^2 + 3^3 + 2^4 + 4^5 + 8^6 = 2 + 36 + 27 + 16 + 1024 + 262144 = 263249

But the pattern re-appears among seven-digit numbers:

2646798 = 2^1 + 6^2 + 4^3 + 6^4 + 7^5 + 9^6 + 8^7 = 2 + 36 + 64 + 1296 + 16807 + 531441 + 2097152 = 2646798

Now try some base behaviour. Some power-sums in base-10 are power-sums in another base:

175 = 1^1 + 7^2 + 5^3 = 1 + 49 + 125 = 175
175 = 6D[b=27] = 6^1 + 13^2 = 6 + 169 = 175

1306 = 1^1 + 3^2 + 0^3 + 6^4 = 1 + 9 + 0 + 1296 = 1306
1306 = A[36][b=127] = 10^1 + 36^2 = 10 + 1296 = 1306

Here is an incomplete list of double-base power-sums:

83 = 1103[b=4] = 1^1 + 1^2 + 0^3 + 3^4 = 1 + 1 + 0 + 81 = 83
83 = 29[b=37] = 2^1 + 9^2 = 2 + 81 = 83

126 = 105[b=11] = 1^1 + 0^2 + 5^3 = 1 + 0 + 125 = 126
126 = 5B[b=23] = 5^1 + 11^2 = 5 + 121 = 126

175 = 1^1 + 7^2 + 5^3 = 1 + 49 + 125 = 175
175 = 6D[b=27] = 6^1 + 13^2 = 6 + 169 = 175

259 = 2014[b=5] = 2^1 + 0^2 + 1^3 + 4^4 = 2 + 0 + 1 + 256 = 259
259 = 3G[b=81] = 3^1 + 16^2 = 3 + 256 = 259

266 = 176[b=13] = 1^1 + 7^2 + 6^3 = 1 + 49 + 216 = 266
266 = AG[b=25] = 10^1 + 16^2 = 10 + 256 = 266

578 = 288[b=15] = 2^1 + 8^2 + 8^3 = 2 + 64 + 512 = 578
578 = 2[24][b=277] = 2^1 + 24^2 = 2 + 576 = 578

580 = 488[b=11] = 4^1 + 8^2 + 8^3 = 4 + 64 + 512 = 580
580 = 4[24][b=139] = 4^1 + 24^2 = 4 + 576 = 580

731 = 209[b=19] = 2^1 + 0^2 + 9^3 = 2 + 0 + 729 = 731
731 = 2[27][b=352] = 2^1 + 27^2 = 2 + 729 = 731

735 = 609[b=11] = 6^1 + 0^2 + 9^3 = 6 + 0 + 729 = 735
735 = 6[27][b=118] = 6^1 + 27^2 = 6 + 729 = 735

1306 = 1^1 + 3^2 + 0^3 + 6^4 = 1 + 9 + 0 + 1296 = 1306
1306 = A[36][b=127] = 10^1 + 36^2 = 10 + 1296 = 1306

1852 = 3BC[b=23] = 3^1 + 11^2 + 12^3 = 3 + 121 + 1728 = 1852
1852 = 3[43][b=603] = 3^1 + 43^2 = 3 + 1849 = 1852

2943 = 3EE[b=29] = 3^1 + 14^2 + 14^3 = 3 + 196 + 2744 = 2943
2943 = [27][54][b=107] = 27^1 + 54^2 = 27 + 2916 = 2943


Previously pre-posted (please peruse):

Narcissarithmetic #1
Narcissarithmetic #2

Narcissarithmetic #2

It’s easy to find patterns like these in base ten:

81 = (8 + 1)^2 = 9^2 = 81

512 = (5 + 1 + 2)^3 = 8^3 = 512
4913 = (4 + 9 + 1 + 3)^3 = 17^3 = 4913
5832 = (5 + 8 + 3 + 2)^3 = 18^3 = 5832
17576 = (1 + 7 + 5 + 7 + 6)^3 = 26^3 = 17576
19683 = (1 + 9 + 6 + 8 + 3)^3 = 27^3 = 19683

2401 = (2 + 4 + 0 + 1)^4 = 7^4 = 2401
234256 = (2 + 3 + 4 + 2 + 5 + 6)^4 = 22^4 = 234256
390625 = (3 + 9 + 0 + 6 + 2 + 5)^4 = 25^4 = 390625
614656 = (6 + 1 + 4 + 6 + 5 + 6)^4 = 28^4 = 614656
1679616 = (1 + 6 + 7 + 9 + 6 + 1 + 6)^4 = 36^4 = 1679616

17210368 = (1 + 7 + 2 + 1 + 0 + 3 + 6 + 8)^5 = 28^5 = 17210368
52521875 = (5 + 2 + 5 + 2 + 1 + 8 + 7 + 5)^5 = 35^5 = 52521875
60466176 = (6 + 0 + 4 + 6 + 6 + 1 + 7 + 6)^5 = 36^5 = 60466176
205962976 = (2 + 0 + 5 + 9 + 6 + 2 + 9 + 7 + 6)^5 = 46^5 = 205962976

1215766545905692880100000000000000000000 = (1 + 2 + 1 + 5 + 7 + 6 + 6 + 5 + 4 + 5 + 9 + 0 + 5 + 6 + 9 + 2 + 8 + 8 + 0 + 1 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0)^20 = 90^20 = 1215766545905692880100000000000000000000

Patterns like this are much rarer:

914457600 = (9 x 1 x 4 x 4 x 5 x 7 x 6)^2 = 30240^2 = 914457600

3657830400 = (3 x 6 x 5 x 7 x 8 x 3 x 4)^2 = 60480^2 = 3657830400

I haven’t found a cube like that in base ten, but base six supplies them:

2212 = (2 x 2 x 1 x 2)^3 = 12^3 = 2212 (b=6) = 8^3 = 512 (b=10)
325000 = (3 x 2 x 5)^3 = 50^3 = 325000 (b=6) = 30^3 = 27000 (b=10)
411412 = (4 x 1 x 1 x 4 x 1 x 2)^3 = 52^3 = 411412 (b=6) = 32^3 = 32768 (b=10)

And base nine supplies a fourth and fifth power:

31400 = (3 x 1 x 4)^4 = 13^4 = 31400 (b=9) = 12^4 = 20736 (b=10)
11600 = (1 x 1 x 6)^5 = 6^5 = 11600 (b=9) = 6^5 = 7776 (b=10)

Then base ten is rich in patterns like these:

81 = (8^1 + 1^1) x (8 + 1) = 9 x 9 = 81

133 = (1^2 + 3^2 + 3^2) x (1 + 3 + 3) = 19 x 7 = 133
315 = (3^2 + 1^2 + 5^2) x (3 + 1 + 5) = 35 x 9 = 315
803 = (8^2 + 0^2 + 3^2) x (8 + 0 + 3) = 73 x 11 = 803
1148 = (1^2 + 1^2 + 4^2 + 8^2) x (1 + 1 + 4 + 8) = 82 x 14 = 1148
1547 = (1^2 + 5^2 + 4^2 + 7^2) x (1 + 5 + 4 + 7) = 91 x 17 = 1547
2196 = (2^2 + 1^2 + 9^2 + 6^2) x (2 + 1 + 9 + 6) = 122 x 18 = 2196

1215 = (1^3 + 2^3 + 1^3 + 5^3) x (1 + 2 + 1 + 5) = 135 x 9 = 1215
3700 = (3^3 + 7^3 + 0^3 + 0^3) x (3 + 7 + 0 + 0) = 370 x 10 = 3700
11680 = (1^3 + 1^3 + 6^3 + 8^3 + 0^3) x (1 + 1 + 6 + 8 + 0) = 730 x 16 = 11680
13608 = (1^3 + 3^3 + 6^3 + 0^3 + 8^3) x (1 + 3 + 6 + 0 + 8) = 756 x 18 = 13608
87949 = (8^3 + 7^3 + 9^3 + 4^3 + 9^3) x (8 + 7 + 9 + 4 + 9) = 2377 x 37 = 87949

182380 = (1^4 + 8^4 + 2^4 + 3^4 + 8^4 + 0^4) x (1 + 8 + 2 + 3 + 8 + 0) = 8290 x 22 = 182380
444992 = (4^4 + 4^4 + 4^4 + 9^4 + 9^4 + 2^4) x (4 + 4 + 4 + 9 + 9 + 2) = 13906 x 32 = 444992

41500 = (4^5 + 1^5 + 5^5 + 0^5 + 0^5) x (4 + 1 + 5 + 0 + 0) = 4150 x 10 = 41500
3508936 = (3^5 + 5^5 + 0^5 + 8^5 + 9^5 + 3^5 + 6^5) x (3 + 5 + 0 + 8 + 9 + 3 + 6) = 103204 x 34 = 3508936
3828816 = (3^5 + 8^5 + 2^5 + 8^5 + 8^5 + 1^5 + 6^5) x (3 + 8 + 2 + 8 + 8 + 1 + 6) = 106356 x 36 = 3828816
4801896 = (4^5 + 8^5 + 0^5 + 1^5 + 8^5 + 9^5 + 6^5) x (4 + 8 + 0 + 1 + 8 + 9 + 6) = 133386 x 36 = 4801896
5659875 = (5^5 + 6^5 + 5^5 + 9^5 + 8^5 + 7^5 + 5^5) x (5 + 6 + 5 + 9 + 8 + 7 + 5) = 125775 x 45 = 5659875


Previously pre-posted (please peruse):

Narcissarithmetic

Narcissarithmetic

Why is 438,579,088 a beautiful number? Simple: it may seem entirely arbitrary, but it’s actually self-empowered:

438,579,088 = 4^4 + 3^3 + 8^8 + 5^5 + 7^7 + 9^9 + 0^0 + 8^8 + 8^8 = 256 + 27 + 16777216 + 3125 + 823543 + 387420489 + 0 + 16777216 + 16777216 (usually 0^0 = 1, but the rule is slightly varied here)

438,579,088 is so beautiful, in fact, that it’s in love with itself as a narcissistic number, or number that can be generated by manipulation of its own digits. 89 = 8^1 + 9^2 = 8 + 81 and 135 = 1^1 + 3^2 + 5^3 = 1 + 9 + 125 are different kinds of narcissistic number. 3435 is self-empowered again:

3435 = 3^3 + 4^4 + 3^3 + 5^5 = 27 + 256 + 27 + 3125

But that’s your lot: there are no more numbers in base-10 that are equal to the sum of their self-empowered digits (apart from the trivial 0 and 1). To prove this, start by considering that there is a limit to the size of a self-empowered number. 9^9 is 387,420,489, which is nine digits long. The function autopower(999,999,999) = 387,420,489 x 9 = 3,486,784,401, which is ten digits long. But autopower(999,999,999,999) = 387,420,489 x 12 = 4,649,045,868, also ten digits long.

The Metamorphosis of Narcissus by Salvador Dalí

Salvador Dalí, La Metamorfosis de Narciso (1937)

So you don’t need to check numbers above a certain size. There still seem a lot of numbers to check: 438,579,088 is a long way above 3435. However, the search is easy to shorten if you consider that checking 3-3-4-5 is equivalent to checking 3-4-3-5, just as checking 034,578,889 is equivalent to checking 438,579,088. If you self-empower a number and the result has the same digits as the original number, you’ve found what you’re looking for. The order of digits in the original number doesn’t matter, because the result has automatically sorted them for you. The function autopower(3345) produces 3435, therefore 3435 must be self-empowered.

So the rule is simple: Check only the numbers in which any digit is greater than or equal to all digits to its left. In other words, you check 12 and skip 21, check 34 and skip 43, check 567 and skip 576, 657, 675, 756 and 765. That reduces the search-time considerably: discarding numbers is computationally simpler than self-empowering them. It’s also computationally simple to vary the base in which you’re searching. Base-10 produces only two self-empowered numbers, but its neighbours base-9 and base-11 are much more fertile:

30 = 3^3 + 0^0 = 30 + 0 (b=9)
27 = 27 + 0 (b=10)

31 = 3^3 + 1^1 = 30 + 1 (b=9)
28 = 27 + 1 (b=10)

156262 = 1^1 + 5^5 + 6^6 + 2^2 + 6^6 + 2^2 = 1 + 4252 + 71000 + 4 + 71000 + 4 (b=9)
96446 = 1 + 3125 + 46656 + 4 + 46656 + 4 (b=10)

1647063 = 1^1 + 6^6 + 4^4 + 7^7 + 0^0 + 6^6 + 3^3 = 1 + 71000 + 314 + 1484617 + 0 + 71000 + 30 (b=9)
917139 = 1 + 46656 + 256 + 823543 + 0 + 46656 + 27 (b=10)

1656547 = 1^1 + 6^6 + 5^5 + 6^6 + 5^5 + 4^4 + 7^7 = 1 + 71000 + 4252 + 71000 + 4252 + 314 + 1484617 (b=9)
923362 = 1 + 46656 + 3125 + 46656 + 3125 + 256 + 823543 (b=10)

34664084 = 3^3 + 4^4 + 6^6 + 6^6 + 4^4 + 0^0 + 8^8 + 4^4 = 30 + 314 + 71000 + 71000 + 314 + 0 + 34511011 + 314 (b=9)
16871323 = 27 + 256 + 46656 + 46656 + 256 + 0 + 16777216 + 256 (b=10)

66500 = 6^6 + 6^6 + 5^5 + 0^0 + 0^0 = 32065 + 32065 + 2391 + 0 + 0 (b=11)
96437 = 46656 + 46656 + 3125 + 0 + 0 (b=10)

66501 = 6^6 + 6^6 + 5^5 + 0^0 + 1^1 = 32065 + 32065 + 2391 + 0 + 1 (b=11)
96438 = 46656 + 46656 + 3125 + 0 + 1 (b=10)

517503 = 5^5 + 1^1 + 7^7 + 5^5 + 0^0 + 3^3 = 2391 + 1 + 512816 + 2391 + 0 + 25 (b=11)
829821 = 3125 + 1 + 823543 + 3125 + 0 + 27 (b=10)

18453278 = 1^1 + 8^8 + 4^4 + 5^5 + 3^3 + 2^2 + 7^7 + 8^8 = 1 + 9519A75 + 213 + 2391 + 25 + 4 + 512816 + 9519A75 (b=11)
34381388 = 1 + 16777216 + 256 + 3125 + 27 + 4 + 823543 + 16777216 (b=10)

18453487 = 1^1 + 8^8 + 4^4 + 5^5 + 3^3 + 4^4 + 8^8 + 7^7 = 1 + 9519A75 + 213 + 2391 + 25 + 213 + 9519A75 + 512816 (b=11)
34381640 = 1 + 16777216 + 256 + 3125 + 27 + 256 + 16777216 + 823543 (b=10)

It’s easy to extend the concept of self-empowered narcisso-numbers. The prime 71 = 131 in base-7 and the prime 83 = 146 in base-7. If 131[b=7] is empowered to the digits of 146[b=7], you get 146[b=7]; and if 146[b=7] is empowered to the digits of 131[b=7], you get 131[b=7], like this:

71 = 131[b=7] → 1^1 + 3^4 + 1^6 = 1 + 81 + 1 = 83 = 146[b=7]

83 = 146[b=7] → 1^1 + 4^3 + 6^1 = 1 + 64 + 6 = 71 = 131[b=7]

But it’s not easy to find more examples. Are there other-empowering pairs like that in base-10? I don’t know.

More Multi-Magic

The answer, I’m glad to say, is yes. The question is: Can a prime magic-square nest inside a second prime magic-square that nests inside a third prime magic-square? I asked this in Multi-Magic, where I described how a magic square is a square of numbers where all rows, all columns and both diagonals add to the same number, or magic total. This magic square consists entirely of prime numbers, or numbers divisible only by themselves and 1:

43 | 01 | 67
61 | 37 | 13
07 | 73 | 31

Base = 10, magic total = 111

It nests inside this prime magic-square, whose digit-sums in base-97 re-create it:

0619  =  [06][37] | 0097  =  [01][00] | 1123  =  [11][56]
1117  =  [11][50] | 0613  =  [06][31] | 0109  =  [01][12]
0103  =  [01][06] | 1129  =  [11][62] | 0607  =  [06][25]

Base = 97, magic total = 1839

And that prime magic-square nests inside this one:

2803  =  [1][0618] | 2281  =  [1][0096] | 3307  =  [1][1122]
3301  =  [1][1116] | 2797  =  [1][0612] | 2293  =  [1][0108]
2287  =  [1][0102] | 3313  =  [1][1128] | 2791  =  [1][0606]

Base = 2185, magic total = 8391

I don’t know whether that prime magic-square nests inside a fourth square, but a 3-nest is good for 3×3 magic squares. On the other hand, this famous 3×3 magic square is easy to nest inside an infinite series of other magic squares:

6 | 1 | 8
7 | 5 | 3
2 | 9 | 4

Base = 10, magic total = 15

It’s created by the digit-sums of this square in base-9 (“14 = 15” means that the number 14 is represented as “15” in base-9):

14 = 15 → 6 | 09 = 10 → 1 | 16 = 17 → 8
15 = 16 → 7 | 13 = 14 → 5 | 11 = 12 → 3
10 = 11 → 2 | 17 = 18 → 9 | 12 = 13 → 4

Base = 9, magic total = 39


And that square in base-9 is created by the digit-sums of this square in base-17:

30 = 1[13] → 14 | 25 = 00018 → 09 | 32 = 1[15] → 16
31 = 1[14] → 15 | 29 = 1[12] → 13 | 27 = 1[10] → 11
26 = 00019 → 10 | 33 = 1[16] → 17 | 28 = 1[11] → 12

Base = 17, magic total = 87

And so on:

62 = 1[29] → 30 | 57 = 1[24] → 25 | 64 = 1[31] → 32
63 = 1[30] → 31 | 61 = 1[28] → 29 | 59 = 1[26] → 27
58 = 1[25] → 26 | 65 = 1[32] → 33 | 60 = 1[27] → 28

Base = 33, magic total = 183

126 = 1[61] → 62 | 121 = 1[56] → 57 | 128 = 1[63] → 64
127 = 1[62] → 63 | 125 = 1[60] → 61 | 123 = 1[58] → 59
122 = 1[57] → 58 | 129 = 1[64] → 65 | 124 = 1[59] → 60

Base = 65, magic total = 375

Previously Pre-Posted (please peruse):

Multi-Magic