Tag Archives: narcissistic numbers

Narcischism

by in Binary numbers, Mathematics, Narcissistic numbers and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

What have bits to do with splits? A lot. Suppose you take the digits 12345, split them in all possible ways, then sum the results, like this:

12345 → (1234 + 5) + (123 + 45) + (123 + 4 + 5) + (12 + 345) + (12 + 34 + 5) + (12 + 3 + 45) + (12 + 3 + 4 + 5) + (1 + 2345) + (1 + 234 + 5) + (1 + 23 + 45) + (1 + 23 + 4 + 5) + (1 + 2 + 345) + (1 + 2 + 34 + 5) + (1 + 2 + 3 + 45) + (1 + 2 + 3 + 4 + 5) = 5175.

That’s a sum in base 10, but base 2 is at work below the surface, because each set of numbers is the answer to a series of binary questions: split or not? There are four possible places to split the digits 12345: after the 1, after the 2, after the 3 and after the 4. In (1 + 2 + 3 + 4 + 5), the binary question “Split or not?” is answered SPLIT every time. In (1234 + 5) and (1 + 2345) it’s answered SPLIT only once.

So the splits are governed by a four-digit binary number ranging from 0001 to 1111. When the binary digit is 1, split; when the binary digit is 0, don’t split. In binary, 0001 to 1111 = 01 to 15 in base 10 = 2^4-1. That’s for a five-digit number, so the four-digit 1234 will have 2^3-1 = 7 sets of sums:

1234 → (123 + 4) + (12 + 34) + (12 + 3 + 4) + (1 + 234) + (1 + 23 + 4) + 110 (1 + 2 + 34) + (1 + 2 + 3 + 4) = 502.

And the six-digit number 123456 will have 2^5-1 = 31 sets of sums. By now, an exciting question may have occurred to some readers. Does any number in base 10 equal the sum of all possible numbers formed by splitting its digits?

The exciting answer is: 0. In other words: No. To see why not, examine a quick way of summing the split-bits of 123,456,789, with nine digits. The long way is to find all possible sets of split-bits. There are 2^8-1 = 255 of them. The quick way is to sum these equations:

1 * 128 + 10 * 64 + 100 * 32 + 1000 * 16 + 10000 * 8 + 100000 * 4 + 1000000 * 2 + 10000000 * 1
2 * 128 + 20 * 64 + 200 * 32 + 2000 * 16 + 20000 * 8 + 200000 * 4 + 2000000 * 2 + 20000000 * 1
3 * 128 + 30 * 64 + 300 * 32 + 3000 * 16 + 30000 * 8 + 300000 * 4 + 3000000 * 3
4 * 128 + 40 * 64 + 400 * 32 + 4000 * 16 + 40000 * 8 + 400000 * 7
5 * 128 + 50 * 64 + 500 * 32 + 5000 * 16 + 50000 * 15
6 * 128 + 60 * 64 + 600 * 32 + 6000 * 31
7 * 128 + 70 * 64 + 700 * 63
8 * 128 + 80 * 127
9 * 255

Sum = 52,322,283.

52,322,283 has eight digits. If you use the same formula for the nine-digit number 999,999,999, the sum is 265,621,761, which has nine digits but is far smaller than 999,999,999. If you adapt the formula for the twenty-digit 19,999,999,999,999,999,999 (starting with 1), the split-bit sum is 16,562,499,999,987,400,705. In base 10, as far as I can see, numbers increase too fast and digit-lengths too slowly for the binary governing the split-sums to keep up. That’s also true in base 9 and base 8:

Num = 18,888,888,888,888,888,888 (b=9)
Sum = 16,714,201,578,038,328,760

Num = 17,777,777,777,777,777,777 (b=8)
Sum = 17,070,707,070,625,000,001

So what about base 7? Do the numbers increase slowly enough and the digit-lengths fast enough for the binary to keep up? The answer is: 1. In base 7, this twenty-digit number is actually smaller than its split-bit sum:

Num = 16,666,666,666,666,666,666 (b=7)
Sum = 20,363,036,303,404,141,363

And if you search below that, you can find a number that is equal to its split-bit sum:

166512 → (1 + 6 + 6 + 5 + 1 + 2) + (16 + 6 + 5 + 1 + 2) + (1 + 66 + 5 + 1 + 2) + (166 + 5 + 1 + 2) + (1 + 6 + 65 + 1 + 2) + (16 + 65 + 1 + 2) + (1 + 665 + 1 + 2) + (1665 + 1 + 2) + (1 + 6 + 6 + 51 + 2) + (16 + 6 + 51 + 2) + (1 + 66 + 51 + 2) + (166 + 51 + 2) + (1 + 6 + 651 + 2) + (16 + 651 + 2) + (1 + 6651 + 2) + (16651 + 2) + (1 + 6 + 6 + 5 + 12) + (16 + 6 + 5 + 12) + (1 + 66 + 5 + 12) + (166 + 5 + 12) + (1 + 6 + 65 + 12) + (16 + 65 + 12) + (1 + 665 + 12) + (1665 + 12) + (1 + 6 + 6 + 512) + (16 + 6 + 512) + (1 + 66 + 512) + (166 + 512) + (1 + 6 + 6512) + (16 + 6512) + (1 + 66512) = 166512[b=7] = 33525[b=10].

So 33525 in base 7 is what might be called a narcischist: it can gaze into the split-bits of its own digits and see itself gazing back. In base 6, 1940 is a narcischist:

12552 → (1 + 2 + 5 + 5 + 2) + (12 + 5 + 5 + 2) + (1 + 25 + 5 + 2) + (125 + 5 + 2) + (1 + 2 + 55 + 2) + (12 + 55 + 2) + (1 + 255 + 2) + (1255 + 2) + (1 + 2 + 5+ 52) + (12 + 5 + 52) + (1 + 25 + 52) + (125 + 52) + (1 + 2 + 552) + (12 + 552) + (1 + 2552) = 12552[b=6] = 1940[b=10].

In base 5, 4074 is a narcischist:

112244 → (1 + 1 + 2 + 2 + 4 + 4) + (11 + 2 + 2 + 4 + 4) + (1 + 12 + 2 + 4 + 4) + (112 + 2 + 4 + 4) + (1 + 1 + 22 + 4 + 4) + (11 + 22 + 4 + 4) + (1 + 122 + 4 + 4) + (1122 + 4 + 4) + (1 + 1 + 2 + 24 + 4) + (11 + 2 + 24 + 4) + (1 + 12 + 24 + 4) + (112 + 24 + 4) + (1 + 1 + 224 + 4) + (11 + 224 + 4) + (1 + 1224 + 4) + (11224 + 4) + (1 + 1 + 2 + 2 + 44) + (11 + 2 + 2 + 44) + (1 + 12 + 2 + 44) + (112 + 2 + 44) + (1 + 1 + 22 + 44) + (11 + 22 + 44) + (1 + 122 + 44) + (1122 + 44) + (1 + 1 + 2 + 244) + (11 + 2 + 244) + (1 + 12 + 244) + (112 + 244) + (1 + 1 + 2244) + (11 + 2244) + (1 + 12244) = 112244[b=5] = 4074.

And in base 4, 27 is:

123 → (1 + 2 + 3) + (12 + 3) + (1 + 23) = 123[b=4] = 27.

And in base 3, 13 and 26 are:

111 → (1 + 1 + 1) + (11 + 1) + (1 + 11) = 111[b=3] = 13.

222 → (2 + 2 + 2) + (22 + 2) + (2 + 22) = 222[b=3] = 26.

There are many more narcischists in all these bases, even if you exclude numbers with zeroes in them, like these in base 4:

1022 → (1 + 0 + 2 + 2) + (10 + 2 + 2) + (1 + 02 + 2) + (102 + 2) + (1 + 0 + 22) + (10 + 22) + (1 + 022) = 1022[b=4] = 74.

1030 → (1 + 0 + 3 + 0) + (10 + 3 + 0) + (1 + 03 + 0) + (103 + 0) + (1 + 0 + 30) + (10 + 30) + (1 + 030) = 1030[b=4] = 76.

1120 → (1 + 1 + 2 + 0) + (11 + 2 + 0) + (1 + 12 + 0) + (112 + 0) + (1 + 1 + 20) + (11 + 20) + (1 + 120) = 1120[b=4] = 88.

Miss This

by in Base Behavior, Mathematics, Narcissistic numbers and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

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)

Persist List

by in Base Behavior, Mathematics, Narcissistic numbers and tagged , , , , , , , , , , , , , , , , , , , , , , , | 1 Comment

Multiplicative persistence is a complex term but a simple concept. Take a number, multiply its digits, repeat. Sooner or later the result is a single digit:

25 → 2 x 5 = 10 → 1 x 0 = 0 (mp=2)
39 → 3 x 9 = 27 → 2 x 7 = 14 → 1 x 4 = 4 (mp=3)

So 25 has a multiplicative persistence of 2 and 39 a multiplicative persistence of 3. Each is the smallest number with that m.p. in base-10. Further records are set by these numbers:

77 → 49 → 36 → 18 → 8 (mp=4)
679 → 378 → 168 → 48 → 32 → 6 (mp=5)
6788 → 2688 → 768 → 336 → 54 → 20 → 0 (mp=6)
68889 → 27648 → 2688 → 768 → 336 → 54 → 20 → 0 (mp=7)
2677889 → 338688 → 27648 → 2688 → 768 → 336 → 54 → 20 → 0 (mp=8)
26888999 → 4478976 → 338688 → 27648 → 2688 → 768 → 336 → 54 → 20 → 0 (mp=9)
3778888999 → 438939648 → 4478976 → 338688 → 27648 → 2688 → 768 → 336 → 54 → 20 → 0 (mp=10)

Now here’s base-9:

25[b=9] → 11 → 1 (mp=2)
38[b=9] → 26 → 13 → 3 (mp=3)
57[b=9] → 38 → 26 → 13 → 3 (mp=4)
477[b=9] → 237 → 46 → 26 → 13 → 3 (mp=5)
45788[b=9] → 13255 → 176 → 46 → 26 → 13 → 3 (mp=6)
2577777[b=9] → 275484 → 13255 → 176 → 46 → 26 → 13 → 3 (mp=7)

And base-11:

26[b=11] → 11 → 1 (mp=2)
3A[b=11] → 28 → 15 → 5 (mp=3)
69[b=11] → 4A → 37 → 1A → A (=10b=10) (mp=4)
269[b=11] → 99 → 74 → 26 → 11 → 1 (mp=5)
3579[b=11] → 78A → 46A → 1A9 → 82 → 15 → 5 (mp=6)
26778[b=11] → 3597 → 78A → 46A → 1A9 → 82 → 15 → 5 (mp=7)
47788A[b=11] → 86277 → 3597 → 78A → 46A → 1A9 → 82 → 15 → 5 (mp=8)
67899AAA[b=11] → 143A9869 → 299596 → 2A954 → 2783 → 286 → 88 → 59 → 41 → 4 (mp=9)
77777889999[b=11] → 2AA174996A → 143A9869 → 299596 → 2A954 → 2783 → 286 → 88 → 59 → 41 → 4 (mp=10)

I was also interested in the narcissism of multiplicative persistence. That is, are any numbers equal to the sum of the numbers created while calculating their multiplicative persistence? Yes:

86 = (8 x 6 = 48) + (4 x 8 = 32) + (3 x 2 = 6)

I haven’t found any more in base-10 (apart from the trivial 0 to 9) and can’t prove that this is the only one. Base-9 offers this:

78[b=9] = 62 + 13 + 3

I can’t find any at all in base-11, but here are base-12 and base-27:

57[b=12] = 2B + 1A + A
A8[b=12] = 68 + 40 + 0

4[23][b=27] = 3B + 16 + 6
7[24][b=27] = 66 + 19 + 9
A[18][b=27] = 6[18] + 40 + 0
[26][24][b=27] = [23]3 + 2F + 13 + 3
[26][23][26][b=27] = [21]8[23] + 583 + 4C + 1[21] + [21]

But the richest base I’ve found so far is base-108, with fourteen narcissistic multiplicative-persistence sums:

4[92][b=108] = 3[44] + 1[24] + [24]
5[63][b=108] = 2[99] + 1[90] + [90]
7[96][b=108] = 6[24] + 1[36] + [36]
A[72][b=108] = 6[72] + 40 + 0
[19][81][b=108] = E[27] + 3[54] + 1[54] + [54]
[26][96][b=108] = [23]C + 2[60] + 1C + C
[35][81][b=108] = [26][27] + 6[54] + 30 + 0
[37][55][b=108] = [18][91] + F[18] + 2[54] + 10 + 0
[73][60][b=108] = [40][60] + [22][24] + 4[96] + 3[60] + 1[72] + [72]
[107][66][b=108] = [65][42] + [25][30] + 6[102] + 5[72] + 3[36] + 10 + 0
[71][84][b=108] = [55][24] + C[24] + 2[72] + 1[36] + [36]
[107][99][b=108] = [98]9 + 8[18] + 1[36] + [36]
5[92][96][b=108] = 3[84][96] + 280 + 0
8[107][100][b=108] = 7[36][64] + 1[41][36] + D[72] + 8[72] + 5[36] + 1[72] + [72]

Update (10/ii/14): The best now is base-180 with eighteen multiplicative-persistence sums.

5[105][b=180] = 2[165] + 1[150] + [150]
7[118][b=180] = 4[106] + 2[64] + [128]
7[160][b=180] = 6[40] + 1[60] + [60]
8[108][b=180] = 4[144] + 3[36] + [108]
A[120][b=180] = 6[120] + 40 + 0 (s=5)
[19][135][b=180] = E[45] + 3[90] + 1[90] + [90]
[21][108][b=180] = C[108] + 7[36] + 1[72] + [72]
[26][160][b=180] = [23][20] + 2[100] + 1[20] + [20]
[31][98][b=180] = [16][158] + E8 + [112]
[35][135][b=180] = [26][45] + 6[90] + 30 + 0 (s=10)
[44][96][b=180] = [23][84] + A[132] + 7[60] + 2[60] + [120]
[71][140][b=180] = [55][40] + C[40] + 2[120] + 1[60] + [60]
[73][100][b=180] = [40][100] + [22][40] + 4[160] + 3[100] + 1[120] + [120]
[107][110][b=180] = [65][70] + [25][50] + 6[170] + 5[120] + 3[60] + 10 + 0
[107][165][b=180] = [98]F + 8[30] + 1[60] + [60] (s=15)
[172][132][b=180] = [126][24] + [16][144] + C[144] + 9[108] + 5[72] + 20 + 0
5[173][145][b=180] = 3[156][145] + 2[17]0 + 0
E[170][120][b=180] = 8[146][120] + 4[58][120] + [154][120] + [102][120] + [68]0 + 0

DeVil to Power

by in Digit-sums, Mathematics, Narcissistic numbers, Powers, Repdigits and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

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(π1,π4) = 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

by in Base Behavior, Digit-sums, Mathematics, Narcissistic numbers, Prime numbers and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

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

by in Digit-sums, Mathematics, Narcissistic numbers, Powers and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

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

by in Base Behavior, Digit-sums, Mathematics, Narcissistic numbers, Powers and tagged , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

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

by in Base Behavior, Digit-sums, Mathematics, Narcissistic numbers, Powers and tagged , , , , , , , , , , , , , , , , , , , , , | Leave a comment

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

by in Base Behavior, Digit-sums, Mathematics, Powers and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

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.