# Autonomata

“Describe yourself.” You can say it to people. And you can say it to numbers too. For example, here’s the number 3412 describing the positions of its own digits, starting at 1 and working upward:

3412 – the 1 is in the 3rd position, the 2 is in the 4th position, the 3 is in the 1st position, and the 4 is in the 2nd position.

In other words, the positions of the digits 1 to 4 of 3412 recreate its own digits:

3412 → (3,4,1,2) → 3412

The number 3412 describes itself – it’s autonomatic (from Greek auto, “self” + onoma, “name”). So are these numbers:

1
21
132
2143
52341
215634
7243651
68573142
321654798

More precisely, they’re panautonomatic numbers, because they describe the positions of all their own digits (Greek pan or panto, “all”). But what if you use the positions of only, say, the 1s or the 3s in a number? In base ten, only one number describes itself like that: 1. But we’re not confined to base 10. In base 2, the positions of the 1s in 110 (= 6) are 1 and 10 (= 2). So 110 is monautonomatic in binary (Greek mono, “single”). 10 is also monautonomatic in binary, if the digit being described is 0: it’s in 2nd position or position 10 in binary. These numbers are monoautonomatic in binary too:

110100 = 52 (digit = 1)
10100101111 = 1327 (d=0)

In 110100, the 1s are in 1st, 2nd and 4th position, or positions 1, 10, 100 in binary. In 10100101111, the 0s are in 2nd, 4th, 5th and 7th position, or positions 10, 100, 101, 111 in binary. Here are more monautonomatic numbers in other bases:

21011 in base 4 = 581 (digit = 1)
11122122 in base 3 = 3392 (d=2)
131011 in base 5 = 5131 (d=1)
2101112 in base 4 = 9302 (d=1)
11122122102 in base 3 = 91595 (d=2)
13101112 in base 5 = 128282 (d=1)
210111221 in base 4 = 148841 (d=1)

For example, in 131011 the 1s are in 1st, 3rd, 5th and 6th position, or positions 1, 3, 10 and 11 in quinary. But these numbers run out quickly and the only monautonomatic number in bases 6 and higher is 1. However, there are infinitely long monoautonomatic integer sequences in all bases. For example, in binary this sequence at the Online Encyclopedia of Integer Sequences describes itself using the positions of its 1s:

A167502: 1, 10, 100, 111, 1000, 1001, 1010, 1110, 10001, 10010, 10100, 10110, 10111, 11000, 11010, 11110, 11111, 100010, 100100, 100110, 101001, 101011, 101100, 101110, 110000, 110001, 110010, 110011, 110100, 111000, 111001, 111011, 111101, 11111, …

In base 10, it looks like this:

A167500: 1, 2, 4, 7, 8, 9, 10, 14, 17, 18, 20, 22, 23, 24, 26, 30, 31, 34, 36, 38, 41, 43, 44, 46, 48, 49, 50, 51, 52, 56, 57, 59, 61, 62, 63, 64, 66, 67, 68, 69, 70, 71, 75, 77, 80, 83, 86, 87, 89, 91, 94, 95, 97, 99, 100, 101, 103, 104, 107, 109, 110, 111, 113, 114, 119, 120, 124, … (see A287515 for a similar sequence using 0s)

In any base, you can find some sequence of integers describing the positions of any of the digits in that base – for example, the 1s or the 7s. But the numbers in the sequence get very large very quickly in higher bases. For example, here are some opening sequences for the digits 0 to 9 in base 10:

3, 10, 1111110, … (d=0)
1, 3, 10, 200001, … (d=1)
3, 12, 100000002, … (d=2)
2, 3, 30, 10000000000000000000000003, … (d=3)
2, 4, 14, 1000000004, … (d=4)
2, 5, 105, … (d=5)
2, 6, 1006, … (d=6)
2, 7, 10007, … (d=7)
2, 8, 100008, … (d=8)
2, 9, 1000009, … (d=9)

In the sequence for d=0, the first 0 is in the 3rd position, the second 0 is in the 10th position, and the third 0 is in the 1111110th position. That’s why I’ve haven’t written the next number – it’s 1,111,100 digits long (= 1111110 – 10). But it’s theoretically possible to write the number. In the sequence for d=3, the next number is utterly impossible to write, because it’s 9,999,999,999,999,999,999,999,973 digits long (= 10000000000000000000000003 – 30). In the sequence for d=5, the next number is this:

1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000005 (100 digits long = 105 – 5).

And in fact there are an infinite number of such sequences for any digit in any base – except for d=1 in binary. Why is binary different? Because 1 is the only digit that can start a number in that base. With 0, you can invent a sequence starting like this:

111, 1110, 1111110, …

Or like this:

1111, 11111111110, …

Or like this:

11111, 1111111111111111111111111111110, …

And so on. But with 1, there’s no room for manoeuvre.

# Narcischism

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

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

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

# 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.

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)

# 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

# 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

# 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

# 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