A046017 Least k > 1 with k = sum of digits of k^n, or 0 if no such k exists.
2, 9, 8, 7, 28, 18, 18, 46, 54, 82, 98, 108, 20, 91, 107, 133, 80, 172, 80, 90, 90, 90, 234, 252, 140, 306, 305, 90, 305, 396, 170, 388, 170, 387, 378, 388, 414, 468, 449, 250, 432, 280, 461, 280, 360, 360, 350, 370, 270, 685, 360, 625, 648, 370, 677, 684, 370, 667, 370, 694, 440, 855, 827, 430, 818
Offset: 1
Examples
a(3) = 8 since 8^3 = 512 and 5+1+2 = 8; a(5) = 28 because 28 is least number > 1 with 28^5 = 17210368, 1+7+2+1+0+3+6+8 = 28. 53^7 = 1174711139837 -> 1+1+7+4+7+1+1+1+3+9+8+3+7 = 53. a(10) = 82 because 82^10 = 13744803133596058624 and 1 + 3 + 7 + 4 + 4 + 8 + 0 + 3 + 1 + 3 + 3 + 5 + 9 + 6 + 0 + 5 + 8 + 6 + 2 + 4 = 82. a(13) = 20: 20^13=81920000000000000, 8+1+9+2=20. a(17) = 80: 80^17=225179981368524800000000000000000, 2+2+5+1+7+9+9+8+1+3+6+8+5+2+4+8 = 80.
References
- G. Balzarotti and P. P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 208-210.
- Joe Roberts, "Lure of the Integers", The Mathematical Association of America, 1992, p. 172.
Links
- Carole Dubois, Table of n, a(n) for n = 1..4522 (terms 1..1000 from T. D. Noe).
- Carole Dubois, Scatterplot of A046017
Crossrefs
Programs
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Mathematica
a[n_] := For[k = 2, k <= 20*n, k++, Which[k == Total[IntegerDigits[k^n]], Return[k], k == 20*n, Return[0]]]; Table[a[n] , {n, 1, 105}] (* Jean-François Alcover, May 23 2012 *) sdk[n_]:=Module[{k=2},While[k!=Total[IntegerDigits[k^n]],k++];k]; Array[sdk,70] (* Harvey P. Dale, Jan 07 2024 *)
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Python
from itertools import chain def c(k, n): return sum(map(int, str(k**n))) == k def a(n): if n == 0: return False d, lim = 1, 1 while lim < n*9*d: d, lim = d+1, lim*10 m = next(k for k in chain(range(2, lim+1), (0,)) if c(k, n)) return m print([a(n) for n in range(1, 66)]) # Michael S. Branicky, Jul 06 2022
Extensions
More terms from Asher Auel, Jun 01 2000
Comments