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%I #14 Jul 02 2019 18:05:35
%S 2,4,2,8,2,3,2,16,2,4,2,3,2,4,2,32,2,3,2,5,2,4,2,3,2,4,2,7,2,3,2,64,2,
%T 4,2,3,2,4,2,5,2,3,2,8,2,4,2,3,2,4,2,8,2,3,2,7,2,4,2,3,2,4,2,128,2,3,
%U 2,8,2,4,2,3,2,4,2,8,2,3,2,5,2,4,2,3,2,4,2,11,2,3,2,8,2,4,2,3,2,4,2,5
%N a(n) = least number k > 1 such that 1^k + 2^k + ... + k^k == n (mod k).
%H Alois P. Heinz, <a href="/A259111/b259111.txt">Table of n, a(n) for n = 1..8192</a>
%F a(2^n) = 2^(n+1) for n >= 0.
%e Consider n=2:
%e Is k=2? 1^2 + 2^2 == 1 (mod 2). No.
%e Is k=3? 1^3 + 2^3 + 3^3 == 0 (mod 3). No.
%e Is k=4? 1^4 + 2^4 + 3^4 + 4^4 == 2 (mod 4). Yes. So a(2) = 4.
%e (Example corrected by _N. J. A. Sloane_, Jul 02 2019)
%p a:= proc(n) local k; for k from 2 while
%p add(i&^k mod k, i=1..k) mod k <> n mod k do od; k
%p end:
%p seq(a(n), n=1..100); # _Alois P. Heinz_, Jun 18 2015
%t lnk[n_]:=Module[{k=2},While[Mod[Total[Range[k]^k],k]!=Mod[n,k],k++];k]; Array[ lnk,100] (* _Harvey P. Dale_, Jul 02 2019 *)
%o (PARI) vector(100,n,k=2;while(sum(i=1,k,i^k)!=Mod(n,k),k++);k)
%Y Cf. A014117, A226960-A226967.
%K nonn
%O 1,1
%A _Derek Orr_, Jun 18 2015
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