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A277859
Least k > 1 such that 1^(k-1) + 2^(k-1) + 3^(k-1) + … + (k-1)^(k-1) - n == 0 (mod k)
1
2, 3, 2, 4, 2, 7, 2, 3, 2, 11, 2, 4, 2, 3, 2, 4, 2, 19, 2, 3, 2, 23, 2, 4, 2, 3, 2, 4, 2, 31, 2, 3, 2, 5, 2, 4, 2, 3, 2, 4, 2, 9, 2, 3, 2, 47, 2, 4, 2, 3, 2, 4, 2, 5, 2, 3, 2, 59, 2, 4, 2, 3, 2, 4, 2, 45, 2, 3, 2, 15, 2, 4, 2, 3, 2, 4, 2, 9, 2, 3, 2, 83, 2, 4, 2
OFFSET
1,1
COMMENTS
a(2*n-1) = 2.
a(n) = n + 1 for some prime n + 1 congruent to {2, 3} mod 4.
LINKS
EXAMPLE
a(8) = 3 because:
1^(2-1) - 8 = -7 but -7 mod 2 = 1;
1^(3-1) + 2^(3-1) - 8 = -3 and -3 mod 3 = 0;
MAPLE
P:=proc(q) local j, k, n; for n from 1 to q do for k from 2 to q do
if (add(j^(k-1), j=1..k-1)-n) mod k=0 then print(k); break; fi; od; od; end: P(10^3);
CROSSREFS
Sequence in context: A365784 A235912 A339749 * A308566 A288535 A105117
KEYWORD
nonn,easy
AUTHOR
Paolo P. Lava, Nov 02 2016
STATUS
approved