A277859 Least k > 1 such that 1^(k-1) + 2^(k-1) + 3^(k-1) + … + (k-1)^(k-1) - n == 0 (mod k).

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

Author

Paolo P. Lava, Nov 02 2016

Keywords

• nonn
• easy

Comments

a(2*n-1) = 2.

a(n) = n + 1 for some prime n + 1 congruent to {2, 3} mod 4.

Examples

			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;
		

Links

• Paolo P. Lava, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A045326, A191677.

Programs

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);