OFFSET
1,3
COMMENTS
Mestrovic conjectures that a(n) > 0 for all n > 1 (Conjecture 2.11).
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..1000
R. Mestrovic, A congruence modulo n^3 involving two consecutive sums of powers and its applications, arXiv:1211.4570 [math.NT], 2012.
FORMULA
a(n) = A076015(n-1) modulo n^4.
EXAMPLE
For n=5 the sum is 1^3 + 2^3 + 3^3 + 4^3 = 1 + 8 + 27 + 64 = 100; the modulus is 5^4 = 625. So a(5) = 100 mod 625 = 100. - Peter Munn, May 01 2017
MATHEMATICA
Table[Mod[Sum[i^(n - 2), {i, n - 1}], n^4], {n, 39}] (* Michael De Vlieger, Apr 05 2017 *)
PROG
(PARI) a(n) = lift(Mod(sum(i=1, n-1, i^(n-2)), n^4))
(PARI) a(n)=my(m=n^4, e=n-2); lift(sum(i=1, n-1, Mod(i, m)^e)) \\ Charles R Greathouse IV, Apr 07 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Felix Fröhlich, Apr 02 2017
STATUS
approved