A284760 a(n) = Sum_{i=1..n-1}(i^(n-2)) mod n^4.

0, 1, 3, 14, 100, 979, 196, 500, 3834, 1333, 2178, 1022, 16731, 12647, 42420, 23912, 23409, 26265, 15162, 79730, 84441, 21723, 28566, 160732, 280625, 329405, 137295, 569702, 74849, 71999, 463202, 715984, 247665, 31873, 1302420, 574170, 807710, 225091, 1377129

Offset: 1

Author

Felix Fröhlich, Apr 02 2017

Keywords

• nonn

Comments

Mestrovic conjectures that a(n) > 0 for all n > 1 (Conjecture 2.11).

Examples

			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
		

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.

Crossrefs

Cf. A076015, A284759.

Programs

Mathematica

Table[Mod[Sum[i^(n - 2), {i, n - 1}], n^4], {n, 39}] (* Michael De Vlieger, Apr 05 2017 *)

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

Formula

a(n) = A076015(n-1) modulo n^4.