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a(n) = Sum_{i=1..n-1}(i^(n-2)) mod n^4.
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%I #21 May 02 2017 11:46:02

%S 0,1,3,14,100,979,196,500,3834,1333,2178,1022,16731,12647,42420,23912,

%T 23409,26265,15162,79730,84441,21723,28566,160732,280625,329405,

%U 137295,569702,74849,71999,463202,715984,247665,31873,1302420,574170,807710,225091,1377129

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

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

%H Seiichi Manyama, <a href="/A284760/b284760.txt">Table of n, a(n) for n = 1..1000</a>

%H R. Mestrovic, <a href="https://arxiv.org/abs/1211.4570">A congruence modulo n^3 involving two consecutive sums of powers and its applications</a>, arXiv:1211.4570 [math.NT], 2012.

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

%e 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

%t Table[Mod[Sum[i^(n - 2), {i, n - 1}], n^4], {n, 39}] (* _Michael De Vlieger_, Apr 05 2017 *)

%o (PARI) a(n) = lift(Mod(sum(i=1, n-1, i^(n-2)), n^4))

%o (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

%Y Cf. A076015, A284759.

%K nonn

%O 1,3

%A _Felix Fröhlich_, Apr 02 2017