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%I #53 Sep 19 2020 17:10:23
%S 0,1,0,0,0,3,0,0,0,5,0,0,0,7,0,0,0,9,0,0,0,11,0,0,0,13,0,0,0,15,0,0,0,
%T 17,0,0,0,19,0,0,0,21,0,0,0,23,0,0,0,25,0,0,0,27,0,0,0,29,0,0,0,31,0,
%U 0,0,33,0,0,0,35,0,0,0,37,0,0,0,39
%N a(n) = n/2 if n mod 4 = 2, 0 otherwise.
%C If S(j,n) = Sum_{k=1..n} k^j then, for any odd j, S(j,n) mod n = a(n). - _Gary Detlefs_, Oct 26 2011
%C Odd numbers A005408, with 3 zeros between them. - _T. D. Noe_, Oct 27 2011
%H Seiichi Manyama, <a href="/A195938/b195938.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,2,0,0,0,-1).
%F Euler transform of length 8 sequence [ 0, 0, 0, 3, 0, 0, 0, -1]. - _Michael Somos_, Oct 29 2011
%F a(n) = -a(-n) for all n in Z. - _Michael Somos_, Oct 29 2011
%F a(n) = (Sum_{k=1..n} k^(2*j-1)) mod n, for any j.
%F a(n) = (n/2)*floor((1/2)*cos((n+2)*Pi/2) + 1/2).
%F G.f.: (1+x^4)*x^2/(1-x^4)^2. - _Philippe Deléham_, Oct 27 2011
%F a(n) = binomial(n^2,3)/4 mod n. - _Gary Detlefs_, May 04 2013
%F a(n) = n*(1 - i^n)*(1 + i^(2*n))/8, where i=sqrt(-1). - _Ammar Khatab_, Aug 25 2020
%e G.f. = x + 3*x^5 + 5*x^9 + 7*x^13 + 9*x^17 + 11*x^21 + 13*x^25 + ...
%p S:=(j,n)-> sum(k^j,k=1..n):seq(S(3,n) mod n, n=1..70);
%t a[n_] := If[Mod[n, 4] == 2, n/2, 0]; Table[a[n], {n, 80}] (* _Alonso del Arte_, Oct 26 2011 *)
%o (PARI) a(n)=if(n%4==2,n/2) \\ _Charles R Greathouse IV_, Oct 26 2011
%K nonn,easy
%O 1,6
%A _Gary Detlefs_, Oct 26 2011