login
a(n) = A048739(n-1) mod A000129(floor(n/2)).
1

%I #12 Mar 29 2016 23:40:18

%S 0,0,0,1,4,3,0,1,28,27,0,1,168,167,0,1,984,983,0,1,5740,5739,0,1,

%T 33460,33459,0,1,195024,195023,0,1,1136688,1136687,0,1,6625108,

%U 6625107,0,1,38613964,38613963,0,1,225058680,225058679,0,1,1311738120,1311738119,0,1,7645370044,7645370043,0,1

%N a(n) = A048739(n-1) mod A000129(floor(n/2)).

%C It appears that a(4*n+1) = 1. - _Michel Marcus_, Mar 23 2016

%F Empirical g.f.: x^5*(1+3*x-6*x^4+6*x^5+x^8-x^9) / ((1-x)*(1+x^2)*(1+2*x^2-x^4)*(1-2*x^2-x^4)). - _Colin Barker_, Mar 22 2016

%e a(7) = 3 because a(7) = A048739(6) mod A000129(floor(7/2)) = (1 + 2 + 5 + 12 + 29 + 70 + 169) mod 5 = 288 mod 5 = 3.

%e a(8) = 0 because a(8) = A048739(7) mod A000129(floor(8/2)) = (1 + 2 + 5 + 12 + 29 + 70 + 169 + 408) mod 12 = 0.

%e a(9) = 1 because a(9) = A048739(8) mod A000129(floor(9/2)) = (1 + 2 + 5 + 12 + 29 + 70 + 169 + 408 + 985) mod 12 = 1.

%o (PARI) a000129(n) = ([2, 1; 1, 0]^n)[2, 1];

%o for(n=2, 1e2, print1(sum(k=1, n, a000129(k)) % a000129(n\2),", "));

%Y Cf. A000129 (Pell numbers), A048739 (partial sums of Pell numbers).

%K nonn

%O 2,5

%A _Altug Alkan_, Mar 22 2016