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Number of subsets of {1, 3, ..., 2*n-1} which sum to 0 modulo 2*n-1.
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%I #10 May 12 2020 11:55:15

%S 2,2,2,2,4,6,10,18,30,54,98,178,328,608,1130,2114,3974,7490,14170,

%T 26890,51150,97542,186420,356962,684784,1315870,2532410,4880646,

%U 9418806,18199014,35204650,68174116,132152842,256415958,497967282,967879954,1882725390,3665038872

%N Number of subsets of {1, 3, ..., 2*n-1} which sum to 0 modulo 2*n-1.

%H Robert Israel, <a href="/A334125/b334125.txt">Table of n, a(n) for n = 1..1000</a>

%F If 2*k - 1 is a prime, then a(k) = (2^k - 2*(-1)^floor(k/2))/(2*k - 1).

%F Conjecture: a(n) = 2*abs(A178738(n)).

%e a(5) = 4 because there are 4 subsets of {1, 3, 5, 7, 9} which sum to 0 modulo 9: {}, {9}, {1, 3, 5}, {1, 3, 5, 9}.

%p f:= proc(n) local V, k;

%p V:= Vector(2*n-1);

%p V[2*n-1]:= 1;

%p for k from 1 to 2*n-1 by 2 do

%p V:= V + V[[$(k+1)..(2*n-1),$1..k]]

%p od;

%p V[2*n-1]

%p end proc:

%p map(f, [$1..40]); # _Robert Israel_, May 12 2020

%o (PARI) a(n) = {my(v=Vec(prod(i=1, n, x^(2*i-1)+1))); sum(i=0, n^2\(2*n-1), v[n^2+1-i*(2*n-1)]); }

%Y Cf. A063776, A178738.

%K nonn

%O 1,1

%A _Jinyuan Wang_, Apr 30 2020