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%I #19 Feb 05 2024 18:07:41
%S 1,1,1,2,5,12,27,69,178,457,1194,3178,8538,23062,62726,171804,473069,
%T 1308397,3634075,10133154,28352421,79575702,223981549,632101856,
%U 1788172541,5069879063,14403962756,41001479103,116921037003,333971884899,955443681814,2737387314548,7853533625522,22560919253095,64890249175438,186854616134794
%N Number of solutions to k_1 + 2*k_2 + ... + n*k_n = n, where k_i are from {-1,0,1}, i=1..n.
%C a(n) is the coefficient of both x^n and 1/x^n in Product_{k=1..n} (1/x^k + 1 + x^k), while A007576 gives the constant term in the symmetric product.
%H Vaclav Kotesovec, <a href="/A316706/b316706.txt">Table of n, a(n) for n = 0..500</a>
%F a(n) = [x^n] Product_{k=1..n} (1/x^k + 1 + x^k).
%F a(n) = [x^(n*(n-1)/2)] Product_{k=1..n} (1 + x^k + x^(2*k)).
%F a(n) = [x^(n*(n+3)/2)] Product_{k=1..n} (1 + x^k + x^(2*k)).
%F a(n) ~ 3^(n + 1) / (2 * sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Jul 11 2018
%t nmax = 40; p = 1; Flatten[{1, Table[Coefficient[p = Expand[p*(1/x^n + 1 + x^n)], x^n], {n, 1, nmax}]}] (* _Vaclav Kotesovec_, Jul 11 2018 *)
%o (PARI) {a(n) = polcoeff( prod(k=1,n, 1/x^k + 1 + x^k) + x*O(x^n),n)}
%o for(n=0,40, print1(a(n),", "))
%o (Python)
%o from collections import Counter
%o def A316706(n):
%o c = {0:1}
%o for k in range(1,n+1):
%o b = Counter(c)
%o for j in c:
%o a = c[j]
%o b[j+k] += a
%o b[j-k] += a
%o c = b
%o return c[n] # _Chai Wah Wu_, Feb 05 2024
%Y Cf. A007576.
%K nonn
%O 0,4
%A _Paul D. Hanna_, Jul 10 2018
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