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%I #13 Apr 28 2020 11:27:10
%S 1,0,2,1,0,2,0,0,2,1,0,4,0,0,0,0,0,2,3,0,2,2,0,0,0,0,2,2,0,0,1,0,4,0,
%T 0,2,2,0,2,0,0,2,0,0,0,1,0,2,2,0,4,0,0,4,0,0,0,0,0,2,0,0,2,3,0,2,0,0,
%U 2,0,0,2,2,0,0,2,0,2,0,0,2,4,0,0,1,0,2,0,0,0,2,0,2,0,0,4,0,0,2,0,0,2,4,0,0
%N Number of representations of n as a sum of a twice a square and three times a triangular number.
%C The greedy inverse (first occurrence of n) starts 1, 0, 2, 18, 11, 900, 116, 44118, 515, 3105, 5702, ... - _R. J. Mathar_, Apr 28 2020
%H G. C. Greubel, <a href="/A112608/b112608.txt">Table of n, a(n) for n = 0..1000</a>
%H M. D. Hirschhorn, <a href="http://dx.doi.org/10.1016/j.disc.2004.08.045">The number of representations of a number by various forms</a>, Discrete Mathematics 298 (2005), 205-211
%F a(n) = d_{1, 3}(8n+3) - d_{2, 3}(8n+3) where d_{a, m}(n) equals the number of divisors of n which are congruent to a mod m.
%F Euler transform of period 24 sequence [0, 2, 1, -3, 0, 1, 0, -1, 1, 2, 0, -4, 0, 2, 1, -1, 0, 1, 0, -3, 1, 2, 0, -2, ...]. - _Michael Somos_, Jan 01 2006
%F Expansion of q^(-3/8)*(eta(q^4)^5*eta(q^6)^2)/(eta(q^2)^2*eta(q^3)*eta(q^8)^2) in powers of q.
%F a(n) = A002324(8*n+3).
%e a(11) = 4 since we can write 11 = 2*(2)^2 + 3*1 = 2*(-2)^2 + 3*1 = 2*(1)^2 + 3*3 = 2*(-1)^2 + 3*3
%t eta[x_] := x^(1/24)*QPochhammer[x]; A112608[n_] := SeriesCoefficient[ q^(-3/8)*(eta[q^4]^5*eta[q^6]^2)/(eta[q^2]^2*eta[q^3]*eta[q^8]^2), {q, 0, n}]; Table[A112608[n], {n, 0, 50}] (* _G. C. Greubel_, Sep 25 2017 *)
%o (PARI) {a(n)=local(A); if(n<0, 0, A=x*O(x^n); polcoeff( eta(x^4+A)^5*eta(x^6+A)^2/ eta(x^2+A)^2/eta(x^3+A)/eta(x^8)^2, n))} /* _Michael Somos_, Jan 01 2006 */
%K nonn
%O 0,3
%A _James A. Sellers_, Dec 21 2005
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