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%I #13 Mar 11 2021 11:45:15
%S 1,2,2,1,2,2,2,1,2,2,2,0,2,4,2,0,1,2,2,2,2,2,2,0,3,2,2,0,2,2,2,2,2,2,
%T 0,2,2,4,2,-1,2,2,2,0,2,2,4,0,2,4,2,1,0,0,2,2,2,4,2,0,2,2,2,0,2,2,2,2,
%U 4,2,0,0,1,4,2,0,2,2,2,2,2,0,2,2,2,4
%N G.f.: Sum_{n>=0} x^n/(1 - x^(5*n+1)) - x^3*Sum_{n>=0} x^(4*n)/(1 - x^(5*n+4)).
%C The g.f. of this sequence equals the denominator of George E. Andrews' expression for the cube of Ramanujan's continued fraction. See references given in A007325.
%F G.f.: Product_{n>=0} (1 - x^(n+1)) * (1 - x^(5*n+5)) / ( (1 - x^(5*n+1))^3 * (1 - x^(5*n+4))^3 ).
%F G.f.: Product_{n>=0} (1 - x^(5*n+5))^2 * (1 - x^(5*n+2))*(1 - x^(5*n+3)) / ( (1 - x^(5*n+1))^2*(1 - x^(5*n+4))^2 ).
%F G.f.: [ Sum_{n>=0} x^(2*n)/(1 - x^(5*n+1)) - x^2 * Sum_{n>=0} x^(3*n)/(1 - x^(5*n+4)) ] / R(x), where R(q) is the expansion of Ramanujan's continued fraction (A007325).
%e G.f.: Q(q) = 1 + 2*q + 2*q^2 + q^3 + 2*q^4 + 2*q^5 + 2*q^6 + q^7 + 2*q^8 + 2*q^9 + 2*q^10 + 2*q^12 + 4*q^13 + 2*q^14 + q^16 + 2*q^17 + 2*q^18 + 2*q^19 + 2*q^20 + ...
%e Given the g.f. of this sequence
%e Q(q) = Sum_{n>=0} q^n/(1 - q^(5*n+1)) - q^3*Sum_{n>=0} q^(4*n)/(1 - q^(5*n+4))
%e and the g.f. of A340455,
%e P(q) = Sum_{n>=0} q^(2*n)/(1 - q^(5*n+2)) - q*Sum_{n>=0} q^(3*n)/(1 - q^(5*n+3))
%e then
%e R(q)^3 = P(q)/Q(q) where
%e P(q) = 1 - q + 2*q^2 + 2*q^6 - 2*q^7 + 2*q^8 + q^9 + q^12 - 2*q^13 + 2*q^14 + 2*q^16 + 2*q^18 + ...
%e R(q)^3 = 1 - 3*q + 6*q^2 - 7*q^3 + 3*q^4 + 6*q^5 - 17*q^6 + 24*q^7 - 21*q^8 + 6*q^9 + 21*q^10 - 54*q^11 + 77*q^12 - 72*q^13 + 24*q^14 + 64*q^15 + ...
%e here, R(q) is the expansion of Ramanujan's continued fraction (A007325).
%o (PARI) {a(n) = my(A = prod(m=0,n\5+1, (1-x^(5*m+5) +x*O(x^n))^2 * (1-x^(5*m+2))*(1-x^(5*m+3)) / ( (1-x^(5*m+1))^2*(1-x^(5*m+4))^2 +x*O(x^n) ) ));polcoeff(A,n)}
%o for(n=0,100,print1(a(n),", "))
%o (PARI) {S(j,k,n) = sum(m=0,n, x^(j*m)/(1 - x^(5*m+k) +x*O(x^n)) ) }
%o {a(n) = polcoeff( S(1,1,n) - x^3*S(4,4,n), n)}
%o for(n=0,100,print1(a(n),", "))
%Y Cf. A007325, A055102, A340455, A340453, A340454.
%K sign
%O 0,2
%A _Paul D. Hanna_, Jan 20 2021
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