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G.f.: A(x) = ( Sum_{n>=0} 3^n*(2*n+1) * x^(n*(n+1)/2) )^(1/3).
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%I #10 Feb 27 2021 21:30:08

%S 1,3,-9,60,-360,2457,-18036,138429,-1093500,8833140,-72622224,

%T 605563452,-5108366277,43512281460,-373690245420,3232056818511,

%U -28126143258444,246080268205092,-2163254305208580,19097478037041840,-169235311045503708,1504837859547132468

%N G.f.: A(x) = ( Sum_{n>=0} 3^n*(2*n+1) * x^(n*(n+1)/2) )^(1/3).

%C Compare to the q-series identity:

%C eta(x)^3 = Sum_{n>=0} (-1)^n*(2*n+1) * x^(n*(n+1)/2),

%C where eta(x) is the Dedekind eta(q) function without the q^(1/24) factor.

%F Conjecture: a(5*n+4) == 0 (mod 5) (checked up to n = 200). - _Peter Bala_, Feb 26 2021

%e G.f.: A(x) = 1 + 3*x - 9*x^2 + 60*x^3 - 360*x^4 + 2457*x^5 - 18036*x^6 +...

%e where

%e A(x)^3 = 1 + 9*x + 45*x^3 + 189*x^6 + 729*x^10 + 2673*x^15 + 9477*x^21 +...+ 3^n*(2*n+1)*x^(n*(n+1)/2) +...

%o (PARI) {a(n)=polcoeff(sum(m=0,sqrtint(2*n+1),3^m*(2*m+1)*(x)^(m*(m+1)/2)+x*O(x^n))^(1/3),n)}

%Y Cf. A193236.

%K sign

%O 0,2

%A _Paul D. Hanna_, Dec 14 2011