|
%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
| 