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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A384829 G.f. satisfies A(x) = Sum_{n>=0} x^(n*(n+1)/2) * A(x)^(n*(n+1)).

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%I A384829 #10 Jun 13 2025 07:01:45
%S A384829 1,1,2,6,22,87,359,1535,6758,30431,139442,648001,3046730,14467286,
%T A384829 69281190,334211603,1622568398,7921905397,38871120255,191586353683,
%U A384829 948083155952,4708743978840,23463673225988,117271827518778,587744334759630,2953138645722287,14872864243128300,75066312240321173
%N A384829 G.f. satisfies A(x) = Sum_{n>=0} x^(n*(n+1)/2) * A(x)^(n*(n+1)).
%H A384829 Paul D. Hanna, <a href="/A384829/b384829.txt">Table of n, a(n) for n = 0..1000</a>
%F A384829 G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
%F A384829 (1) A(x) = Sum_{n>=0} x^(n*(n+1)/2) * A(x)^(n*(n+1)).
%F A384829 (2) A(x) = sqrt( (1/x) * Series_Reversion( x*eta(x)^2/eta(x^2)^4 ) ), where eta(x) is Dedekind's eta(q) function without the q^(1/24) factor.
%F A384829 (3) A(x) = Product_{n>=1} (1 + (x*A(x)^2)^n) * (1 - (x*A(x)^2)^(2*n)).
%F A384829 (4) A(x) = exp( Sum_{n>=1} ( (x*A(x)^2)^n / (1 + (x*A(x)^2)^n) )/n ).
%F A384829 (5) A(x)^4 = Sum_{n>=0} (2*n+1) * (x*A(x)^2)^n / (1 - (x*A(x)^2)^(2*n+1)).
%F A384829 (6) A(x^2)^2 = Sum_{n>=0} (x*A(x^2))^n / (1 + (x*A(x^2))^(2*n+1)).
%F A384829 a(n) ~ c * d^n / n^(3/2), where d = 5.33733388876021052204016376282654316742329168165380444126... and c = 0.24712373554952847890961627688964866920906379264976188659... - _Vaclav Kotesovec_, Jun 13 2025
%e A384829 G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 22*x^4 + 87*x^5 + 359*x^6 + 1535*x^7 + 6758*x^8 + 30431*x^9 + 139442*x^10 + 648001*x^11 + 3046730*x^12 + ...
%e A384829 where A(x) = 1 + x*A(x)^2 + x^3*A(x)^6 + x^6*A(x)^12 + x^10*A(x)^20 + x^15*A(x)^30 + ... + x^(n*(n+1)/2) * A(x)^(n*(n+1)) + ...
%t A384829 (* Calculation of constants {d,c}: *) {1/r, 2*(s/Sqrt[Pi*(-5 + 8*r^(7/8)*s^(3/4) * Derivative[0, 0, 2][EllipticTheta][2, 0, Sqrt[r]*s])])} /. FindRoot[{2*r^(1/8)*s^(5/4) == EllipticTheta[2, 0, Sqrt[r]*s], 5*s^(1/4) == 2*r^(3/8) * Derivative[0, 0, 1][EllipticTheta][2, 0, Sqrt[r]*s]}, {r, 1/4}, {s, 1}, WorkingPrecision -> 120] (* _Vaclav Kotesovec_, Jun 13 2025 *)
%o A384829 (PARI) {a(n) = my(A = sqrt( (1/x) * serreverse( x*eta(x +x^2*O(x^n))^2/eta(x^2 +x^2*O(x^n))^4 ) ) ); polcoef(A,n)}
%o A384829 for(n=0,30, print1(a(n),", "))
%Y A384829 Cf. A106336, A109085.
%K A384829 nonn
%O A384829 0,3
%A A384829 _Paul D. Hanna_, Jun 13 2025