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G.f. A(x) satisfies: x^2*A(x) = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n+1)/2) * A(x)^n.
6

%I #6 Jul 20 2022 08:44:31

%S 1,0,1,3,11,34,110,350,1147,3800,12836,43929,152285,533205,1883187,

%T 6698612,23974179,86258459,311811314,1131863444,4124127216,

%U 15078422405,55301519095,203405409935,750122683729,2773048061073,10274442343829,38147288401915

%N G.f. A(x) satisfies: x^2*A(x) = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n+1)/2) * A(x)^n.

%C Equals the antidiagonal sums of A355360; a(n) = Sum_{k=0..n} A355360(n-k,k).

%H Paul D. Hanna, <a href="/A355364/b355364.txt">Table of n, a(n) for n = 0..400</a>

%F G.f. A(x) satisfies:

%F (1) x^2*A(x) = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n+1)/2) * A(x)^n.

%F (2) -x^2*A(x)^2 = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n+1)/2) / A(x)^n.

%F (3) x^2*A(x)*P(x) = Product_{n>=1} (1 - x^n*A(x)) * (1 - x^(n-1)/A(x)), where P(x) = Product_{n>=1} 1/(1 - x^n) is the partition function (A000041), due to the Jacobi triple product identity.

%e G.f.: A(x) = 1 + x^2 + 3*x^3 + 11*x^4 + 34*x^5 + 110*x^6 + 350*x^7 + 1147*x^8 + 3800*x^9 + 12836*x^10 + 43929*x^11 + 152285*x^12 + ...

%e where

%e x^2*A(x) = ... - x^10/A(x)^5 + x^6/A(x)^4 - x^3/A(x)^3 + x/A(x)^2 - 1/A(x) + 1 - x*A(x) + x^3*A(x)^2 - x^6*A(x)^3 + x^10*A(x)^4 -+ ... + (-1)^n * x^(n*(n+1)/2) * A(x)^n + ...

%o (PARI) {a(n) = my(A=[1,0],t); for(i=1,n, A=concat(A,0); t = ceil(sqrt(2*n+9));

%o A[#A] = polcoeff( x^2*Ser(A) - sum(m=-t,t, (-1)^m*x^(m*(m+1)/2)*Ser(A)^m ), #A-1));A[n+1]}

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A355360, A355361, A355362, A355363, A355365.

%K nonn

%O 0,4

%A _Paul D. Hanna_, Jul 19 2022