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A356777 G.f.: Sum_{n=-oo..+oo} x^(n^2) * C(x)^(2*n-1), where C(x) = 1 + x*C(x)^2 is a g.f. of the Catalan numbers (A000108).

%I A356777 #8 Sep 13 2022 04:42:16
%S A356777 1,1,-3,0,1,-5,5,0,0,1,-7,14,-7,0,0,0,1,-9,27,-30,9,0,0,0,0,1,-11,44,
%T A356777 -77,55,-11,0,0,0,0,0,1,-13,65,-156,182,-91,13,0,0,0,0,0,0,1,-15,90,
%U A356777 -275,450,-378,140,-15,0,0,0,0,0,0,0,1,-17,119,-442,935,-1122,714,-204,17,0,0,0,0,0,0,0,0,1,-19,152,-665,1729,-2717,2508,-1254,285,-19
%N A356777 G.f.: Sum_{n=-oo..+oo} x^(n^2) * C(x)^(2*n-1), where C(x) = 1 + x*C(x)^2 is a g.f. of the Catalan numbers (A000108).
%H A356777 Paul D. Hanna, <a href="/A356777/b356777.txt">Table of n, a(n) for n = 0..2555</a>
%F A356777 G.f. A(x) = Sum_{n>=0} a(n)*x^n may be obtained from the following expressions; here, C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).
%F A356777 (1) A(x) = Sum_{n=-oo..+oo} x^(n^2) * C(x)^(2*n-1).
%F A356777 (2) A(x) = 1/C(x) * Product_{n>=1} (1 + x^(2*n-1)*C(x)^2) * (1 + x^(2*n-1)/C(x)^2) * (1 - x^(2*n)), by the Jacobi triple product identity.
%F A356777 (3) A(x) = 1/C(x) + Sum_{n>=1} x^(n^2) * (C(x)^(2*n-1) + 1/C(x)^(2*n+1)).
%F A356777 (4) A(x) = Sum_{n>=0} Sum_{k=0..n} (-1)^k * binomial(2*n-k, k) * (2*n+1)/(2*n-2*k+1) * x^(n^2 + k).
%e A356777 G.f.: A(x) = 1 + x - 3*x^2 + x^4 - 5*x^5 + 5*x^6 + x^9 - 7*x^10 + 14*x^11 - 7*x^12 + x^16 - 9*x^17 + 27*x^18 - 30*x^19 + 9*x^20 + x^25 - 11*x^26 + 44*x^27 - 77*x^28 + 55*x^29 - 11*x^30 + x^36 - 13*x^37 + 65*x^38 - 156*x^39 + 182*x^40 - 91*x^41 + 13*x^42 + x^49 - 15*x^50 + 90*x^51 - 275*x^52 + 450*x^53 - 378*x^54 + 140*x^55 - 15*x^56 + ...
%e A356777 such that
%e A356777 A(x) = ... + x^16/C(x)^9 + x^9/C(x)^7 + x^4/C(x)^5 + x/C(x)^3 + 1/C(x) + x*C(x) + x^4*C(x)^3 + x^9*C(x)^5 + x^16*C(x)^7 + x^25*C(x)^9 + ... + x^(n^2)*C^(2*n-1) + ...
%e A356777 where the Catalan function C(x) = (1 - sqrt(1-4*x))/(2*x) begins
%e A356777 C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + 1430*x^8 + 4862*x^9 + ... + A000108(n)*x^n + ...
%e A356777 RELATED TABLE.
%e A356777 This sequence may be written in the form of an irregular triangle:
%e A356777 1,
%e A356777 1, -3, 0,
%e A356777 1, -5, 5, 0, 0,
%e A356777 1, -7, 14, -7, 0, 0, 0,
%e A356777 1, -9, 27, -30, 9, 0, 0, 0, 0,
%e A356777 1, -11, 44, -77, 55, -11, 0, 0, 0, 0, 0,
%e A356777 1, -13, 65, -156, 182, -91, 13, 0, 0, 0, 0, 0, 0,
%e A356777 1, -15, 90, -275, 450, -378, 140, -15, 0, 0, 0, 0, 0, 0, 0,
%e A356777 1, -17, 119, -442, 935, -1122, 714, -204, 17, 0, 0, 0, 0, 0, 0, 0, 0,
%e A356777 1, -19, 152, -665, 1729, -2717, 2508, -1254, 285, -19, 0, 0, 0, 0, 0, 0, 0, 0, 0,
%e A356777 ...
%e A356777 Compare the above construction to triangle A082985.
%o A356777 (PARI) /* By Definition: */
%o A356777 {a(n) = my(A, C = 1/x*serreverse(x-x^2 +O(x^(n+2))), M=ceil(sqrt(n+1)));
%o A356777 A = sum(m=-M, M, x^(m^2) * C^(2*m-1) ); polcoeff(A, n)}
%o A356777 for(n=0, 90, print1(a(n), ", "))
%o A356777 (PARI) /* Without Using Catalan Series */
%o A356777 {a(n) = my(A, M=ceil(sqrt(n+1)));
%o A356777 A = sum(m=0, M, sum(k=0, 2*m, (-1)^k*binomial(2*m-k, k)*(2*m+1)/(2*m-2*k+1) * x^(m^2 + k) ) +x*O(x^n)); polcoeff(A, n)}
%o A356777 for(n=0, 90, print1(a(n), ", "))
%Y A356777 Cf. A355341, A355342, A355345, A356778, A082985, A000108.
%K A356777 sign
%O A356777 0,3
%A A356777 _Paul D. Hanna_, Sep 08 2022