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A357220 a(n) = coefficient of x^n in Sum_{n>=0} x^n/(1 - x*C(x)^n), where C(x) = 1/(1 - x*C(x)) is a g.f. of the Catalan numbers (A000108). 1

%I #12 Dec 03 2022 12:06:30

%S 1,2,3,5,11,31,101,355,1304,4938,19155,75857,306075,1256782,5248018,

%T 22278742,96141427,421787510,1881594580,8537257714,39408291543,

%U 185114771571,885043068109,4307374572585,21340519926034,107627435856554,552473684683454,2885909702592788

%N a(n) = coefficient of x^n in Sum_{n>=0} x^n/(1 - x*C(x)^n), where C(x) = 1/(1 - x*C(x)) is a g.f. of the Catalan numbers (A000108).

%C Related Identity due to George E. Andrews: Sum_{n>=0} x^(k*n)/(1 - x^k*q^n) = Sum_{n>=0} q^(n^2) * x^(2*k*n) * (1 + x^k*q^n)/(1 - x^k*q^n), which holds for positive integer k.

%H Paul D. Hanna, <a href="/A357220/b357220.txt">Table of n, a(n) for n = 0..400</a>

%F Generating function A(x) = Sum_{n>=0} a(n)*x^n may be expressed by the following formulas, where C(x) = 1/(1 - x*C(x)).

%F (1) A(x) = Sum_{n>=0} x^n/(1 - x*C(x)^n).

%F (2) A(x) = Sum_{n>=0} C(x)^(n^2) * x^(2*n) * (1 + x*C(x)^n)/(1 - x*C(x)^n).

%F (3) A(x-x^2) = Sum_{n>=0} x^n * (1-x)^(2*n) / ((1-x)^n - x*(1-x)).

%e G.f.: A(x) = 1 + 2*x + 3*x^2 + 5*x^3 + 11*x^4 + 31*x^5 + 101*x^6 + 355*x^7 + 1304*x^8 + 4938*x^9 + 19155*x^10 + 75857*x^11 + 306075*x^12 + ...

%e such that

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

%e also

%e A(x) = (1+x)/(1-x) + C(x)*x^2*(1+x*C(x))/(1-x*C(x)) + C(x)^4*x^4*(1+x*C(x)^2)/(1-x*C(x)^2) + C(x)^9*x^6*(1+x*C(x)^3)/(1-x*C(x)^3) + C(x)^16*x^8*(1+x*C(x)^4)/(1-x*C(x)^4) + ... + C(x)^(n^2)*x^(2*n)*(1+x*C(x)^n)/(1-x*C(x)^n) + ...

%e where C(x) = 1/(1 - x*C(x)) begins

%e 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 + 16796*x^10 + A000108(n)*x^n + ...

%o (PARI) {a(n) = my(A, C = 1/x*serreverse(x - x^2 + O(x^(n+2))));

%o A = sum(m=0,n+1, x^m/(1 - x*C^m)); polcoeff(A,n)}

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

%Y Cf. A000108.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Oct 16 2022

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