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a(n) = A361540(2*n,n) / binomial(2*n,n) for n >= 0.
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%I #6 Mar 24 2023 09:03:54

%S 1,2,71,10915,4063645,2842101221,3255178907803,5605980824208871,

%T 13710496284516264953,45746570903514799640905,

%U 202291094041887013214628871,1160411497892246920315488823067,8496377826955803443098054623140629,78398366060939693412478828210386035725

%N a(n) = A361540(2*n,n) / binomial(2*n,n) for n >= 0.

%C E.g.f. F(x,y) of triangle A361540 satisfies the following.

%C (1) F(x,y) = Sum_{n>=0} (F(x,y)^n + y)^n * x^n/n!.

%C (2) F(x,y) = Sum_{n>=0} F(x,y)^(n^2) * exp(y*x*F(x,y)^n) * x^n/n!.

%C This sequence equals the central terms of triangle A361540 divided by the central binomial coefficients A000984.

%H Paul D. Hanna, <a href="/A361688/b361688.txt">Table of n, a(n) for n = 0..25</a>

%e E.g.f. A(x) = 1 + 2*x + 71*x^2/2! + 10915*x^3/3! + 4063645*x^4/4! + 2842101221*x^5/5! + 3255178907803*x^6/6! + 5605980824208871*x^7/7! + 13710496284516264953*x^8/8! + ... + a(n)*x^n/n! + ...

%o (PARI) /* E.g.f. of triangle A361540 is F(x,y) = Sum_{n>=0} (F(x,y)^n + y)^n * x^n/n! */

%o {A361540(n,k) = my(F = 1); for(i=1,n, F = sum(m=0, n, (F^m + y +x*O(x^n))^m * x^m/m! )); n!*polcoeff(polcoeff(F, n,x),k,y)}

%o {a(n) = A361540(2*n,n)/binomial(2*n,n)}

%o for(n=0, 15, print1(a(n), ", "))

%Y Cf. A361540, A000984.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Mar 20 2023