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A136728 E.g.f.: A(x) = (exp(x)/(4 - 3*exp(x)))^(1/4). 7

%I #27 Nov 15 2023 08:03:49

%S 1,1,4,31,349,5146,93799,2036161,51283894,1470035101,47250248569,

%T 1683031711516,65800765032589,2801364476781781,129003301751229364,

%U 6389120632590635971,338644807090096148809,19126604338708282552186

%N E.g.f.: A(x) = (exp(x)/(4 - 3*exp(x)))^(1/4).

%H Seiichi Manyama, <a href="/A136728/b136728.txt">Table of n, a(n) for n = 0..373</a>

%F E.g.f. A(x) satisfies: A(x) = 1 + integral( A(x)^5 * exp(-x) ).

%F O.g.f.: 1/(1 - x/(1-3*x/(1 - 5*x/(1-6*x/(1 - 9*x/(1-9*x/(1 - 13*x/(1-12*x/(1 - 17*x/(1-15*x/(1 - ...)))))))))), a continued fraction.

%F G.f.: 1/G(0) where G(k) = 1 - x*(4*k+1)/( 1 - 3*x*(k+1)/G(k+1) ); (continued fraction ). - _Sergei N. Gladkovskii_, Mar 23 2013

%F a(n) ~ n! * Gamma(3/4)/(sqrt(2)*3^(1/4)*n^(3/4)*Pi*log(4/3)^(n+1/4)). - _Vaclav Kotesovec_, Jun 15 2013

%F a(n) = 1 + 3 * Sum_{k=1..n-1} (binomial(n,k) - 1) * a(k). - _Ilya Gutkovskiy_, Jul 09 2020

%F From _Seiichi Manyama_, Nov 15 2023: (Start)

%F a(n) = Sum_{k=0..n} (-1)^(n-k) * (Product_{j=0..k-1} (4*j+1)) * Stirling2(n,k).

%F a(0) = 1; a(n) = Sum_{k=1..n} (-1)^k * (3*k/n - 4) * binomial(n,k) * a(n-k).

%F a(0) = 1; a(n) = a(n-1) + 3*Sum_{k=1..n-1} binomial(n-1,k) * a(n-k). (End)

%t CoefficientList[Series[(E^x/(4-3*E^x))^(1/4), {x, 0, 20}], x]* Range[0, 20]! (* _Vaclav Kotesovec_, Jun 15 2013 *)

%o (PARI) a(n)=n!*polcoeff((exp(x +x*O(x^n))/(4-3*exp(x +x*O(x^n))))^(1/4),n)

%o (PARI) /* As solution to integral equation: */ a(n)=local(A=1+x+x*O(x^n));for(i=0,n,A=1+intformal(A^4*exp(-x+x*O(x^n))));n!*polcoeff(A,n)

%Y Cf. A201354, variants: A014307, A136727, A136729.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jan 24 2008

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