A273953 E.g.f. satisfies A(x) = Sum_{n>=0} x^n/n! * exp(n/2*x) * A(x)^(n/2).

1, 1, 3, 13, 77, 581, 5347, 58213, 732937, 10487737, 168217811, 2990748509, 58397418037, 1242643927357, 28627000014355, 709933328752981, 18859531958840273, 534365880859577777, 16087267158157316323, 512844446937529664173, 17259468942471032848861, 611530055485070740134901, 22755171133646348369448323, 887228501593124485460914373, 36173480392953890421156056665, 1539307965110263598673884269801, 68247672532254821767545000249907

Offset: 0

Author

Paul D. Hanna, Jun 14 2016

Keywords

• nonn

Examples

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 13*x^3/3! + 77*x^4/4! + 581*x^5/5! + 5347*x^6/6! + 58213*x^7/7! + 732937*x^8/8! + 10487737*x^9/9! + 168217811*x^10/10! + 2990748509*x^11/11! + 58397418037*x^12/12! +...
such that
A(x) = 1 + x*exp(x/2)*A(x)^(1/2) + x^2/2!*exp(x)*A(x) + x^3/3!*exp(3*x/2)*A(x)^(3/2) + x^4/4!*exp(2*x)*A(x)^2 + x^5/5!*exp(5*x/2)*A(x)^(5/2) + x^6/6!*exp(3*x)*A(x)^3 +...
The logarithm of A(x) begins:
log(A(x)) = x + 2*x^2/2! + 6*x^3/3! + 28*x^4/4! + 180*x^5/5! + 1476*x^6/6! + 14728*x^7/7! + 173216*x^8/8! + 2346480*x^9/9! + 35981200*x^10/10! + 616111056*x^11/11! + 11652662880*x^12/12! +...+ A100526(n)*x^n/n! +...
which equals -2*LambertW(-x*exp(x/2)/2).
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..411
• Eric Weisstein's World of Mathematics, Lambert W-Function.

Crossrefs

Cf. A273954, A100526.

Programs

Mathematica

CoefficientList[Series[4*LambertW[-x/2*E^(x/2)]^2 / (x^2*E^x), {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Jun 23 2016 *)

PARI

{a(n) = my(A=1+x); for(i=1,n, A = sum(m=0,n,x^m/m!*exp(m/2*x +x*O(x^n))*A^(m/2)) ); n!*polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

PARI

a(n) = sum(k=0, n, k^(n-k)*(k+2)^(k-1)*binomial(n, k))/2^(n-1); \\ Seiichi Manyama, Feb 11 2023

Formula

E.g.f.: 4*LambertW(-x/2*exp(x/2))^2 / (x^2*exp(x)).

E.g.f.: exp( L(x) ) where L(x) = -2*LambertW(-x*exp(x/2)/2) is the e.g.f. of A100526.

a(n) ~ sqrt(1+LambertW(exp(-1)))*n^(n-1)/(2^(n-1)*exp(n-2)*LambertW(exp(-1))^n). - Vaclav Kotesovec, Jun 23 2016

From Seiichi Manyama, Feb 11 2023: (Start)

E.g.f. satisfies A(x) = exp( x * ( exp(x) * A(x) )^(1/2) ).

a(n) = (1/2^(n-1)) * Sum_{k=0..n} k^(n-k) * (k+2)^(k-1) * binomial(n,k). (End)