A209937 E.g.f. A(x) satisfies: A( x - Sum_{n>=2} x^n/(n*(n-1)/2) ) = x.

1, 2, 14, 164, 2692, 56832, 1466656, 44735392, 1574507104, 62807331520, 2800211567936, 137988945383808, 7447519816062848, 436913348544200192, 27682538499602786816, 1883880221782019929088, 137046014280583363879936, 10612885049611654523670528

Offset: 1

Author

Paul D. Hanna, Mar 15 2012

Keywords

• nonn

Comments

Compare e.g.f. to the identity: let W(x) = Sum_{n>=1} (n-1)^(n-1)*x^n/n!, then W( x - Sum_{n>=2} x^n/(n*(n-1)) ) = x.

Examples

			E.g.f.: A(x) = x + 2*x^2/2! + 14*x^3/3! + 164*x^4/4! + 2692*x^5/5! + 56832*x^6/6! + 1466656*x^7/7! + 44735392*x^8/8! +...
Let R(x) be the series reversion of e.g.f. A(x), then R(x) begins:
R(x) = x - x^2/1 - x^3/3 - x^4/6 - x^5/10 - x^6/15 - x^7/21 - x^8/28 -...
...
Compare to the series reversion of the function W(x) defined by:
W(x) = x + x^2/2! + 2^2*x^3/3! + 3^3*x^4/4! + 4^4*x^5/5! + 5^5*x^6/6! +...
where W(x - x^2/2 - x^3/6 - x^4/12 - x^5/20 - x^6/30 - x^7/42 -...) = x.
		

Crossrefs

Cf. A209923, A075856.

Programs

Maple

A209937_list := proc(len) local A, n; A[1] := 1; for n from 2 to len do
A[n] := (n-2)*A[n-1] + add(binomial(n,j)*A[j]*A[n-j], j=1..n-1) od:
convert(A,list) end: A209937_list(18); # Peter Luschny, May 24 2017

Mathematica

Rest[CoefficientList[InverseSeries[Series[-x-2*(1-x)*Log[1-x], {x, 0, 20}], x],x]*Range[0, 20]!] (* Vaclav Kotesovec, Jan 23 2014 *)

PARI

{a(n)=n!*polcoeff(serreverse(x-sum(m=2, n, x^m/(m*(m-1)/2)) +x*O(x^n)), n)}

PARI

{a(n)=n!*polcoeff(serreverse(-x-2*(1-x)*log(1-x +x*O(x^n))), n)}
for(n=1, 25, print1(a(n), ", "))

PARI

/* From a formula of Michael Somos for triangle A075856: */
{A075856(n, k)=if(k<1|nA075856(n-1,k)+(n+k-1)*A075856(n-1,k-1)))}
{a(n)=if(n<1,0,if(n==1,1,sum(k=1,n-1,A075856(n-1, k)*2^k)))}
for(n=1,20,print1(a(n),", "))

Formula

E.g.f. A(x) satisfies: A( -x - 2*(1-x)*log(1-x) ) = x.

a(n) = Sum_{k=1..n-1} A075856(n-1,k)*2^k for n>1 with a(1)=1.

a(n) ~ n^(n-1) / (sqrt(2) * (2-exp(1/2))^(n-1/2) * exp(n/2+1/2)). - Vaclav Kotesovec, Jan 23 2014

a(n) = (n-2)*a(n-1) + Sum_{j=1..n-1} binomial(n,j)*a(j)*a(n-j) for n>1, a(1)=1. - Peter Luschny, May 24 2017