A092635 Coefficients in asymptotic expansion of (Pi*n/16^n)*binomial(2n,n)^2 in powers of 1/(8n).

1, -2, 2, 4, -10, -92, 212, 4744, -10330, -440492, 926972, 64905784, -133989700, -13958115992, 28507072232, 4129849252624, -8382195909370, -1609609189734092, 3254514296768492, 799400112196210264, -1612314122867559340

Offset: 0

Author

Michael Somos, Mar 02 2004

Keywords

• sign

Examples

			binomial(2*n,n)^2*Pi*n/16^n ~ 1 -2/(8n) +2/(8n)^2 +4/(8n)^3 + ...
From _Paul D. Hanna_, Apr 02 2011: (Start)
O.g.f.: A(x) = 1 - 2*x + 2*x^2 + 4*x^3 - 10*x^4 - 92*x^5 + 212*x^6 + ...
E.g.f.: E(x) = 1 - 2*x + 2*x^2/2! + 4*x^3/3! - 10*x^4/4! - 92*x^5/5! + 212*x^6/6! + ...
Note that E(x)*exp(2*x) = sqrt(E(x)*E(-x)) is an even function:
E(x)*exp(2*x) = 1 - 2*x^2/2! + 22*x^4/4! - 692*x^6/6! + 45862*x^8/8! + ... (End)
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..300

Crossrefs

Cf. A002894, A012393, A188514.

Programs

Maple

#Uses the conjectural recurrence equation
A092635 := proc (n) option remember; if n = 0 then 1 else add((-4)^(n-k)*euler(n-k, 1)*A092635(k), k = 0 .. n-1)/n end if; end proc:
seq(A092635(n), n = 0 .. 20); # Peter Bala, Mar 12 2015

Mathematica

m = 21;
1/(1+2x/(1-x+ContinuedFractionK[(4i^2-1)x^2, 1, {i, 1, m/2 // Floor}]))+O[x]^m // CoefficientList[#, x]& (* Jean-François Alcover, Nov 02 2019 *)
CoefficientList[Normal[Series[Pi*n*Binomial[n/4, n/8]^2 / 2^(n/2+3), {n, Infinity, 20}]], 1/n] (* Vaclav Kotesovec, Aug 18 2024 *)

PARI

a(n)=if(n<0,0,polcoeff(exp(intformal(serlaplace(-1/cosh(x*2+x*O(x^n))^2*2))),n))

PARI

a(n)=local(A);if(n<1,n==0,A=1; for(m=1,n,A+=x^2*O(x^m);A=Pol(A)+polcoeff(subst(A,x,x/(1+8*x))/A*(1+8*x)/(1+4*x)^2,m+1)/8/m*x^m);polcoeff(A,n))

PARI

{B(x,n) = prod(k=1,n, (1 + 4*k*x)^((-1)^k*binomial(n,k)) )}
{a(n) = polcoeff( A = prod(m=1,n, B(x  +x*O(x^n), m)^(1/2^m)), n)}
for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Mar 03 2024

Formula

E.g.f. E(x) satisfies: E(x) = E(-x)*exp(-4*x). - Paul D. Hanna, Apr 02 2011

O.g.f.: A(x) = 1/(1 + 2*x/(1-x + 3*x^2/(1 + 15*x^2/(1 + 35*x^2/(1 + 63*x^2/(1 + 99*x^2/(1 + ...+ (4*n^2-1)*x^2/(1 + ...)))))))) (continued fraction). - Paul D. Hanna, Apr 02 2011

From Peter Bala, Mar 12 2015: (Start)

Let 2/(1 + exp(4*x)) = Sum_{n >= 0} b(n)*x^n/n!, so that b(n) = (-4)^n*E(n,1), where E(n,x) is the n-th Euler polynomial.

Then exp( Sum_{n >= 1} b(n)*x^n/n ) = 1 - 2*x + 2*x^2 + 4*x^3 - 10*x^4 - 92*x^5 + ... appears to be the o.g.f. for this sequence.

Assuming this is true we have the recurrence a(0) = 1 and a(n) = 1/n*Sum_{k = 0..n-1} (-4)^(n-k)*E(n-k,1)*a(k) for n >= 1. (End)

O.g.f.: A(x) = Product_{n>=1} B(x,n)^(1/2^n) where B(x,n) = Product{k=1..n} (1 + 4*k*x)^((-1)^k*binomial(n,k)). Compare to Product_{n>=1} B(x,n) = 1 - 4*x and Product_{n>=1} B(x,n)^(1/n) = exp(-4*x). - Paul D. Hanna, Mar 03 2024

Conjecture from Paul D. Hanna, Aug 15 2024: (Start)

Let A(x) = Sum_{n>=0} a(n)*x^(n+1), then A(x) satisfies:

A(x) = -x^2 / A(-x),

A(x) = -A(-x/(1+4*x)),

A(x) = x^2 / A(x/(1-4*x)). (End)