OFFSET
0,3
COMMENTS
This expansion has the form 1/(1-arctan(1/sqrt(2)*x)^2) = Sum_{n>=0} a(2*n)*x^(2*n)/(2n)!.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..238
FORMULA
a(n) = (2*n)!/2^n*Sum_{k=1..2*n} ((2*k)!*(-1)^((n+k))*Sum_{i=0..2*n-2*k} (2^(i)*Stirling1(i+2*k,2*k)*binomial(2*n-1,i+2*k-1))/(i+2*k)!)), n > 0, a(0)=1.
a(n) ~ -(-1)^n * (2*n)! / (n * (log(2*n))^3 * 2^(n-3)) * (1 - 3*(gamma + log(2)) / log(2*n)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Oct 11 2018
MAPLE
seq(coeff(series(factorial(n)*(1-arctan(1/sqrt(2)*x)^2)^(-1), x, n+1), x, n), n = 0 .. 30, 2); # Muniru A Asiru, Oct 07 2018
MATHEMATICA
Join[{1}, Table[((2*n)!/2^n)*Sum[(2*k)!*(-1)^(n + k)*Sum[(2^j* StirlingS1[j + 2*k, 2*k]*Binomial[2*n - 1, j + 2*k - 1])/(j + 2*k)!, {j, 0, 2*(n - k)}], {k, 1, 2*n}], {n, 1, 50}]] (* G. C. Greubel, Oct 05 2018 *)
PROG
(Maxima)
a(n):=if n=0 then 1 else (2*n)!/2^n*sum((2*k)!*(-1)^((n+k))*sum((2^(i)*stirling1(i+2*k, 2*k)*binomial(2*n-1, i+2*k-1))/(i+2*k)!, i, 0, 2*n-2*k), k, 1, 2*n);
(PARI) for(n=0, 30, print1(if(n==0, 1, ((2*n)!/2^n)*sum(k=1, 2*n, sum(j=0, 2*n-2*k, (2*k)!*(-1)^(n + k)*2^j*stirling(j + 2*k, 2*k, 1)*binomial(2*n - 1, j + 2*k - 1)/(j + 2*k)!) )), ", ")) \\ G. C. Greubel, Oct 05 2018
CROSSREFS
KEYWORD
sign
AUTHOR
Vladimir Kruchinin, Nov 23 2011
STATUS
approved