login
A125054
Central terms of triangle A125053.
5
1, 3, 21, 327, 9129, 396363, 24615741, 2068052367, 225742096209, 31048132997523, 5252064083753061, 1071525520294178007, 259439870666594250489, 73542221109962636293083, 24125551094579137082039181, 9068240688454120376775401247, 3871645204706420218816959159969
OFFSET
0,2
COMMENTS
Triangle A125053 is a variant of triangle A008301 (enumeration of binary trees) such that the leftmost column is the secant numbers (A000364).
Right edge of triangle A210108.
Apparently all terms (except the initial 1) have 3-valuation 1. - F. Chapoton, Aug 02 2021
FORMULA
Binomial transform of A000182 (e.g.f. tan(x)).
a(n) = Sum_{k=0..n} A130847(n,k)*2^k. - Philippe Deléham, Jul 22 2007
G.f.: 1/(1-sqrt(x))/Q(0), where Q(k)= 1 + sqrt(x) - x*(2*k+1)*(2*k+2)/(1 - sqrt(x) - x*(2*k+2)*(2*k+3)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 27 2013
G.f.: Q(0)/(1-3*x), where Q(k) = 1 - 4*x^2*(2*k+1)*(2*k+3)*(k+1)^2/( 4*x^2*(2*k+1)*(2*k+3)*(k+1)^2 - (1 - 8*x*k^2 - 8*x*k -3*x)*(1 - 8*x*k^2 - 24*x*k -19*x)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 23 2013
G.f.: Q(0)/(1-1*x), where Q(k) = 1 - (2*k+1)*(2*k+2)*x/(x*(2*k+1)*(2*k+2) - (1-x)/(1 - (2*k+2)*(2*k+3)*x/(x*(2*k+2)*(2*k+3) - (1-x)/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2013
a(n) ~ 2^(4*n+5) * n^(2*n+3/2) / (exp(2*n) * Pi^(2*n+3/2)). - Vaclav Kotesovec, May 30 2015
From Peter Bala, May 11 2017: (Start)
O.g.f. as an S-fraction: A(x) = 1/(1 - 3*x/(1 - 4*x/(1 - 15*x/(1 - 16*x/(1 - 35*x/(1 - 36*x/(1 - ...))))))), where the unsigned coefficients in the partial numerators [3, 4, 15, 16, 35, 36, ...] come in pairs of the form 4*n^2 - 1, 4*n^2 for n = 1,2,....
A(x) = 1/(1 + 3*x - 6*x/(1 - 2*x/(1 + 3*x - 20*x/(1 - 12*x/(1 + 3*x - 42*x/(1 - 30*x/(1 + 3*x - ...))))))), , where the unsigned coefficients in the partial numerators [6, 2, 20, 12, 42, 30, ...] are obtained from the sequence [2, 6, 12, 20, ..., n*(n + 1), ...] by swapping adjacent terms. (End)
MATHEMATICA
b[n_]:=n!*SeriesCoefficient[Tan[x], {x, 0, n}]; Table[Sum[Binomial[n, k]*b[2*k+1], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 30 2015 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Nov 21 2006
STATUS
approved