A125054 - OEIS

%I #37 Aug 02 2021 06:46:03

%S 1,3,21,327,9129,396363,24615741,2068052367,225742096209,

%T 31048132997523,5252064083753061,1071525520294178007,

%U 259439870666594250489,73542221109962636293083,24125551094579137082039181,9068240688454120376775401247,3871645204706420218816959159969

%N Central terms of triangle A125053.

%C Triangle A125053 is a variant of triangle A008301 (enumeration of binary trees) such that the leftmost column is the secant numbers (A000364).

%C Right edge of triangle A210108.

%C Apparently all terms (except the initial 1) have 3-valuation 1. - _F. Chapoton_, Aug 02 2021

%H Vincenzo Librandi, <a href="/A125054/b125054.txt">Table of n, a(n) for n = 0..100</a>

%H Peter Bala, <a href="/A002439/a002439.pdf">Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)</a>

%F Binomial transform of A000182 (e.g.f. tan(x)).

%F a(n) = Sum_{k=0..n} A130847(n,k)*2^k. - _Philippe Deléham_, Jul 22 2007

%F 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

%F 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

%F 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

%F a(n) ~ 2^(4*n+5) * n^(2*n+3/2) / (exp(2*n) * Pi^(2*n+3/2)). - _Vaclav Kotesovec_, May 30 2015

%F From _Peter Bala_, May 11 2017: (Start)

%F 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,....

%F 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)

%t 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 *)

%Y Cf. A125053, A125055, A000182.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Nov 21 2006