A118351 Central terms of pendular triangle A118350.

1, 1, 6, 42, 325, 2688, 23286, 208659, 1918314, 17994264, 171542460, 1657212768, 16188521454, 159634359415, 1586932321578, 15886925400954, 160026976985205, 1620715748715648, 16493797802077032, 168583560794745684

Offset: 0

Author

Paul D. Hanna, Apr 26 2006

Keywords

• nonn

Links

• G. C. Greubel, Table of n, a(n) for n = 0..500

Crossrefs

Cf. A118350, A118352, A118353, A118354.

Programs

Magma

R:=PowerSeriesRing(Rationals(), 30);
[1] cat Coefficients(R!( Reversion( x/((1+x)*(1+5*x+x^2)) ) )); // G. C. Greubel, Feb 18 2021

Mathematica

T[n_, k_, p_]:= T[n,k,p] = If[nG. C. Greubel, Feb 18 2021 *)
Join[{1}, Rest@CoefficientList[InverseSeries[Series[ x/((1+x)*(1+5*x+x^2)), {x,0,30}]], x]] (* G. C. Greubel, Feb 18 2021 *)

PARI

{a(n)=polcoeff((serreverse(x*(1-3*x+sqrt((1-3*x)*(1-7*x)+x*O(x^n)))/2/(1-3*x))/x),n)}
for(n=0,30,print1(a(n),", "))

PARI

{a(n)=polcoeff(1 + serreverse( x/((1+x)*(1+5*x+x^2 +x*O(x^n)))),n)}
for(n=0,30,print1(a(n),", "))

Sage

def S_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (x/((1+x)*(1+5*x+x^2))).reverse() ).list()
a=S_list(31); [1]+a[1:] # G. C. Greubel, Feb 18 2021

Formula

G.f. A=A(x) satisfies: A = 1 - 3*x*A + 3*x*A^2 + x*A^3.

G.f.: 1 + Series_Reversion( x/((1+x)*(1+5*x+x^2)) ).

G.f.: (1/x)*Series_Reversion( x*(1-3*x+sqrt((1-3*x)*(1-7*x)))/2/(1-3*x) ).

For n>0: a(n) = 1/n*sum(j=0..n, C(n,j) *sum(i=0..(n-1), C(j,i)*C(n-j,2*j-n-i-1) *6^(2*n-3*j+2*i+1))). - Vladimir Kruchinin, Dec 26 2010

a(n) ~ s^(3/2) / (3*sqrt(2*Pi*(1 + 3*s + 3*s^2)) * n^(3/2) * r^(n+1)), where s = 2*sin(Pi/6 + arctan(sqrt(7)/3)/3) - 1, r = 2*s/(9 - 12*sin(Pi/6 - 2*arctan(sqrt(7)/3)/3)). - Vaclav Kotesovec, Feb 18 2021