A118346 Central terms of pendular triangle A118345.

1, 1, 5, 30, 201, 1445, 10900, 85128, 682505, 5585115, 46461437, 391743850, 3340361700, 28755475180, 249572076200, 2181469638880, 19186562661273, 169677521094215, 1507881643936015, 13458730170115778, 120599648894147185

Offset: 0

Author

Paul D. Hanna, Apr 26 2006

Keywords

• nonn

Comments

Also, g.f. A(x) = (1/x)*series_reversion of x/(1 + x*GF(A005572)), where GF(A005572) is the g.f. of A005572, which is the number of walks on cubic lattice starting and finishing on the xy plane and never going below it.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A118345, A118347, A118348, A118349.

Programs

Magma

R:=PowerSeriesRing(Rationals(), 30);
[1] cat Coefficients(R!( Reversion( x/((1+x)*(1+4*x+x^2)) ) )); // G. C. Greubel, Mar 17 2021

Mathematica

CoefficientList[1 +InverseSeries[Series[x/((1+x)*(1+4*x+x^2)), {x,0,30}]], x] (* G. C. Greubel, Mar 17 2021 *)

PARI

{a(n) = polcoeff(serreverse( x*(1-2*x+sqrt((1-2*x)*(1-6*x)+x*O(x^n)))/(2*(1-2*x)))/x, n)}

Sage

def A118346_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( ( x/((1+x)*(1+4*x+x^2)) ).reverse() ).list()
a=A118346_list(31); [1]+a[1:] # G. C. Greubel, Mar 17 2021

Formula

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

G.f.: A(x) = 1 + series_reversion( x/((1+x)*(1+4*x+x^2)) ).

G.f.: A(x) = (1/x)*series_reversion( x*(1-2*x + sqrt((1-2*x)*(1-6*x)))/(2*(1-2*x)) ).

For n>0: a(n) = (1/n)*Sum_{j=0..n} Sum_{i=0..n-1} ( binomial(n,j) * binomial(j,i) * binomial(n-j,2*j-n-i-1) * 5^(2*n-3*j+2*i+1) ). -Vladimir Kruchinin, Dec 26 2010