A119370 G.f. satisfies A(x) = 1 + x*A(x)^2 + x^2*(A(x)^2 - A(x)).

1, 1, 2, 6, 19, 64, 225, 816, 3031, 11473, 44096, 171631, 675130, 2679728, 10719237, 43168826, 174885089, 712222799, 2914150406, 11973792218, 49385167369, 204386777160, 848530495383, 3532844222611, 14747626307436, 61712139464939

Offset: 0

Author

Paul D. Hanna, May 16 2006

Keywords

• nonn
• easy

Comments

Equals base sequence of pendular trinomial triangle A119369; iterated convolutions of this sequence with the central terms (A119371) generates all diagonals of A119369. For example: A119372 = A119370 * A119371; A119373 = A119370^2 * A119371.

Diagonal sums of number array A133336. - Philippe Deléham, Nov 09 2009

Examples

			A(x) = 1 + x + 2*x^2 + 6*x^3 + 19*x^4 + 64*x^5 + 225*x^6 + 816*x^7 +...
x*A(x)^2 = x + 2*x^2 + 5*x^3 + 16*x^4 + 54*x^5 + 190*x^6 + 690*x^7 +...
x^2*( A(x)^2 - A(x) ) = 1*x^3 + 3*x^4 + 10*x^5 + 35*x^6 + 126*x^7 +...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Stoyan Dimitrov, On permutation patterns with constrained gap sizes, arXiv:2002.12322 [math.CO], 2020.
• Hannah Golab, Pattern avoidance in Cayley permutations, Master's Thesis, Northern Arizona Univ. (2024). See p. 41.

Crossrefs

Cf. A104547, A119369, A119371, A119372, A119373, A119374, A119375, A119376.

Programs

Magma

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

Maple

m:= 30;
S:= series( (1+x^2 -sqrt(1-4*x-2*x^2+x^4))/(2*x*(1+x)), x, m+1);
seq(coeff(S, x, j), j = 0..m); # G. C. Greubel, Mar 17 2021

Mathematica

CoefficientList[Series[((1+x^2)-Sqrt[(1+x^2)^2-4*x*(1+x)])/(2*x*(1+x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 11 2013 *)

PARI

{a(n)=polcoeff(2/((1+x^2)+sqrt((1+x^2)^2-4*x*(1+x)+x*O(x^n))),n)}

Sage

def A119370_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( (1+x^2 -sqrt(1-4*x-2*x^2+x^4))/(2*x*(1+x)) ).list()
A119370_list(30) # G. C. Greubel, Mar 17 2021

Formula

G.f.: A(x) = ((1+x^2) - sqrt( (1+x^2)^2 - 4*x*(1+x) ))/(2*x*(1+x)). Equals the inverse binomial transform of A104547.

Recurrence: (n+1)*a(n) = 3*(n-1)*a(n-1) + 6*(n-1)*a(n-2) + 2*(n-2)*a(n-3) - (n-5)*a(n-4) - (n-5)*a(n-5). - Vaclav Kotesovec, Sep 11 2013

a(n) ~ sqrt(-z^2-3*z+1)*(4+2*z-z^3)^(n+1)*(-z^3+z^2+z+3) / (8*sqrt(Pi) * n^(3/2)), where z = 1/(2*sqrt(3/(4+(280-24*sqrt(129))^(1/3) + 2*(35 + 3*sqrt(129))^(1/3)))) - 1/2*sqrt(8/3-1/3*(280-24*sqrt(129))^(1/3) - 2/3*(35+3*sqrt(129))^(1/3) + 8*sqrt(3/(4+(280-24*sqrt(129))^(1/3) + 2*(35 + 3*sqrt(129))^(1/3)))) = 0.225270426... is the root of the equation 1-2*z^2+z^4-4*z=0. - Vaclav Kotesovec, Sep 11 2013

G.f.: 1/G(0) where G(k) = 1 - q/(1 - (q + q^2) / G(k+1) ). - Joerg Arndt, Dec 06 2014

a(n) = Sum_{k=0..floor(n/2)} binomial(n-k-1,k) * binomial(2*n-3*k+1,n-2*k)/(2*n-3*k+1). - Seiichi Manyama, Aug 28 2023

Conjecture: A(x) = 1 + x*exp(Sum_{n >= 1} g(n, x)*x^n/n), where g(n, x) = Sum_{k = 0..n} binomial(n, k)^2*(1 + x)^k. Cf. A105633 and A167638. - Peter Bala, Sep 10 2024