A119370 - OEIS

%I #36 Sep 11 2024 05:49:47

%S 1,1,2,6,19,64,225,816,3031,11473,44096,171631,675130,2679728,

%T 10719237,43168826,174885089,712222799,2914150406,11973792218,

%U 49385167369,204386777160,848530495383,3532844222611,14747626307436,61712139464939

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

%C 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.

%C Diagonal sums of number array A133336. - _Philippe Deléham_, Nov 09 2009

%H G. C. Greubel, <a href="/A119370/b119370.txt">Table of n, a(n) for n = 0..1000</a>

%H Stoyan Dimitrov, <a href="https://arxiv.org/abs/2002.12322">On permutation patterns with constrained gap sizes</a>, arXiv:2002.12322 [math.CO], 2020.

%H Hannah Golab, <a href="http://danaernst.com/scholarship/GolabThesis.pdf">Pattern avoidance in Cayley permutations</a>, Master's Thesis, Northern Arizona Univ. (2024). See p. 41.

%F 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.

%F 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

%F 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

%F G.f.: 1/G(0) where G(k) = 1 - q/(1 - (q + q^2) / G(k+1) ). - _Joerg Arndt_, Dec 06 2014

%F 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

%F 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

%e A(x) = 1 + x + 2*x^2 + 6*x^3 + 19*x^4 + 64*x^5 + 225*x^6 + 816*x^7 +...

%e x*A(x)^2 = x + 2*x^2 + 5*x^3 + 16*x^4 + 54*x^5 + 190*x^6 + 690*x^7 +...

%e x^2*( A(x)^2 - A(x) ) = 1*x^3 + 3*x^4 + 10*x^5 + 35*x^6 + 126*x^7 +...

%p m:= 30;

%p S:= series( (1+x^2 -sqrt(1-4*x-2*x^2+x^4))/(2*x*(1+x)), x, m+1);

%p seq(coeff(S, x, j), j = 0..m); # _G. C. Greubel_, Mar 17 2021

%t 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 *)

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

%o (Sage)

%o def A119370_list(prec):

%o     P.<x> = PowerSeriesRing(QQ, prec)

%o     return P( (1+x^2 -sqrt(1-4*x-2*x^2+x^4))/(2*x*(1+x)) ).list()

%o A119370_list(30) # _G. C. Greubel_, Mar 17 2021

%o (Magma)

%o R<x>:=PowerSeriesRing(Rationals(), 30);

%o Coefficients(R!( (1+x^2 -Sqrt(1-4*x-2*x^2+x^4))/(2*x*(1+x)) )); // _G. C. Greubel_, Mar 17 2021

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

%K nonn,easy

%O 0,3

%A _Paul D. Hanna_, May 16 2006