A171416 A sequence with Somos-4 Hankel transform.

1, 1, 2, 3, 7, 13, 31, 65, 156, 351, 849, 1993, 4866, 11733, 28921, 70997, 176560, 438979, 1100302, 2761797, 6969909, 17625015, 44742636, 113822415, 290416803, 742486655, 1902767481, 4885201701, 12567065582, 32382099109, 83580301371

Offset: 0

Author

Paul Barry, Dec 08 2009

Keywords

• easy
• nonn

Comments

Hankel transform is the Somos-4 sequence A006720(n+2).

The generating function A(x) satisfies A(x) = 1 + x + x^2*A(x) + (x*A(x))^2.

BINOMIAL transform is A087626. HANKEL transform with a(0) omitted is A051138(n+2). - Michael Somos, Jan 11 2013

Examples

			G.f. = 1 + x + 2*x^2 + 3*x^3 + 7*x^4 + 13*x^5 + 31*x^6 + 65*x^7 + 156*x^8 + ...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Paul Barry, Riordan arrays, the A-matrix, and Somos 4 sequences, arXiv:1912.01126 [math.CO], 2019.

Crossrefs

Cf. A006720, A051138, A087626, A377264.

Programs

Magma

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!((1-x^2-Sqrt(1-6*x^2-4*x^3+x^4))/(2*x^2))); // G. C. Greubel, Sep 22 2018

Maple

m:=30; S:=series((1 -x^2 -sqrt(1-6*x^2-4*x^3+x^4))/(2*x^2), x, m+1): seq(coeff(S, x, j), j=0..m); # G. C. Greubel, Feb 18 2020

Mathematica

CoefficientList[Series[(1-x^2 -Sqrt[1 -6x^2 -4x^3 +x^4])/(2x^2), {x, 0, 30}], x] (* Or *)
a[n_]:= a[n]= a[n-2] + Sum[a[k-1]a[n-k-1], {k, n-1}]; a[0]=a[1]=1; Array[a, 31, 0] (* Robert G. Wilson v, Mar 28 2011 *)

PARI

{a(n) = if( n<0, 0, polcoeff( 2*(1 + x) / (1 - x^2 + sqrt(1 - 6*x^2 - 4*x^3 + x^4 + x*O(x^n))), n))}; /* Michael Somos, Jan 11 2013 */

Sage

def A171416_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1 -x^2 -sqrt(1-6*x^2-4*x^3+x^4))/(2*x^2) ).list()
A171416_list(30) # G. C. Greubel, Feb 18 2020

Formula

G.f.: (1 - x^2 - sqrt(1-6*x^2-4*x^3+x^4))/(2*x^2).

G.f.: (1/(1-x))*c(x^2/((1-x)*(1-x^2))) where c(x) is the g.f. of A000108.

G.f.: 1/(1-x-x^2/(1-x^2-x^2/(1-x-x^2/(1-x^2-x^2/(1-x-x^2/(1-x^2-x^2/(1-...))))))) (continued fraction).

a(n) = a(n-2) + Sum_{k=1..n-1} a(k-1)*a(n-k-1) with a(0)=a(1)=1.

Conjecture: (n+2)*a(n) +(n+1)*a(n-1) +6*(1-n)*a(n-2) +2*(11-5*n)*a(n-3) +(10-3*n)*a(n-4) +(n-5)*a(n-5)=0. - R. J. Mathar, Jul 24 2012

G.f.: 2*(1 + x) / (1 - x^2 + sqrt(1 - 6*x^2 - 4*x^3 + x^4)).

(n+2) * a(n) - (6*n-6) * a(n-2) - (4*n-10) * a(n-3) + (n-4) * a(n-4) = 0 if n>3. - Michael Somos, Jan 11 2013

If we write the generating function as 1/(1-b_{0}*x/(1-c_{0}x/(1-b_{1}*x/(1-c_{1}*x/(1-...))))), then b_{n}*c_{n} = A006720(n+1)*A006720(n+3)/A006720(n+2)^2 = A377264(n)/A006720(n+2)^2. - Thomas Scheuerle, Oct 22 2024