A160702 Sequence such that the Hankel transform of a(n+1) satisfies a generalized Somos-4 recurrence.

1, 1, 1, 5, 19, 79, 333, 1441, 6351, 28451, 129185, 593373, 2752427, 12876343, 60684533, 287857209, 1373286375, 6584979659, 31719337353, 153416338549, 744777567043, 3627787084319

Offset

0,4

Comments

The Hankel transform of a(n+1) satisfies a generalized Somos-4 Hankel determinant recurrence.

Hankel transform of a(n+1) is A160703. In general, we can conjecture that the Hankel transform of

h(n) of a(n+1), where a(n)=if(n=0,1,if(n=1,1,if(n=2,1,r*a(n-1)+s*sum{k=0..n-2, a(k)*a(n-1-k)}))),

satisfies the generalized Somos-4 recurrence

h(n)=(s^2*h(n-1)*h(n-3)+s^3*(2*s+r-2)*h(n-2)^2)/h(n-4).

The case r=s=1 is proved in the Xin reference.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Paul Barry, On the Hurwitz Transform of Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.8.7.

Gouce Xin, Proof of the Somos-4 Hankel determinants conjecture, Advances in Applied Mathematics, Volume 42, Issue 2, February 2009, Pages 152-156.

Formula

a(n) = if(n=0,1,if(n=1,1,if(n=2,1,a(n-1)+2*sum{k=0..n-2, a(k)*a(n-1-k)})))

Recurrence: (n+1)*a(n) = 3*(2*n-1)*a(n-1) - (n-2)*a(n-2) - 8*(2*n-7)*a(n-3). - Vaclav Kotesovec, Nov 20 2012

G.f.: 1/4+(1-sqrt(16*x^3+x^2-6*x+1))/(4*x). - Vaclav Kotesovec, Nov 20 2012

Mathematica

CoefficientList[Series[1/4+(1-Sqrt[16*x^3+x^2-6*x+1])/(4*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Nov 20 2012 *)

Crossrefs

Cf.: A025262, A056010, A006720.

Sequence in context: A149773 A363548 A149774 * A149775 A149776 A149777

Adjacent sequences: A160699 A160700 A160701 * A160703 A160704 A160705

Keyword

easy,nonn

Author

Paul Barry, May 24 2009

Status

approved