A157002 Transform of Catalan numbers whose Hankel transform gives the Somos-4 sequence.

1, 0, 1, 2, 6, 17, 51, 156, 488, 1552, 5006, 16337, 53849, 179015, 599535, 2020924, 6851150, 23344138, 79902364, 274606264, 947240592, 3278404274, 11381240074, 39621423949, 138288477617, 483805404673, 1696318159457, 5959737806635

Offset: 0

Author

Paul Barry, Feb 20 2009

Keywords

• easy
• nonn

Comments

Image of the Catalan numbers A000108 by the Riordan array (1-x,x(1-x^2)). Hankel transform is A006720(n+1).

The sequence a(n)+a(n+1) begins 1,1,3,8,23,68,... which is A056010. The sequence a(n)+a(n-1) begins 1,1,1,3,8,23,68,... which is A025262. This is obtained by applying (1-x^2,x(1-x^2)) to the Catalan numbers.

Hankel transform of a(n+1) is -A051138(n). - Michael Somos, Feb 10 2015

Examples

			G.f. = 1 + x^2 + 2*x^3 + 6*x^4 + 17*x^5 + 51*x^6 + 156*x^7 + 488*x^8 + ...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Gouce Xin, Proof of the Somos-4 Hankel determinants conjecture, Advances in Applied Mathematics, Volume 42, Issue 2, February 2009, Pages 152-156.

Crossrefs

Cf. A000108, A006720, A025262, A051138, A056010.

Programs

Magma

m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (1 -Sqrt(1-4*x*(1-x^2)))/(2*x*(1+x)) )); // G. C. Greubel, Feb 26 2019

Mathematica

CoefficientList[Series[(1-Sqrt[1-4x(1-x^2)])/(2x(1+x)), {x,0,30}], x] (* G. C. Greubel, Feb 26 2019 *)

PARI

{a(n) = if( n<0, -(-1)^n / 2 * (n<-1), polcoeff( (1 - sqrt(1 - 4*x * (1 - x^2) + x^2 * O(x^n))) / (2 * x * (1 + x)), n))}; /* Michael Somos, Feb 10 2015 */

Sage

((1-sqrt(1-4*x*(1-x^2)))/(2*x*(1+x))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 26 2019

Formula

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

a(n) = Sum_{k=0..n} (-1)^floor((n-k+1)/2)*C(k,floor((n-k)/2))*A000108(k).

Conjecture: (n+1)*a(n) +3*(-n+1)*a(n-1) +2*(-2*n+1)*a(n-2) +2*(2*n-7)*a(n-3) +2*(2*n-7)*a(n-4)=0. - R. J. Mathar, Nov 19 2014

0 = a(n)*(+16*a(n+1) + 16*a(n+2) - 64*a(n+3) - 42*a(n+4) + 22*a(n+5)) + a(n+1)*(+16*a(n+1) + 48*a(n+2) - 46*a(n+3) - 56*a(n+4) + 22*a(n+5)) + a(n+2)*(+32*a(n+2) + 34*a(n+3) - 8*a(n+4) - 10*a(n+5)) + a(n+3)*(+18*a(n+3) + 11*a(n+4) - 9*a(n+5)) + a(n+4)*(+3*a(n+4) + a(n+5)) for all n in Z. - Michael Somos, Feb 10 2015