A026585 a(n) = T(n,n), T given by A026584. Also a(n) is the number of integer strings s(0), ..., s(n) counted by T, such that s(n)=0.

1, 0, 2, 2, 8, 14, 40, 86, 222, 518, 1296, 3130, 7770, 19066, 47324, 117094, 291260, 724302, 1806220, 4507230, 11266718, 28188070, 70609316, 177023466, 444231564, 1115639586, 2803975860, 7052132546, 17748069294, 44693162266

Offset: 0

Author

Clark Kimberling

Keywords

• nonn

Comments

The signed sequence 1,0,2,-2,8,-14,... is the inverse binomial transform of A026569. - Paul Barry, Sep 09 2004

Hankel transform of a(n) is 2^n. Hankel transform of a(n+1) is {0, -4, 0, 16, 0, -64, 0, 256, 0, ...} or -2^(n+1)*[x^n](x/(1+x^2)). Hankel transform of a(n+2) is 2^(n+1)*A109613(n+1). - Paul Barry, Mar 23 2011

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.

Crossrefs

Cf. A026584, A026587, A026589, A026590, A026591, A026592, A026593, A026594, A026595, A026596, A026597, A026598, A026599, A027282, A027283, A027284, A027285, A027286.

Cf. A000984, A026569, A109613.

Programs

Magma

[(&+[Binomial(n-j-1, n-2*j)*Binomial(2*j, j): j in [0..Floor(n/2)]]): n in [0..40]]; // G. C. Greubel, Dec 12 2021

Mathematica

CoefficientList[Series[Sqrt[(1-x)/(1-x-4*x^2)], {x, 0, 40}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

Sage

[sum(binomial(n-j-1, n-2*j)*binomial(2*j, j) for j in (0..(n//2))) for n in [0..40]] # G. C. Greubel, Dec 12 2021

Formula

a(n) = A026584(n, n).

G.f.: sqrt((1-x)/(1-x-4*x^2)). - Ralf Stephan, Jan 08 2004

From Paul Barry, Jul 01 2009: (Start)

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

a(0) = 1, a(n) = Sum_{k=0..floor(n/2)} (k/(n-k))*C(n-k,k)*A000984(k). (End)

From Paul Barry, Mar 23 2011: (Start)

a(n) = Sum_{k=0..floor(n/2)} C(n-k-1,n-2*k)*A000984(k).

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

D-finite with recurrence n*a(n) +2*(-n+1)*a(n-1) +(-3*n+2)*a(n-2) +2*(2*n-5)*a(n-3) = 0. - R. J. Mathar, Nov 24 2012

a(n) ~ (sqrt(17)+1)^(n-1/2) / (17^(1/4) * sqrt(Pi*n) * 2^(n-3/2)). - Vaclav Kotesovec, Feb 12 2014