A108081 a(n) = Sum_{i=0..n} binomial(2*n-i, n+i).

1, 2, 7, 25, 92, 344, 1300, 4950, 18955, 72905, 281403, 1089343, 4227273, 16438345, 64037453, 249855417, 976205516, 3818779616, 14954876080, 58623077586, 230007291334, 903164858092, 3549071519462, 13955918890440, 54912972103772, 216194101316654, 851622127750060

Offset: 0

Author

Ralf Stephan, Jun 03 2005

Keywords

• nonn

Comments

A transform of the Fibonacci numbers A000045(n+1) under the mapping g(x)->(1/(c(x)*sqrt(1-4*x)))*g(x*c(x)), c(x) the g.f. of A000108. Hankel transform is the bisection of the Fibonacci numbers F(2*n+2) (A001906(n+1)). - Paul Barry, Sep 28 2007

Diagonal sums of A159965. - Paul Barry, Apr 28 2009

Comment from Li-yao Xia, Oct 22 2015: (Start)

Consider the smallest set X of finite sequences of integer (or words), such that

- 0 belongs to it;

- if a and b are two words in X, let L(a) be the word obtained by reversing a and subtracting 1 from every term, and R(b) be the word obtained by reversing b and adding 1 to every term; then the concatenations L(a).b and a.R(b) belong to X.

Examples of L and R values: L(10,30,20) = 19, 29, 9; R(10,30,20) = 21, 31, 11

List of words of X of lengths 1, 2, 3:

0

0, 1

-1, 0

-1, 0, 1 = L(0), 0, 1 = -1, 0, R(0)

0, 2, 1 = 0, R(0, 1)

1, -1, 0 = L(0), -1, 0

0, 1, 0 = 0, R(-1, 0)

0, -1, 0 = L(0, 1), 0

0, 1, 1 = 0, 1, R(0)

-1, -2, 0 = L(-1, 0), 0

The number of words of length n for n<=12 is given by a(n+1). Is this always true? (End)

Links

• G. C. Greubel and Vincenzo Librandi, Table of n, a(n) for n = 0..1000(terms 0..200 from Vincenzo Librandi)

Crossrefs

Cf. A000045, A000108, A001906, A159965.

Programs

Mathematica

CoefficientList[Series[(1+Sqrt[1-4*x])/(2*Sqrt[1-4*x]*(x+Sqrt[1-4*x])), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 17 2012 *)
Table[Sum[Binomial[2n-i,n+i],{i,0,n}],{n,0,30}] (* Harvey P. Dale, Oct 20 2013 *)

PARI

my(x='x+O('x^66)); Vec((1+sqrt(1-4*x))/(2*sqrt(1-4*x)*(x+sqrt(1-4*x)))) \\ Joerg Arndt, May 15 2013

PARI

a(n) = sum(k=0,n, binomial(n+k-1,k)*fibonacci(n-k+1)); \\ G. C. Greubel, Jan 31 2017

Formula

G.f.: 1/2*(1-5*x+4*x^2+((1-4*x)*(1-5*x)^2)^(1/2))/(1-4*x)/(1-4*x-x^2). - Vladeta Jovovic, Sep 06 2006

G.f.: (1+sqrt(1-4*x))/(2*sqrt(1-4*x)*(x+sqrt(1-4*x))). - Paul Barry, Sep 28 2007

a(n) = Sum_{k=0..n} C(n+k-1,k)*F(n-k+1). - Paul Barry, Sep 28 2007

Recurrence: n*(n+1)*a(n) = 2*(4*n^2 + 3*n - 6)*a(n-1) - (15*n^2 + 7*n - 48)*a(n-2) - 2*(n+2)*(2*n-3)*a(n-3). - Vaclav Kotesovec, Oct 17 2012

a(n) ~ 2^(2*n+1)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 17 2012

a(n) = [x^n] 1/((1-x-x^2) * (1-x)^n). - Seiichi Manyama, Apr 05 2024