login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A108081
a(n) = Sum_{i=0..n} binomial(2*n-i, n+i).
8
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
0,2
COMMENTS
A transform of the Fibonacci numbers A000045(n+1) under the mapping g(x)->(1/(c(x)sqrt(1-4x))g(xc(x)), c(x) the g.f. of A000108. Hankel transform is the bisection of the Fibonacci numbers F(2n+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)
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
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 *)
PROG
(PARI) 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) for(n=0, 25, print1(sum(k=0, n, binomial(n+k-1, k)*fibonacci(n-k+1)), ", ")) \\ G. C. Greubel, Jan 31 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Ralf Stephan, Jun 03 2005
STATUS
approved