A123164 Row sums of A123160.
1, 2, 8, 38, 192, 1002, 5336, 28814, 157184, 864146, 4780008, 26572086, 148321344, 830764794, 4666890936, 26283115038, 148348809216, 838944980514, 4752575891144, 26964373486406, 153196621856192, 871460014012682, 4962895187697048, 28292329581548718, 161439727075246592
Offset: 0
Keywords
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..1000
- A. Laradji and A. Umar, A. Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
- A. Laradji and A. Umar, Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), 04.3.8
- A. Laradji and A. Umar, Asymptotic results for semigroups of order-preserving partial transformations Comm. Algebra 34 (2006), 1071-1075. [From _Abdullahi Umar_, Oct 11 2008]
- Huyile Liang, Yanni Pei, and Yi Wang, Analytic combinatorics of coordination numbers of cubic lattices, arXiv:2302.11856 [math.CO], 2023. See p. 4.
Programs
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Magma
[1] cat [n le 2 select 2*4^(n-1) else (4*(3*(n-1)^2-1)*Self(n-1) - (2*n-1)*(n-2)*Self(n-2))/((2*n-3)*(n)): n in [1..30]]; // G. C. Greubel, Jul 19 2023
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Mathematica
a[n_]:= a[n]= Sum[Binomial[n+k-1,k]*Binomial[n,k], {k,0,n}]; Table[a[n], {n,0,30}]
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SageMath
def A123164(n): return sum(binomial(n,j)*binomial(n+j-1,j) for j in range(n+1)) [A123164(n) for n in range(31)] # G. C. Greubel, Jul 19 2023
Formula
a(n) = A122542(2*n,n). - Philippe Deléham, May 28 2007
a(n) = Sum_{k=0..n} C(n, k)*C(n+k-1, k). - Paul Barry, Aug 22 2007
(2*n-1)*(n+1)*a(n+1) = 4*(3*n^2-1)*a(n) - (2*n+1)*(n-1)*a(n-1) for n >= 1 with a(0) = 1 and a(1) = 2. - Abdullahi Umar, Aug 25 2008
a(n) = Jacobi_P(n, 0, -1, 3). - Paul Barry, Sep 27 2009
G.f.: (1 + x + sqrt(1 - 6*x + x^2))/(2*sqrt(1 - 6*x + x^2)). - Paul Barry, Sep 29 2010
From Abdullahi Umar, Oct 11 2008: (Start)
a(n+1) - a(n) = (2*n + 1)*A006318 (n >= 0);
a(n) = Hypergeometric2F1([-n, n], [1], -1). - Peter Luschny, Aug 02 2014
a(n) ~ (1 + sqrt(2))^(2*n) / (2^(3/4) * sqrt(Pi*n)). - Vaclav Kotesovec, Feb 14 2021
From Peter Bala, Oct 07 2021: (Start)
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k*C(n, k)*C(3*n-2*k-1, n-2*k).
a(p) == 2 (mod p^3) for prime p >= 5.
Conjecture: a(n*p^k) == a(n*p^(k-1)) mod( p^(3*k) ) for prime p >= 5 and all positive integers n and k. (End)
Extensions
Edited by N. J. A. Sloane, Oct 04 2006
Offset changed (a(0)=1) by Michael Somos, Feb 07 2011
Comments