A026378 a(n) = number of integer strings s(0),...,s(n) counted by array T in A026374 that have s(n)=1; also a(n) = T(2n-1,n-1).
1, 4, 17, 75, 339, 1558, 7247, 34016, 160795, 764388, 3650571, 17501619, 84179877, 406020930, 1963073865, 9511333155, 46169418195, 224484046660, 1093097083475, 5329784874185, 26018549129545, 127154354598330, 622031993807565
Offset: 1
Keywords
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Shu-Chiuan Chang, Robert Shrock, Structure of the Partition Function and Transfer Matrices for the Potts Model in a Magnetic Field on Lattice Strips, J. Stat. Physics 137 (2009) 667, table 5.
- D. E. Davenport, L. W. Shapiro and L. C. Woodson, The Double Riordan Group, The Electronic Journal of Combinatorics, 18(2) (2012), #P33. - From _N. J. A. Sloane_, May 11 2012
- Isaac DeJager, Madeleine Naquin, Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
- E. Deutsch, E. Munarini, S. Rinaldi, Skew Dyck paths, J. Stat. Plann. Infer. 140 (8) (2010) 2191-2203.
- Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
- J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.
- Toufik Mansour, Jose Luis Ramirez, Enumration of Fuss-skew paths, Ann. Math. Inform. 55 (2022) 125-136, table 2, l=1.
- László Németh, Tetrahedron trinomial coefficient transform, arXiv:1905.13475 [math.CO], 2019.
Crossrefs
Programs
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Maple
a := n -> (-1)^n*simplify(GegenbauerC(n-2,-n+1,3/2) - GegenbauerC(n-1,-n+1,3/2)): seq(a(n), n=1..23); # Peter Luschny, May 13 2016
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Mathematica
CoefficientList[Series[(1/2)/(5*x^2-x)*(1-5*x-(1-6*x+5*x^2)^(1/2)),{x,0,30}],x] (* Vincenzo Librandi, May 13 2012 *) Table[Hypergeometric2F1[3/2, 1-n, 2, -4], {n, 1, 20}] (* Vladimir Reshetnikov, Apr 25 2016 *)
Formula
G.f.: (1/2)/(5*x^2-x)*(1-5*x-(1-6*x+5*x^2)^(1/2)). E.g.f.: exp(3*x)*(BesselI(0, 2*x)+BesselI(1, 2*x)). - Vladeta Jovovic, Oct 03 2003
G.f.: [(1-z)/sqrt(1-6z+5z^2)-1]/2 = z + 4z^2 + 17z^3 + ... - Emeric Deutsch, Jan 22 2004
a(n) = coefficient of t^n in (1+t)(1+3t+t^2)^(n-1). - Emeric Deutsch, Jan 30 2004
a(n) = A026380(2n-2). - Emeric Deutsch, Feb 18 2004
a(n) = [2(3n-2)a(n-1) - 5(n-2)a(n-2)]/n for n>=2; a(0)=0, a(1)=1. - Emeric Deutsch, Mar 18 2004
a(n+1) = sum(k=0, n, binomial(n, k)*sum(i=0, k, binomial(k+i, i))). - Benoit Cloitre, Aug 06 2004
a(n+1) = sum(k=0, n, binomial(n, k)*binomial(2*k+1, k+1)). - Benoit Cloitre, Aug 06 2004
a(n) = Sum(k*A126182(n-1,k-1),k=1..n). - Emeric Deutsch, Jul 25 2007
From Paul Barry, Jan 13 2009: (Start)
G.f.: (1/(1-5x))*c(-x/(1-5x)), c(x) the g.f. of A000108;
a(n) = sum{k=0..n, C(n,k)*(-1)^k*A000108(k)*5^(n-k)} (offset 0). (End)
G.f. 1/(1 - 3x - x(1 - x)/(1 - x - x(1 - x)/(1 - x - x(1 - x)/(1 - x - x(1 - x)/(1...(continued fraction). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Jul 02 2010
a(n) ~ 5^(n-1/2)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 08 2012
a(n) = hypergeom([3/2, 1-n], [2], -4). - Vladimir Reshetnikov, Apr 25 2016
a(n) = (-1)^n*(GegenbauerC(n-2,-n+1,3/2) - GegenbauerC(n-1,-n+1,3/2)). - Peter Luschny, May 13 2016
Comments