A110170 First differences of the central Delannoy numbers (A001850).
1, 2, 10, 50, 258, 1362, 7306, 39650, 217090, 1196834, 6634890, 36949266, 206549250, 1158337650, 6513914634, 36718533570, 207412854786, 1173779487810, 6653482333450, 37770112857074, 214694383882498, 1221832400430482, 6961037946938250, 39697830840765090, 226596964146630658
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Thomas Selig, Combinatorial aspects of sandpile models on wheel and fan graphs, arXiv:2202.06487 [math.CO], 2022.
- Robert A. Sulanke, Objects Counted by the Central Delannoy Numbers, Journal of Integer Sequences, Volume 6, 2003, Article 03.1.5.
Programs
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Haskell
a110170 0 = 1 a110170 n = a128966 (2 * n) n -- Reinhard Zumkeller, Jul 20 2013
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Maple
with(orthopoly): a:=proc(n) if n=0 then 1 else P(n,3)-P(n-1,3) fi end: seq(a(n),n=0..25); a := n -> `if`(n=0, 1, 2*hypergeom([1 - n, -n], [1], 2)): seq(simplify(a(n)), n=0..24); # Peter Luschny, May 22 2017
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Mathematica
CoefficientList[Series[(1 - x)/Sqrt[1 - 6 * x + x^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 18 2012 *)
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PARI
x='x+O('x^66); Vec((1-x)/sqrt(1-6*x+x^2)) \\ Joerg Arndt, May 16 2013
Formula
G.f.: (1-z)/sqrt(1-6*z+z^2).
a(n) = P_n(3) - P_{n-1}(3) (n >= 1), where P_j is j-th Legendre polynomial.
From Paul Barry, Oct 18 2009: (Start)
G.f.: (1-x)/(1-x-2x/(1-x-x/(1-x-x/(1-x-x/(1-... (continued fraction);
G.f.: 1/(1-2x/((1-x)^2-x/(1-x/((1-x)^2-x/(1-x/((1-x)^2-x/(1-... (continued fraction);
a(n) = Sum_{k = 0..n} (0^(n + k) + C(n + k - 1, 2k - 1)) * C(2k, k) = 0^n + Sum_{k = 0..n} C(n + k - 1, 2k - 1) * C(2k, k). (End)
D-finite with recurrence: n*(2*n-3)*a(n) = 2*(6*n^2-12*n+5)*a(n-1) - (n-2)*(2*n-1)*a(n-2). - Vaclav Kotesovec, Oct 18 2012
a(n) ~ 2^(-1/4)*(3+2*sqrt(2))^n/sqrt(Pi*n). - Vaclav Kotesovec, Oct 18 2012
a(n) = A277919(2n,n). - John P. McSorley, Nov 23 2016
a(n) = 2*hypergeom([1 - n, -n], [1], 2) for n>0. - Peter Luschny, May 22 2017
D-finite with recurrence: n*a(n) +(-7*n+5)*a(n-1) +(7*n-16)*a(n-2) +(-n+3)*a(n-3)=0. - R. J. Mathar, Jan 15 2020
a(0) = 1; a(n) = (2/n) * Sum_{k=0..n-1} (n^2-k^2) * a(k). - Seiichi Manyama, Mar 28 2023
G.f.: Sum_{n >= 0} binomial(2*n, n)*x^n/(1 - x)^(2*n) = 1 + 2*x + 10*x^2 + 50*x^3 + .... - Peter Bala, Apr 17 2024
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