A026010 a(n) = number of (s(0), s(1), ..., s(n)) such that s(i) is a nonnegative integer and |s(i) - s(i-1)| = 1 for i = 1,2,...,n and s(0) = 2. Also a(n) = sum of numbers in row n+1 of array T defined in A026009.
1, 2, 4, 7, 14, 25, 50, 91, 182, 336, 672, 1254, 2508, 4719, 9438, 17875, 35750, 68068, 136136, 260338, 520676, 999362, 1998724, 3848222, 7696444, 14858000, 29716000, 57500460, 115000920, 222981435, 445962870, 866262915, 1732525830, 3370764540
Offset: 0
Keywords
Examples
The a(3) = 7 compositions of 5 in which the even parts appear as often at even positions as at odd positions are (5), (311), (131), (113), (221), (122), (11111). Missing are (41), (14), (32), (23), (212), (2111), (1211), (1121), (1112). - _Gus Wiseman_, Mar 17 2018
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..3326
- Christian Krattenthaler, Daniel Yaqubi, Some determinants of path generating functions, II, arXiv:1802.05990 [math.CO], 2018; Adv. Appl. Math. 101 (2018), 232-265.
Crossrefs
Programs
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Magma
[(&+[Binomial(Floor((n+k)/2), Floor(k/2)): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Nov 08 2018
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Mathematica
Array[Sum[Binomial[Floor[(# + k)/2], Floor[k/2]], {k, 0, #}] &, 34, 0] (* Michael De Vlieger, May 16 2018 *) Table[2^(-1 + n)*(((2 + 3*#)*Gamma[(1 + #)/2])/(Sqrt[Pi]*Gamma[2 + #/2]) &[n + Mod[n, 2]]), {n,0,40}] (* Peter Pein, Nov 08 2018 *) Table[(1/2)^((5 - (-1)^n)/2)*(6*n + 7 - 3*(-1)^n)*CatalanNumber[(2*n + 1 - (-1)^n)/4], {n, 0, 40}] (* G. C. Greubel, Nov 08 2018 *)
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PARI
vector(40, n, n--; sum(k=0,n, binomial(floor((n+k)/2), floor(k/2)))) \\ G. C. Greubel, Nov 08 2018
Formula
a(2*n) = ((3*n + 1)/(2*n + 1))*C(2*n + 1, n)= A051924(1+n), n>=0, a(2*n-1) = a(2*n)/2 = A097613(1+n), n >= 1. - Herbert Kociemba, May 08 2004
a(n) = Sum_{k=0..n} binomial(floor((n+k)/2), floor(k/2)). - Paul Barry, Jul 15 2004
Inverse binomial transform of A005774: (1, 3, 9, 26, 75, 216, ...). - Gary W. Adamson, Oct 22 2007
Conjecture: (n+3)*a(n) - 2*a(n-1) + (-5*n-3)*a(n-2) + 2*a(n-3) + 4*(n-3)*a(n-4) = 0. - R. J. Mathar, Jun 20 2013
a(n) = (1/2)^((5 - (-1)^n)/2)*(6*n + 7 - 3*(-1)^n)*Catalan((2*n + 1 - (-1)^n)/4), where Catalan is the Catalan number = A000108. - G. C. Greubel, Nov 08 2018
Comments