A063886 Number of n-step walks on a line starting from the origin but not returning to it.
1, 2, 2, 4, 6, 12, 20, 40, 70, 140, 252, 504, 924, 1848, 3432, 6864, 12870, 25740, 48620, 97240, 184756, 369512, 705432, 1410864, 2704156, 5408312, 10400600, 20801200, 40116600, 80233200, 155117520, 310235040, 601080390, 1202160780, 2333606220, 4667212440
Offset: 0
Examples
a(4) = 6 because there are six length four walks that do not return to the origin: {-1, -2, -3, -4}, {-1, -2, -3, -2}, {-1, -2, -1, -2}, {1, 2, 1, 2}, {1, 2, 3, 2}, {1, 2, 3, 4}. There are also six such walks that return exactly one time: {-1, -2, -1, 0}, {-1, 0, -1, -2}, {-1, 0, 1, 2}, {1, 0, -1, -2}, {1, 0, 1, 2}, {1, 2, 1, 0}. - _Geoffrey Critzer_, Jan 24 2010 The a(5) = 12 subsets in which the even elements appear as often at even positions as at odd positions: {}, {1}, {3}, {5}, {1,3}, {1,5}, {2,4}, {3,5}, {1,2,4}, {1,3,5}, {2,4,5}, {1,2,4,5}. - _Gus Wiseman_, Mar 17 2018
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- Paul Barry, A Note on Riordan Arrays with Catalan Halves, arXiv:1912.01124 [math.CO], 2019.
- Emeric Deutsch, Problem 11424, The American Mathematical Monthly, Vol. 116, No. 3 (March 2009), p. 277.
- D. Perrin, A conjecture on rational sequences, pp. 267-274 of R. M. Capocelli, ed., Sequences, Springer-Verlag, NY 1990.
Crossrefs
Apart from initial terms, same as A182027.
Cf. A000108, A000246, A000712, A000984, A001405, A001700, A002420, A006232, A007877, A026010, A028329, A045931, A047073, A089849, A097613, A106180, A120617, A130777, A130780, A138364, A164584, A171966, A239241, A300787, A300788, A300789.
Cf. A307768 (complementary event).
Programs
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Magma
[1] cat [2*Binomial(n-1, Floor((n-1)/2)): n in [1..40]]; // G. C. Greubel, Jun 07 2023
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Maple
seq(seq(binomial(2*j,j)*i, i=1..2),j=0..16); # Zerinvary Lajos, Apr 28 2007 # second Maple program: a:= proc(n) option remember; `if`(n<2, n+1, 4*a(n-2) +2*(a(n-1) -4*a(n-2))/n) end: seq(a(n), n=0..40); # Alois P. Heinz, Feb 10 2014 # third program: A063886 := series(BesselI(0, 2*x)*(1 + x*2 + x*Pi*StruveL(1, 2*x)) - Pi*x*BesselI(1, 2*x)*StruveL(0, 2*x), x = 0, 34): seq(n!*coeff(A063886, x, n), n = 0 .. 33); # Mélika Tebni, Jun 17 2024
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Mathematica
Table[Length[Select[Map[Accumulate, Strings[{-1, 1}, n]], Count[ #, 0] == 0 &]], {n, 0, 20}] (* Geoffrey Critzer, Jan 24 2010 *) CoefficientList[Series[Sqrt[(1+2x)/(1-2x)],{x,0,40}],x] (* Harvey P. Dale, Apr 28 2016 *)
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PARI
a(n)=(n==0)+2*binomial(n-1,(n-1)\2)
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PARI
a(n) = 2^n*prod(k=0,n-1,(k/n+1/n)^((-1)^k)); \\ Michel Marcus, Dec 03 2013
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Python
from math import ceil from sympy import binomial def a(n): if n==0: return 1 return 2*binomial(n-1,(n-1)//2) print([a(n) for n in range(18)]) # David Nacin, Feb 29 2012
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SageMath
[2*binomial(n-1, (n-1)//2) + int(n==0) for n in range(41)] # G. C. Greubel, Jun 07 2023
Formula
G.f.: sqrt((1+2*x)/(1-2*x)).
a(n+1) = 2*C(n, floor(n/2)) = 2*A001405(n); a(2n) = C(2n, n) = A000984(n) = 4*a(2n-2)-|A002420(n)| = 4*a(2n-2)-2*A000108(n-1) = 2*A001700(n-1); a(2n+1) = 2*a(2n) = A028329(n).
2*a(n) = A047073(n+1).
a(n) = Sum_{k=0..n} abs(A106180(n,k)). - Philippe Deléham, Oct 06 2006
a(n) = Sum_{k=0..n} (k+1)binomial(n, (n-k)/2) ( 1-cos((k+1)*Pi/2) (1+(-1)^(n-k))/(n+k+2) ). - Paul Barry, Oct 12 2004
G.f.: 1/(1-2*x/(1+x/(1+x/(1-x/(1-x/(1+x/(1+x/(1-x/(1-x/(1+ ... (continued fraction). - Paul Barry, Aug 10 2009
G.f.: 1 + 2*x/(G(0)-x+x^2) where G(k)= 1 - 2*x^2 - x^4/G(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Aug 10 2012
D-finite with recurrence: n*a(n) = 2*a(n-1) + 4*(n-2)*a(n-2). - R. J. Mathar, Dec 03 2012
From Sergei N. Gladkovskii, Jul 26 2013: (Start)
G.f.: 1/G(0), where G(k) = 1 - 2*x/(1 + 2*x/(1 + 1/G(k+1) )); (continued fraction).
G.f.: G(0), where G(k) = 1 + 2*x/(1 - 2*x/(1 + 1/G(k+1) )); (continued fraction).
G.f.: W(0)/2*(1+2*x), where W(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)/(x*(2*k+1))/W(k+1) )), abs(x) < 1/2; (continued fraction). (End)
a(n) = 2^n*Product_{k=0..n-1} (k/n + 1/n)^((-1)^k). - Peter Luschny, Dec 02 2013
G.f.: G(0), where G(k) = 1 + 2*x*(4*k+1)/((2*k+1)*(1+2*x) - (2*k+1)*(4*k+3)*x*(1+2*x)/((4*k+3)*x + (k+1)*(1+2*x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 19 2014
From Peter Bala, Mar 29 2024: (Start)
a(n) = 2^n * Sum_{k = 0..n} (-1)^(n+k)*binomial(1/2, k)*binomial(- 1/2, n-k) = 2^n * A000246(n)/n!.
a(n) = (1/2^n) * binomial(2*n, n) * hypergeom([-1/2, -n], [1/2 - n], -1). (End)
E.g.f.: BesselI(0, 2*x)*(1 + x*(2 + Pi)*StruveL(1, 2*x)) - Pi*x*BesselI(1, 2*x)*StruveL(0, 2*x). - Stefano Spezia, May 11 2024
From Amiram Eldar, Aug 15 2025: (Start)
Sum_{n>=0} 1/a(n) = Pi/(3*sqrt(3)) + 2.
Sum_{n>=0} (-1)^n/a(n) = 2/3 + Pi/(9*sqrt(3)). (End)
Comments