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

Author

Henry Bottomley, Aug 28 2001

Keywords

• nonn
• walk

Comments

A Chebyshev transform of A007877(n+1). The g.f. is transformed to (1+x)/((1-x)(1+x^2)) under the mapping G(x)->(1/(1+x^2))G(1/(1+x^2)). - Paul Barry, Oct 12 2004

a(n-1) = 2*C(n-2, floor((n-2)/2)) is also the number of bit strings of length n in which the number of 00 substrings is equal to the number of 11 substrings. For example, when n = 4 we have 4 such bit strings: 0011, 0101, 1010, and 1100. - Angel Plaza, Apr 23 2009

Hankel transform is A120617. - Paul Barry, Aug 10 2009

The Hankel transform of a(n) is (-2)^C(n+1,2). The Hankel transform of (-1)^C(n+1,2)*a(n) is (-1)^C(n+1,2)*A164584(n). - Paul Barry, Aug 17 2009

For n > 1, a(n) is also the number of n-step walks starting from the origin and returning to it exactly once. - Geoffrey Critzer, Jan 24 2010

-a(n) is the Z-sequence for the Riordan array A130777. (See the W. Lang link under A006232 for A- and Z-sequences for Riordan matrices). - Wolfdieter Lang, Jul 12 2011

Number of subsets of {1,...,n} in which the even elements appear as often at even positions as at odd positions. - Gus Wiseman, Mar 17 2018

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

Magma

[1] cat [2*Binomial(n-1, Floor((n-1)/2)): n in [1..40]]; // G. C. Greubel, Jun 07 2023

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

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 *)

PARI

a(n)=(n==0)+2*binomial(n-1,(n-1)\2)

PARI

a(n) = 2^n*prod(k=0,n-1,(k/n+1/n)^((-1)^k)); \\ Michel Marcus, Dec 03 2013

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

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

a(n) = A089849(n) + A138364(n). - Mélika Tebni, Jun 17 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)