A172060 The number of returns to the origin in all possible one-dimensional walks of length 2n.
0, 2, 14, 76, 374, 1748, 7916, 35096, 153254, 661636, 2831300, 12030632, 50826684, 213707336, 894945944, 3734901296, 15540685574, 64496348516, 267060529364, 1103587381256, 4552196053844, 18747042089816, 77092267322984, 316602500019536, 1298657603761244
Offset: 0
Examples
a(2) = 14 because there are fourteen 0's in the set of all possible walks of length 4: {{-1, -2, -3, -4}, {-1, -2, -3, -2}, {-1, -2, -1, -2}, {-1, -2, -1, 0}, {-1, 0, -1, -2}, {-1, 0, -1, 0}, {-1, 0, 1, 0}, {-1, 0, 1, 2}, {1, 0, -1, -2}, {1, 0, -1, 0}, {1, 0, 1, 0}, {1, 0, 1, 2}, {1, 2, 1, 0}, {1, 2, 1, 2}, {1, 2, 3, 2}, {1, 2, 3, 4}}.
References
- W. Feller, An Introduction to Probability Theory and its Applications, Vol 1, 3rd ed. New York: Wiley, pp. 67-97, 1968.
Links
- M. Bona, Non-overlapping permutation patterns, PU. M. A. Vol. 22 (2012), 99-105. - _N. J. A. Sloane_, Oct 13 2012
Programs
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Magma
[Factorial(2*n+1)/Factorial(n)^2 - 4^n : n in [0..30]]; // Wesley Ivan Hurt, Mar 24 2015
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Magma
[0] cat [n le 1 select 2 else 4*Self(n-1)+ Binomial(2*n,n): n in [1..30]]; // Vincenzo Librandi, Mar 25 2015
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Maple
A172060:=n->(2*n+1)!/(n!)^2 - 4^n: seq(A172060(n), n=0..30); # Wesley Ivan Hurt, Mar 24 2015
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Mathematica
Table[Count[Flatten[Map[Accumulate, Strings[{-1, 1}, n]]], 0], {n, 0, 20, 2}] CoefficientList[Series[(1 - (1 - 4 x)^(1/2)) / (1 - 4 x)^(3/2), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 25 2015 *)
Formula
a(n) = (2n+1)!/(n!)^2 - 4^n.
a(n) = 4*a(n-1) + binomial(2n,n).
O.g.f.: (1-(1-4x)^(1/2))/(1-4x)^(3/2).
a(n) = 2*A000531(n). - R. J. Mathar, Jan 03 2018
Comments