A072819 Variance of time for a random walk starting at 0 to reach one of the boundaries at +n or -n for the first time.

0, 0, 8, 48, 160, 400, 840, 1568, 2688, 4320, 6600, 9680, 13728, 18928, 25480, 33600, 43520, 55488, 69768, 86640, 106400, 129360, 155848, 186208, 220800, 260000, 304200, 353808, 409248, 470960, 539400, 615040, 698368, 789888, 890120, 999600

Offset: 0

Author

Henry Bottomley, Jul 14 2002

Keywords

• nonn
• easy

Examples

			a(2)=8 since for a random walk with absorbing boundaries at +2 or -2, the probability of first reaching a boundary at time t=2 is 1/2, at t=4 is 1/4, at t=6 is 1/8, at t=8 is 1/16, etc., giving a mean of 2/2 + 4/4 + 6/8 + 8/16 + ... = 4 and a variance of 2^2/2 + 4^2/4 + 6^2/8 + 8^2/16 + ... - 4^2 = 24 - 16 = 8.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A000290 (i.e., n^2) for mean time. A072818(n)=sqrt(a(A001079(n))) attempts to identify the integer standard deviations.

Programs

Magma

[n^2*(n^2-1)*2/3: n in [0..40]]; // Vincenzo Librandi, Sep 14 2011

Mathematica

CoefficientList[Series[8 (1 + x) x^2/(1 - x)^5, {x, 0, 35}], x] (* Michael De Vlieger, Jul 02 2019 *)

Formula

a(n) = n^2*(n^2 - 1)*2/3 = 4*A008911(n) = 8*A002415(n) = A069971(n, n).

G.f.: 8*(1 + x)*x^2/(1 - x)^5. - Arkadiusz Wesolowski, Feb 08 2012

E.g.f.: 2*exp(x)*x^2*(6 + 6*x + x^2)/3. - Stefano Spezia, Dec 12 2021

a(n) = 2*n * A007290(n+1). - C.S. Elder, Jan 09 2024