%I #41 Jan 10 2024 07:58:33
%S 0,0,8,48,160,400,840,1568,2688,4320,6600,9680,13728,18928,25480,
%T 33600,43520,55488,69768,86640,106400,129360,155848,186208,220800,
%U 260000,304200,353808,409248,470960,539400,615040,698368,789888,890120,999600
%N Variance of time for a random walk starting at 0 to reach one of the boundaries at +n or -n for the first time.
%H Vincenzo Librandi, <a href="/A072819/b072819.txt">Table of n, a(n) for n = 0..10000</a>
%H Milan Janjić, <a href="https://arxiv.org/abs/1905.04465">On Restricted Ternary Words and Insets</a>, arXiv:1905.04465 [math.CO], 2019.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) = n^2*(n^2 - 1)*2/3 = 4*A008911(n) = 8*A002415(n) = A069971(n, n).
%F G.f.: 8*(1 + x)*x^2/(1 - x)^5. - _Arkadiusz Wesolowski_, Feb 08 2012
%F E.g.f.: 2*exp(x)*x^2*(6 + 6*x + x^2)/3. - _Stefano Spezia_, Dec 12 2021
%F a(n) = 2*n * A007290(n+1). - _C.S. Elder_, Jan 09 2024
%e 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.
%t CoefficientList[Series[8 (1 + x) x^2/(1 - x)^5, {x, 0, 35}], x] (* _Michael De Vlieger_, Jul 02 2019 *)
%o (Magma) [n^2*(n^2-1)*2/3: n in [0..40]]; // _Vincenzo Librandi_, Sep 14 2011
%Y Cf. A000290 (i.e., n^2) for mean time. A072818(n)=sqrt(a(A001079(n))) attempts to identify the integer standard deviations.
%K nonn,easy
%O 0,3
%A _Henry Bottomley_, Jul 14 2002