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%I #27 Apr 21 2020 19:06:50
%S 0,2,6,18,44,110,252,588,1304,2934,6380,14036,30120,65260,138712,
%T 297240,627248,1332902,2796876,5904516,12333320,25899972,53897096,
%U 112693928,233776464,487034300,1007623032,2092755528,4319728784,8948009624,18432890160,38094639664
%N a(n)/2^n is the expected value of the maximum of the number of heads and the number of tails when n fair coins are tossed.
%H Alois P. Heinz, <a href="/A230137/b230137.txt">Table of n, a(n) for n = 0..1000</a>
%H Georg Braun, <a href="https://arxiv.org/abs/2001.09836">On the Growth of a Ballistic Deposition Model on Finite Graphs</a>, arXiv:2001.09836 [math.PR], 2020.
%H D. R. L. Brown, <a href="https://ia.cr/2015/375">Bounds on surmising remixed keys</a>, IACR, Report 2015/375, 2015-2016. See Table 1.
%F a(2n) = 2*Sum_{k=n+1..2n} binomial(2n,k)*k + binomial(2n,n)*n.
%F a(2n+1) = 2*Sum_{k=n+1..2n+1} binomial(2n+1,k)*k.
%F a(n) = 2*n/(n-1)*a(n-1) +4*(n-3)/(n-2)*a(n-2) -8*a(n-3) for n>2, else a(n) = n*(1+n). - _Alois P. Heinz_, Oct 10 2013
%F From _Vaclav Kotesovec_, Jul 20 2019: (Start)
%F a(2*n) = (4^n + binomial(2*n,n))*n.
%F a(2*n+1) = (4^n + binomial(2*n,n))*(2*n+1). (End)
%e a(2) = 6 because there are four possible events when 2 coins are tossed: HH, HT, TH, TT. The maximum of the number of heads and number of tails is respectively: 2 + 1 + 1 + 2 = 6.
%p a:= proc(n) option remember; `if`(n<3, n*(1+n),
%p 2*n/(n-1)*a(n-1) +4*(n-3)/(n-2)*a(n-2) -8*a(n-3))
%p end:
%p seq(a(n), n=0..40); # _Alois P. Heinz_, Oct 10 2013
%t nn=15;even=Table[n 2^(2n)+n Binomial[2n,n],{n,0,nn}];odd=Table[2Sum[ Binomial[2n+1,k]k,{k,n+1,2n+1}],{n,0,nn}];Riffle[even,odd]
%K nonn
%O 0,2
%A _Geoffrey Critzer_, Oct 10 2013
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