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%I #10 Feb 27 2020 22:15:50
%S 1,1,1,1,2,1,1,1,3,1,1,1,3,4,1,1,1,5,2,5,1,1,1,1,20,5,6,1,1,1,1,35,35,
%T 3,7,1,1,1,1,1,1,28,7,8,1,1,1,1,1,9,63,42,4,9,1,1,1,1,1,15,42,35,120,
%U 9,10,1,1,1,1,1,1,231,11,66,55,5,11,1,1,1,1,1,1,396,231,72,45,55,11,12,1,1,1
%N Let T(m,p) be the value of the following game: there are m "minus" balls and p "plus" balls in an urn, for a total of n=m+p balls. You may draw balls from the urn one at a time at random and without replacement until you decide to stop drawing. Each minus ball drawn costs you $1 and each plus ball drawn gets you $1. Sequence gives triangle of denominators of T(n-p,p), 0 <= p <= n, read by rows.
%D L. A. Shepp, Stochastic Processes [Course], Statistics Dept., Rutgers University, 2004.
%F T(m, 0)=0, T(0, p)=p; T(m, p) = max{0, (m/(m+p))*(-1+T(m-1, p))+(p/(m+p))*(1+T(m, p-1))}.
%e Triangle of values T(n-p,p), 0 <= p <= n, begins:
%e [0]
%e [0, 1]
%e [0, 1/2, 2]
%e [0, 0, 4/3, 3]
%e [0, 0, 2/3, 9/4, 4]
%e [0, 0, 1/5, 3/2, 16/5, 5]
%e [0, 0, 0, 17/20, 12/5, 25/6, 6]
%e [0, 0, 0, 12/35, 58/35, 10/3, 36/7, 7]
%e [0, 0, 0, 0, 1, 71/28, 30/7, 49/8, 8]
%p M:=60; for m from 0 to M do T(m,0):=0; od: for p from 0 to M do T(0,p):=p; od: for n from 1 to M do for m from 1 to n-1 do p:=n-m; t1:=(m/(m+p))*(-1+T(m-1,p))+(p/(m+p))*(1+T(m,p-1)); T(m,p):=max(0,t1); od: od:
%t M = 60; Clear[T]; For[m = 0, m <= M, m++, T[m, 0] = 0]; For[p = 0, p <= M, p++, T[0, p] = p]; For[n = 1, n <= M, n++, For[m = 1, m <= n-1, m++, p = n-m; t1 = (m/(m+p))*(-1+T[m-1, p]) + (p/(m+p))*(1+T[m, p-1]); T[m, p] = Max[0, t1]]]; Table[T[n-p, p] // Denominator, {n, 0, 13}, {p, 0, n}] // Flatten (* _Jean-François Alcover_, Jan 09 2014, translated from Maple *)
%Y Cf. A108885. Sequence T(m, m) is A108883/A108884.
%K nonn,tabl,frac
%O 0,5
%A _N. J. A. Sloane_, Jul 16 2005
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