%I
%S 0,0,1,0,1,2,0,0,4,3,0,0,2,9,4,0,0,1,3,16,5,0,0,0,17,12,25,6,0,0,0,12,
%T 58,10,36,7,0,0,0,0,1,71,30,49,8,0,0,0,0,4,113,145,21,64,9,0,0,0,0,1,
%U 47,93,527,56,81,10,0,0,0,0,0,127,21,235,294,36,100,11,0,0,0,0,0,61,284,199,202
%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 numerators of T(np,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(m1, p))+(p/(m+p))*(1+T(m, p1))}.
%e Triangle of values T(np,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 n1 do p:=nm; t1:=(m/(m+p))*(1+T(m1,p))+(p/(m+p))*(1+T(m,p1)); 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 <= n1, m++, p = nm; t1 = (m/(m+p))*(1+T[m1, p]) + (p/(m+p))*(1+T[m, p1]); T[m, p] = Max[0, t1]]]; Table[T[np, p] // Numerator, {n, 0, 12}, {p, 0, n}] // Flatten (* _JeanFrançois Alcover_, Jan 09 2014, translated from Maple *)
%Y Cf. A108886. Sequence T(m, m) is A108883/A108884.
%K nonn,tabl,frac
%O 0,6
%A _N. J. A. Sloane_, Jul 16 2005
