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A108886
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.
4
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, 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, 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
OFFSET
0,5
REFERENCES
L. A. Shepp, Stochastic Processes [Course], Statistics Dept., Rutgers University, 2004.
FORMULA
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))}.
EXAMPLE
Triangle of values T(n-p,p), 0 <= p <= n, begins:
[0]
[0, 1]
[0, 1/2, 2]
[0, 0, 4/3, 3]
[0, 0, 2/3, 9/4, 4]
[0, 0, 1/5, 3/2, 16/5, 5]
[0, 0, 0, 17/20, 12/5, 25/6, 6]
[0, 0, 0, 12/35, 58/35, 10/3, 36/7, 7]
[0, 0, 0, 0, 1, 71/28, 30/7, 49/8, 8]
MAPLE
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:
MATHEMATICA
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 *)
CROSSREFS
Cf. A108885. Sequence T(m, m) is A108883/A108884.
Sequence in context: A115622 A294892 A371211 * A140886 A001492 A347049
KEYWORD
nonn,tabl,frac
AUTHOR
N. J. A. Sloane, Jul 16 2005
STATUS
approved