cp's OEIS Frontend

A300305 Expected rounded number of draws until two persons simultaneously drawing cards with replacement from two decks of cards with m and n cards, respectively, both obtain complete collections, written as triangle T(m,n), 1 <= n <= m.

%I A300305 #20 Mar 09 2018 06:28:22
%S A300305 1,3,4,6,6,7,8,8,9,10,11,11,12,12,14,15,15,15,15,16,18,18,18,18,18,19,
%T A300305 20,22,22,22,22,22,22,23,24,26,25,25,25,26,26,26,27,29,31,29,29,29,29,
%U A300305 29,30,31,32,33,35,33,33,33,33,33,34,34,35,36,38,40,37,37,37,37,37,37,38,38,39,41,42,45
%N A300305 Expected rounded number of draws until two persons simultaneously drawing cards with replacement from two decks of cards with m and n cards, respectively, both obtain complete collections, written as triangle T(m,n), 1 <= n <= m.
%C A300305 This is the two-person version of the coupon collector's problem.
%H A300305 IBM Research, <a href="https://www.research.ibm.com/haifa/ponderthis/solutions/February2018.html">Ponder This Challenge February 2018</a>. Solution for 3 persons from Robert Lang.
%F A300305 T(m,n) = round(1 - Sum_{j=0..m} Sum_{k=0..n} ( (-1)^(m-j+n-k) * binomial(m,j) * binomial(n,k) * j * k / (m*n-j*k) )) excluding term with j=m and k=n in summation.
%e A300305 T(1,1)=1, T(2,1)=3, T(2,2)=round(11/3)=4, T(3,1)=round(11/2)=6, T(3,2)=round(57/10)=6, T(3,3)=round(1909/280)=7.
%e A300305 The triangle starts:
%e A300305    1
%e A300305    3   4
%e A300305    6   6   7
%e A300305    8   8   9  10
%e A300305   11  11  12  12  14
%e A300305   15  15  15  15  16  18
%e A300305   18  18  18  18  19  20  22
%e A300305   22  22  22  22  22  23  24  26
%e A300305   25  25  25  26  26  26  27  29  31
%e A300305   ...
%Y A300305 Cf. A073593, A090582, A135736 (first column in triangle), A300306 (diagonal in triangle).
%K A300305 nonn,tabl
%O A300305 1,2
%A A300305 _Hugo Pfoertner_, Mar 07 2018