reviewed

approved

proposed

reviewed

editing

proposed

The number of pairs of independent outcomes when rolling an n-sided die. Or in other terms, words, the number of pairs of proper subsets A,B of a set S, such that #A/#S * #B/#S = #(A \intersect B)/#S.

approved

editing

editing

approved

Sum[Total[s!/(c!(#-c)!(s c/#-c)!(s - # - s c/# + c)!) &/@Select[Divisors[s c], c <= # <= s &]], {c, 1, s}] - <a href="http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_cs; action=display; num=1234635667#3(* ">Eigenray</a>, ", see link, Feb 15th 2009 *)

approved

editing

The number of pairs of independent outcomes when rolling an n-sided die. Or in other terms, the number of pairs of proper subsets A,B of a set S, such that #A/#S * #B/#S = #(A \intersect B)/#S.

1, 5, 13, 53, 61, 845, 253, 7509, 16141, 128045, 4093, 1785965, 16381, 23576285, 55921333, 274696789, 262141, 5338300157, 1048573, 63028146573, 117924207421, 995274180125, 16777213, 15265519672173, 14283159085861

1,2

<a href="http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_cs;action=display;num=1234635667">wwu riddle forum thread on the problem</a>

For N=4 we have 53 solutions, because {1,2,3,4} together with any proper subset yields 2*15-1 = 29 valid pairs, and a further 24 pairs can be obtained from {1,2} & {1,3}, by substituting the numbers with any permutation of (1,2,3,4).

Sum[Total[s!/(c!(#-c)!(s c/#-c)!(s - # - s c/# + c)!) &/@Select[Divisors[s c], c <= # <= s &]], {c, 1, s}] - <a href="http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_cs; action=display; num=1234635667#3">Eigenray</a>, Feb 15th 2009

nonn

Harmen Wassenaar (towr(AT)ai.rug.nl), Mar 16 2009

approved