|
%I #13 Nov 23 2016 12:32:52
%S 1,0,31,7682,7931709,24843464324,193342583284315,3250662144028779654,
%T 106536051676371091349113,6291424280473807580386161416,
%U 629175403160580417773688864819351
%N Number of coincidence-free length n lists of 5-tuples with all numbers 1,...,n in tuple position k, for k=1..5.
%C a(n) enumerates (ordered) lists of n 5-tuples such that every number from 1 to n appears once at each of the five tuple positions and the j-th list member is not the tuple (j,j,j,j,j), for every j=1,..,n. Called coincidence-free 5-tuple lists of length n. See the Charalambides reference for this combinatorial interpretation.
%D Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 187, Exercise 13.(a), for r=5.
%H G. C. Greubel, <a href="/A135811/b135811.txt">Table of n, a(n) for n = 0..100</a>
%F a(n) = Sum_{j=0..n} ( ((-1)^(n-j))*binomial(n,j)*(j!)^5 ). See the Charalambides reference a(n)=B_{n,5}.
%e 5-tuple combinatorics: a(1)=0 because the only list of 5-tuples composed of 1 is [(1,1,1,1,1)] and this is a coincidence for j=1.
%e 5-tuple combinatorics: from the 2^5 possible 5-tuples of numbers 1 and 2 all except (1,1,1,1,1) appear as first members of the length 2 lists. The second members are the 5-tuples obtained by interchanging 1 and 2 in the first member. E.g. one of the a(2)=2^5-1 =31 lists is [(1,1,1,1,2),(2,2,2,2,1)]. The list [(1,1,1,1,1),(2,2,2,2,2) does not qualify because it has in fact two coincidences, those for j=1 and j=2.
%t Table[Sum[(-1)^(n - k)*Binomial[n, k]*(k!)^5, {k, 0, n}], {n,0,25}] (* _G. C. Greubel_, Nov 23 2016 *)
%Y Cf. A135810 (coincidence-free 4-tuples). A135812 (coincidence-free 6-tuples).
%K nonn,easy
%O 0,3
%A _Wolfdieter Lang_, Jan 21 2008
|
