

A233744


Numbers p = a(n) such that p divided by (n1)! is equal to the average number of elements of partition sets of n elements excluding sets with a singleton.


2



1, 2, 8, 36, 200, 1300, 9720, 82180, 775520, 8082180, 92205800, 1143084580, 15302486160, 220019440420, 3381685263320, 55333244924100, 960361672886720, 17622501030879940, 340893902373527880, 6933456765092580580, 147919915357498809200, 3303011756746128625380
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OFFSET

2,2


COMMENTS

a(n)/(n1)! is the average number of loops when n players are randomly taking the name of another player (excluding their own name). This is just referring to a game called the "Guardian angel" proposed by a former colleague at work.
For n>2, a(n) appears to be divisible by 2n4 and, for n>5, additionally by either 10 or 30.  Ralf Stephan, Dec 17 2013


LINKS



FORMULA

a(n) = (n1)! * S(n), with S(n) = 1 + 1/(n1) * sum of previous S(i) with i in (2, n2).


EXAMPLE

For example, if, among 5 players A,B,C,D,E, A takes C and C takes B and B takes A, one loop would be ACB and the other loop would be DE. No single element loop is allowed.
For n = 2, the only possible loop is a 2 elements loop AB ==> a(2) = 1.
for n = 3, the only possible loop is a 3 elements loop ABC or ACB ==> a(3) = 2 as a(3) / 2! = 1.
for n = 4, there are two types of loop, the ABCD loop and AB + CD loops, there is 2 chances out of 3 to get the ABCD type loop and 1 chance out of 3 to get the "AB + CD" configuration. The average number of loop is therefore 2/3 X 1 + 1/3 X 2 = 4/3 = 8 / 6 = a(4)/3!.
for n = 5, there are two types of loop, the ABCDE loop and ABC + DE loops, there is 2 chances out of 4 to get the ABCDE type loop and 2 chance out of 4 to get the "ABC + DE" configuration. The average number of loop is therefore 2/4 X 1 + 2/4 X 2 = 6/4 = 36 / 24 = a(5)/4!.


MATHEMATICA

S[1] = 1;
S[n_] := S[n] = 1 + 1/(n1) (S /@ Range[2, n2] // Total);
a[n_] := (n1)! S[n];


PROG

(Python)
from sympy import factorial, Integer
angel=[0, 0, 1, 1]
n = 20
for i in range(4, n):
new = 1+sum(angel[:1])/Integer(i1)
angel.append(new)
(PARI) S(n)=if(n<2, 1, 1+sum(i=2, n2, S(i))/(n1)); a(n)=(n1)!*S(n) \\ Ralf Stephan, Dec 17 2013


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



