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%I #31 May 02 2021 21:44:33
%S 1,2,8,36,200,1300,9720,82180,775520,8082180,92205800,1143084580,
%T 15302486160,220019440420,3381685263320,55333244924100,
%U 960361672886720,17622501030879940,340893902373527880,6933456765092580580,147919915357498809200,3303011756746128625380
%N Numbers p = a(n) such that p divided by (n-1)! is equal to the average number of elements of partition sets of n elements excluding sets with a singleton.
%C a(n)/(n-1)! 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.
%C For n>2, a(n) appears to be divisible by 2n-4 and, for n>5, additionally by either 10 or 30. - _Ralf Stephan_, Dec 17 2013
%H Martin Y. Champel, <a href="/A233744/b233744.txt">Table of n, a(n) for n = 2..99</a>
%F a(n) = (n-1)! * S(n), with S(n) = 1 + 1/(n-1) * sum of previous S(i) with i in (2, n-2).
%e 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.
%e For n = 2, the only possible loop is a 2 elements loop AB ==> a(2) = 1.
%e for n = 3, the only possible loop is a 3 elements loop ABC or ACB ==> a(3) = 2 as a(3) / 2! = 1.
%e 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!.
%e 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!.
%t S[1] = 1;
%t S[n_] := S[n] = 1 + 1/(n-1) (S /@ Range[2, n-2] // Total);
%t a[n_] := (n-1)! S[n];
%t a /@ Range[2, 99] (* _Jean-François Alcover_, Sep 19 2020 *)
%o (Python)
%o from sympy import factorial, Integer
%o angel=[0,0,1,1]
%o A233744=[1,2]
%o n = 20
%o for i in range(4,n):
%o new = 1+sum(angel[:-1])/Integer(i-1)
%o angel.append(new)
%o A233744.append(new*factorial(i-1))
%o print(A233744)
%o (PARI) S(n)=if(n<2,1,1+sum(i=2,n-2,S(i))/(n-1)); a(n)=(n-1)!*S(n) \\ _Ralf Stephan_, Dec 17 2013
%Y Cf. A000041, A000142, A000774, A002865, A145574.
%K nonn,easy
%O 2,2
%A _Martin Y. Champel_, Dec 15 2013
%E More terms from _Ralf Stephan_, Dec 17 2013
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