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A102262 Numerators of probabilities in gift exchange problem with n people.

%I A102262 #30 Dec 11 2021 02:34:50
%S A102262 0,1,5,19,203,4343,63853,58129,160127,8885501,1500518539,404156337271,
%T A102262 16040576541971,1694200740145637,24047240650458731,22823917472900053,
%U A102262 2511014355032164231,143734030512459889193,49611557898193759558813,950601970122346247310883
%N A102262 Numerators of probabilities in gift exchange problem with n people.
%C A102262 This is a version of the Secret Santa game.
%C A102262 n friends organize a gift exchange. The n names are put into a hat and the first person draws one. If she picks her own name, then she returns it to the bag and draws again, repeating until she has a name that is not her own. Then the second person draws, again returning his own name if it is drawn. This continues down the line. What is the probability p(n) that when the n-th person draws, only her own name will be left in the bag?
%C A102262 I heard about the problem from Gary Thompson at Grove City College in PA.
%H A102262 Jon E. Schoenfield, <a href="/A102262/b102262.txt">Table of n, a(n) for n = 2..389</a>
%H A102262 Math Forum at Drexel, <a href="http://mathforum.org/kb/message.jspa?messageID=6667330&amp;tstart=0">A variant on the "Secret Santa"</a>
%F A102262 From _Jon E. Schoenfield_, Sep 30 2006: (Start)
%F A102262 p(n) = Sum_{i=1..n-2} t(n,i)/(n-1)!^2
%F A102262 where
%F A102262   t(n,i) = (n-2)*i^2/(i-1)*t(n-1,i-1) - (n-i-2)*t(n-1,i) for 1 < i < n-1;
%F A102262   t(n,1) = (-1)^(n-1)*(n-1)!/2 for i = 1 and n > 2;
%F A102262   t(n,i) = 0 otherwise.
%F A102262 (End)
%F A102262 Based on the values of p(n) for n <= 1000, it seems plausible that, as n increases, p(n) approaches 1/(n + log(n) + EulerGamma), where EulerGamma = 0.5772156649015... (the Euler-Mascheroni constant). - _Jon E. Schoenfield_, Dec 11 2021
%e A102262 p(2) through p(10) are 0, 1/4, 5/36, 19/144, 203/1800, 4343/43200, 63853/705600, 58129/705600, 160127/2116800.
%o A102262 (Magma) N:=21; a:=[]; row:=[]; T:=[]; for n in [2..N] do row[n-1]:=0; T[n]:=row; T[n][1]:=(-1)^(n-1)*Factorial(n-1) div 2; for i in [2..n-2] do T[n][i]:=(n-2)*i^2/(i-1)*T[n-1][i-1]-(n-i-2)*T[n-1][i]; end for; p:=0; for i in [1..n-2] do p+:=T[n][i]/Factorial(n-1)^2; end for; a[#a+1]:=Numerator(p); end for; a; // _Jon E. Schoenfield_, Dec 10 2021
%Y A102262 Cf. A102263, A136300.
%K A102262 nonn,frac
%O A102262 2,3
%A A102262 _Jerrold Grossman_, Feb 17 2005
%E A102262 More terms from _Jon E. Schoenfield_, Sep 30 2006