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%I #52 Dec 16 2025 13:49:12
%S 1,0,0,0,4,24,168,1280,10860,101976,1053136,11881152,145510740,
%T 1923678680,27313300344,414633520704,6702860119228,114974897260440,
%U 2085904412222880,39909278145297536,803157866412577956,16960527261105495192,375011130469825988680,8664636644578485432960
%N Number of derangements of a set of n elements where 2 specific elements cannot appear in each other's positions.
%C The sequence was originally generated using Python to exhaustively enumerate permutations describing the secret Santa setup for n people where 2 specific people cannot receive each other's names. Derangements, A000166, describe a restriction-free secret Santa setup and are related to this sequence.
%C If a derangement is not included, then both of the two friends must be in the same permutation cycle and must be adjacent in this cycle. The second friend can be either immediately before or immediately after the first giving two possibilities (except when the cycle contains only the two friends). These considerations lead to a formula for a(n). - _Andrew Howroyd_, Jan 07 2020
%C There are three distinct types of derangement that must be excluded, (1) friend 1 and friend 2 receive each other's names, (2) friend 2 receives friend 1's name but friend 1 does not receive friend 2's name, and (3) friend 1 receives friend 2's name but friend 2 does not receive friend 1's name. - _William P. Orrick_, Jul 25 2020
%H Andrew Howroyd, <a href="/A331007/b331007.txt">Table of n, a(n) for n = 0..500</a>
%H J. Touchard, <a href="https://gallica.bnf.fr/ark:/12148/bpt6k31506/f631.item.zoom">Sur un problème de permutations</a>, Comptes Rendus Acad. Sci. Paris, 198 (1934) 631-633.
%F a(n) = A000166(n) + A000166(n-2) - 2*Sum_{k=0, n-2} binomial(n-2,k)*A000166(k)*(n-k-2)! for n >= 2. - _Andrew Howroyd_, Jan 07 2020
%F a(n) = (n-3)*(n-2)*A055790(n-3) for n > 2. - _Jon E. Schoenfield_, Jan 07 2020
%F a(n) = !n - 2 * !(n-1) - !(n-2) for n >= 2, where !n = A000166(n). - _William P. Orrick_, Jul 25 2020
%F a(n) = A335391(n-2, 2) for n >= 2 (Touchard). - _William P. Orrick_, Jul 25 2020
%F D-finite with recurrence: (n-4)*a(n) = (n-2)*(n-3)*(a(n-1) + a(n-2)), a(0)=1, a(1)=0, a(2)=0, a(3)=0, a(4)=4. - _Georg Fischer_, Jun 12 2021
%e For a group of 4 friends, the number of possible permutations of their names in a secret Santa draw in which neither friend number 1 nor friend number 2 can draw the other one's name is 4. The permutations are 3412, 3421, 4312, 4321.
%e For a group of 6 friends, the number of possible permutations of their names in a secret Santa draw in which neither friend number 1 nor friend number 2 can draw the other one's name is 168.
%p f:=gfun:-rectoproc({(n-4)*a(n) = (n-2)*(n-3)*(a(n-1) + a(n-2)), a(0)=1, a(1)=0, a(2)=0, a(3)=0, a(4)=4}, a(n), remember): map(f,[$0..23]); # _Georg Fischer_, Jun 12 2021
%t f[n_] := If[n < 0, 0, Subfactorial[n]];
%t a[n_] := f[n] + f[n-2] - 2 Sum[Binomial[n-2, k]*f[k]*(n-k-2)!, {k, 0, n-2}];
%t Table[a[n], {n, 0, 23}] (* _Jean-François Alcover_, Sep 27 2022, after _Andrew Howroyd_ *)
%o (Python)
%o def permutation(n):
%o permutations = [[]]
%o for i in range(1,n + 1):
%o new_permutations = []
%o for p in permutations:
%o for j in range(0, len(p) + 1):
%o n = p.copy()
%o n.insert(j, i)
%o new_permutations.append(n)
%o permutations = new_permutations
%o return permutations
%o def check_secret_santa(permutations):
%o num_valid = 0
%o for perm in permutations:
%o valid = True
%o for i, p in enumerate(perm):
%o if i == p - 1 or (i == 0 and p == 2) or (i == 1 and p == 1):
%o valid = False
%o break
%o if valid:
%o num_valid += 1
%o return num_valid
%o (PARI) seq(n) = {my(g=exp(-x + O(x*x^n)) / (1 - x), b(k)=k!*polcoef(g,k)); vector(n+1, n, n--; if(n<=1, n==0, b(n) + b(n-2) - 2*sum(k=0, n-2, binomial(n-2, k)*b(k)*(n-k-2)!)))} \\ _Andrew Howroyd_, Jan 07 2020
%Y Cf. A000166 (number of derangements), A055790, A335391.
%K nonn
%O 0,5
%A _Anthony Susevski_, Jan 06 2020
%E Terms a(11) and beyond from _Andrew Howroyd_, Jan 07 2020