Search: a047865 -id:a047865
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A000166
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Subfactorial or rencontres numbers, or derangements: number of permutations of n elements with no fixed points.
(Formerly M1937 N0766)
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+10
524
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1, 0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961, 14684570, 176214841, 2290792932, 32071101049, 481066515734, 7697064251745, 130850092279664, 2355301661033953, 44750731559645106, 895014631192902121, 18795307255050944540, 413496759611120779881, 9510425471055777937262
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OFFSET
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0,4
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COMMENTS
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Euler not only gives the first ten or so terms of the sequence, he also proves both recurrences a(n) = (n-1)*(a(n-1) + a(n-2)) and a(n) = n*a(n-1) + (-1)^n.
a(n) is the permanent of the matrix with 0 on the diagonal and 1 elsewhere. - Yuval Dekel, Nov 01 2003
a(n) is the number of desarrangements of length n. A desarrangement of length n is a permutation p of {1,2,...,n} for which the smallest of all the ascents of p (taken to be n if there are no ascents) is even. Example: a(3) = 2 because we have 213 and 312 (smallest ascents at i = 2). See the J. Désarménien link and the Bona reference (p. 118). - Emeric Deutsch, Dec 28 2007
a(n) is the number of deco polyominoes of height n and having in the last column an even number of cells. A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. - Emeric Deutsch, Dec 28 2007
Attributed to Nicholas Bernoulli in connection with a probability problem that he presented. See Problem #15, p. 494, in "History of Mathematics" by David M. Burton, 6th edition. - Mohammad K. Azarian, Feb 25 2008
a(n) is the number of permutations p of {1,2,...,n} with p(1)!=1 and having no right-to-left minima in consecutive positions. Example a(3) = 2 because we have 231 and 321. - Emeric Deutsch, Mar 12 2008
a(n) is the number of permutations p of {1,2,...,n} with p(n)! = n and having no left to right maxima in consecutive positions. Example a(3) = 2 because we have 312 and 321. - Emeric Deutsch, Mar 12 2008
Number of wedged (n-1)-spheres in the homotopy type of the Boolean complex of the complete graph K_n. - Bridget Tenner, Jun 04 2008
The only prime number in the sequence is 2. - Howard Berman (howard_berman(AT)hotmail.com), Nov 08 2008
a(n) is the number of permutations of {1,2,...,n} having exactly one small ascent. A small ascent in a permutation (p_1,p_2,...,p_n) is a position i such that p_{i+1} - p_i = 1. (Example: a(3) = 2 because we have 312 and 231; see the Charalambides reference, pp. 176-180.) [See also David, Kendall and Barton, p. 263. - N. J. A. Sloane, Apr 11 2014]
a(n) is the number of permutations of {1,2,...,n} having exactly one small descent. A small descent in a permutation (p_1,p_2,...,p_n) is a position i such that p_i - p_{i+1} = 1. (Example: a(3)=2 because we have 132 and 213.) (End)
a(n) is the sum of the values of the largest fixed points of all non-derangements of length n-1. Example: a(4)=9 because the non-derangements of length 3 are 123, 132, 213, and 321, having largest fixed points 3, 1, 3, and 2, respectively.
a(n) is the number of non-derangements of length n+1 for which the difference between the largest and smallest fixed point is 2. Example: a(3) = 2 because we have 1'43'2 and 32'14'; a(4) = 9 because we have 1'23'54, 1'43'52, 1'53'24, 52'34'1, 52'14'3, 32'54'1, 213'45', 243'15', and 413'25' (the extreme fixed points are marked).
(End)
a(n), n >= 1, is also the number of unordered necklaces with n beads, labeled differently from 1 to n, where each necklace has >= 2 beads. This produces the M2 multinomial formula involving partitions without part 1 given below. Because M2(p) counts the permutations with cycle structure given by partition p, this formula gives the number of permutations without fixed points (no 1-cycles), i.e., the derangements, hence the subfactorials with their recurrence relation and inputs. Each necklace with no beads is assumed to contribute a factor 1 in the counting, hence a(0)=1. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010). - Wolfdieter Lang, Jun 01 2010
a(n) is the number of permutations of {1,2,...,n, n+1} starting with 1 and having no successions. A succession in a permutation (p_1,p_2,...,p_n) is a position i such that p_{i+1} - p_i = 1. Example: a(3)=2 because we have 1324 and 1432.
a(n) is the number of permutations of {1,2,...,n} that do not start with 1 and have no successions. A succession in a permutation (p_1,p_2,...,p_n) is a position i such that p_{i+1} - p_i = 1. Example: a(3)=2 because we have 213 and 321.
(End)
Increasing colored 1-2 trees with choice of two colors for the rightmost branch of nonleave except on the leftmost path, there is no vertex of outdegree one on the leftmost path. - Wenjin Woan, May 23 2011
a(n) is the maximal number of totally mixed Nash equilibria in games of n players, each with 2 pure options. - Raimundas Vidunas, Jan 22 2014
The number of interior lattice points of the subpolytope of the n-dimensional permutohedron whose vertices correspond to permutations avoiding 132 and 312. - Robert Davis, Oct 05 2016
Consider n circles of different radii, where each circle is either put inside some bigger circle or contains a smaller circle inside it (no common points are allowed). Then a(n) gives the number of such combinations. - Anton Zakharov, Oct 12 2016
If we partition the permutations of [n+1] in A000240 according to their starting digit, we will get (n+1) equinumerous classes each of size a(n), i.e., A000240(n+1) = (n+1)*a(n), hence a(n) is the size of each class of permutations of [n+1] in A000240. For example, for n = 4 we have 45 = 5*9. - Enrique Navarrete, Jan 10 2017
Call d_n1 the permutations of [n] that have the substring n1 but no substring in {12,23,...,(n-1)n}. If we partition them according to their starting digit, we will get (n-1) equinumerous classes each of size A000166(n-2) (the class starting with the digit 1 is empty since we must have the substring n1). Hence d_n1 = (n-1)*A000166(n-2) and A000166(n-2) is the size of each nonempty class in d_n1. For example, d_71 = 6*44 = 264, so there are 264 permutations in d_71 distributed in 6 nonempty classes of size A000166(5) = 44. (To get permutations in d_n1 recursively from more basic ones see the link "Forbidden Patterns" below.) - Enrique Navarrete, Jan 15 2017
Also the number of maximum matchings and minimum edge covers in the n-crown graph. - Eric W. Weisstein, Jun 14 and Dec 24 2017
The sequence a(n) taken modulo a positive integer k is periodic with exact period dividing k when k is even and dividing 2*k when k is odd. This follows from the congruence a(n+k) = (-1)^k*a(n) (mod k) for all n and k, which in turn is easily proved by induction making use of the recurrence a(n) = n*a(n-1) + (-1)^n. - Peter Bala, Nov 21 2017
a(n) is the number of distinct possible solutions for a directed, no self loop containing graph (not necessarily connected) that has n vertices, and each vertex has an in- and out-degree of exactly 1. - Patrik Holopainen, Sep 18 2018
a(n) is the dimension of the kernel of the random-to-top and random-to-random shuffling operators over a collection of n objects (in a vector space of size n!), as noticed by M. Wachs and V. Reiner. See the Reiner, Saliola and Welker reference below. - Nadia Lafreniere, Jul 18 2019
a(n) is the number of distinct permutations for a Secret Santa gift exchange with n participants. - Patrik Holopainen, Dec 30 2019
a(2*n+1) is even. More generally, a(m*n+1) is divisible by m*n, which follows from a(n+1) = n*(a(n) + a(n-1)) = n*A000255(n-1) for n >= 1. a(2*n) is odd; in fact, a(2*n) == 1 (mod 8). Other divisibility properties include a(6*n) == 1 (mod 24), a(9*n+4) == a(9*n+7) == 0 (mod 9), a(10*n) == 1 (mod 40), a(11*n+5) == 0 (mod 11) and a(13*n+8 ) == 0 (mod 13). - Peter Bala, Apr 05 2022
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REFERENCES
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U. Abel, Some new identities for derangement numbers, Fib. Q., 56:4 (2018), 313-318.
M. Bona, Combinatorics of Permutations, Chapman & Hall/CRC, Boca Raton, Florida, 2004.
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 32.
R. A. Brualdi and H. J. Ryser: Combinatorial Matrix Theory, 1992, Section 7.2, p. 202.
Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 182.
Florence Nightingale David and D. E. Barton, Combinatorial Chance. Hafner, NY, 1962, p. 168.
Florence Nightingale David, Maurice George Kendall, and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 263, Table 7.5.1, row 1.
P. R. de Montmort, On the Game of Thirteen (1713), reprinted in Annotated Readings in the History of Statistics, ed. H. A. David and A. W. F. Edwards, Springer-Verlag, 2001, pp. 25-29.
J. M. de Saint-Martin, "Le problème des rencontres" in Quadrature, No. 61, pp. 14-19, 2006, EDP-Sciences Les Ulis (France).
H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 19.
Leonhart Euler, Solution quaestionis curiosae ex doctrina combinationum, Mémoires Académie sciences St. Pétersburg 3 (1809/1810), 57-64; also E738 in his Collected Works, series I, volume 7, pages 435-440.
J. M. Gandhi, On logarithmic numbers, Math. Student, 31 (1963), 73-83.
A. Hald, A History of Probability and Statistics and Their Applications Before 1750, Wiley, NY, 1990 (Chapter 19).
Irving Kaplansky, John Riordan, The problème des ménages. Scripta Math. 12 (1946), 113-124. See Eq(1).
Arnold Kaufmann, "Introduction à la combinatorique en vue des applications." Dunod, Paris, 1968. See p. 92.
Florian Kerschbaum and Orestis Terzidis, Filtering for Private Collaborative Benchmarking, in Emerging Trends in Information and Communication Security, Lecture Notes in Computer Science, Volume 3995/2006.
E. Lozansky and C. Rousseau, Winning Solutions, Springer, 1996; see p. 152.
P. A. MacMahon, Combinatory Analysis, 2 vols., Chelsea, NY, 1960, see p. 102.
M. S. Petković, "Non-attacking rooks", Famous Puzzles of Great Mathematicians, pp. 265-268, Amer. Math. Soc.(AMS), 2009.
V. Reiner, F. Saliola and V. Welker. Spectra of Symmetrized Shuffling Operators, Memoirs of the American Mathematical Society, vol. 228, Amer. Math. Soc., Providence, RI, 2014, pp. 1-121. See section VI.9.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 65.
H. J. Ryser, Combinatorial Mathematics. Mathematical Association of America, Carus Mathematical Monograph 14, 1963, p. 23.
T. Simpson, Permutations with unique fixed and reflected points. Ars Combin. 39 (1995), 97-108.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 82.
H. S. Wilf, Generatingfunctionology, Academic Press, NY, 1990, p. 147, Eq. 5.2.9 (q=1).
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LINKS
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R. K. Guy and R. J. Nowakowski, Mousetrap, Preprint, Feb 10 1993. [Annotated scanned copy]
Eric Weisstein's World of Mathematics, Edge Cover.
Eric Weisstein's World of Mathematics, Matching.
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FORMULA
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a(n) + A003048(n+1) = 2*n!. - D. G. Rogers, Aug 26 2006
a(n) = {(n-1)!/exp(1)}, n > 1, where {x} is the nearest integer function. - Simon Plouffe, March 1993 [This uses offset 1, see below for the version with offset 0. - Charles R Greathouse IV, Jan 25 2012]
a(0) = 1, a(n) = round(n!/e) = floor(n!/e + 1/2) for n > 0.
a(n) = n!*Sum_{k=0..n} (-1)^k/k!.
D-finite with recurrence a(n) = (n-1)*(a(n-1) + a(n-2)), n > 0.
a(n) = n*a(n-1) + (-1)^n.
E.g.f.: exp(-x)/(1-x).
a(n) = Sum_{k=0..n} binomial(n, k)*(-1)^(n-k)*k! = Sum_{k=0..n} (-1)^(n-k)*n!/(n-k)!. - Paul Barry, Aug 26 2004
The e.g.f. y(x) satisfies y' = x*y/(1-x).
In Maple notation, representation as n-th moment of a positive function on [-1, infinity]: a(n)= int( x^n*exp(-x-1), x=-1..infinity ), n=0, 1... . a(n) is the Hamburger moment of the function exp(-1-x)*Heaviside(x+1). - Karol A. Penson, Jan 21 2005
a(n) = Integral_{x=0..oo} (x-1)^n*exp(-x) dx. - Gerald McGarvey, Oct 14 2006
a(n) = Sum_{k=2,4,...} T(n,k), where T(n,k) = A092582(n,k) = k*n!/(k+1)! for 1 <= k < n and T(n,n)=1. - Emeric Deutsch, Feb 23 2008
a(n) = n!/e + (-1)^n*(1/(n+2 - 1/(n+3 - 2/(n+4 - 3/(n+5 - ...))))). Asymptotic result (Ramanujan): (-1)^n*(a(n) - n!/e) ~ 1/n - 2/n^2 + 5/n^3 - 15/n^4 + ..., where the sequence [1,2,5,15,...] is the sequence of Bell numbers A000110. - Peter Bala, Jul 14 2008
From William Vaughn (wvaughn(AT)cvs.rochester.edu), Apr 13 2009: (Start)
a(n) = Integral_{p=0..1} (log(1/(1-p)) - 1)^n dp.
Proof: Using the substitutions 1=log(e) and y = e(1-p) the above integral can be converted to ((-1)^n/e) Integral_{y=0..e} (log(y))^n dy.
From CRC Integral tables we find the antiderivative of (log(y))^n is (-1)^n n! Sum_{k=0..n} (-1)^k y(log(y))^k / k!.
Using the fact that e(log(e))^r = e for any r >= 0 and 0(log(0))^r = 0 for any r >= 0 the integral becomes n! * Sum_{k=0..n} (-1)^k / k!, which is line 9 of the Formula section. (End)
a(n) = exp(-1)*Gamma(n+1,-1) (incomplete Gamma function). - Mark van Hoeij, Nov 11 2009
G.f.: 1/(1-x^2/(1-2x-4x^2/(1-4x-9x^2/(1-6x-16x^2/(1-8x-25x^2/(1-... (continued fraction). - Paul Barry, Nov 27 2009
a(n) = Sum_{p in Pano1(n)} M2(p), n >= 1, with Pano1(n) the set of partitions without part 1, and the multinomial M2 numbers. See the characteristic array for partitions without part 1 given by A145573 in Abramowitz-Stegun (A-S) order, with A002865(n) the total number of such partitions. The M2 numbers are given for each partition in A-St order by the array A036039. - Wolfdieter Lang, Jun 01 2010
a(n) = ((-1)^n)*(n-1)*hypergeom([-n+2, 2], [], 1), n>=1; 1 for n=0. - Wolfdieter Lang, Aug 16 2010
a(n) = (-1)^n * hypergeom([ -n, 1], [], 1), n>=1; 1 for n=0. From the binomial convolution due to the e.g.f. - Wolfdieter Lang, Aug 26 2010
Integral_{x=0..1} x^n*exp(x) = (-1)^n*(a(n)*e - n!).
O.g.f.: Sum_{n>=0} n^n*x^n/(1 + (n+1)*x)^(n+1). - Paul D. Hanna, Oct 06 2011
G.f.: hypergeom([1,1],[],x/(x+1))/(x+1). - Mark van Hoeij, Nov 07 2011
From Sergei N. Gladkovskii, Nov 25 2011, Jul 05 2012, Sep 23 2012, Oct 13 2012, Mar 09 2013, Mar 10 2013, Oct 18 2013: (Start)
Continued fractions:
In general, e.g.f. (1+a*x)/exp(b*x) = U(0) with U(k) = 1 + a*x/(1-b/(b-a*(k+1)/U(k+1))). For a=-1, b=-1: exp(-x)/(1-x) = 1/U(0).
E.g.f.: (1-x/(U(0)+x))/(1-x), where U(k) = k+1 - x + (k+1)*x/U(k+1).
E.g.f.: 1/Q(0) where Q(k) = 1 - x/(1 - 1/(1 - (k+1)/Q(k+1))).
G.f.: 1/U(0) where U(k) = 1 + x - x*(k+1)/(1 - x*(k+1)/U(k+1)).
G.f.: Q(0)/(1+x) where Q(k) = 1 + (2*k+1)*x/((1+x)-2*x*(1+x)*(k+1)/(2*x*(k+1)+(1+x)/ Q(k+1))).
G.f.: 1/Q(0) where Q(k) = 1 - 2*k*x - x^2*(k + 1)^2/Q(k+1).
G.f.: T(0) where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2-(1-2*x*k)*(1-2*x-2*x*k)/T(k+1)). (End)
0 = a(n)*(a(n+1) + a(n+2) - a(n+3)) + a(n+1)*(a(n+1) + 2*a(n+2) - a(n+3)) + a(n+2)*a(n+2) if n>=0. - Michael Somos, Jan 25 2014
a(n) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(k + x)^k*(k + x + 1)^(n-k) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(k + x)^(n-k)*(k + x - 1)^k, for arbitrary x. - Peter Bala, Feb 19 2017
a(n) = Sum_{j=0..n} Sum_{k=0..n} binomial(-j-1, -n-1)*abs(Stirling1(j, k)).
a(n) = Sum_{k=0..n} (-1)^(n-k)*Pochhammer(n-k+1, k) (cf. A008279). (End)
a(n) = n! - Sum_{j=0..n-1} binomial(n,j) * a(j). - Alois P. Heinz, Jan 23 2019
a(n) = (-1)^n*Sum_{k=0..n} Bell(k)*Stirling1(n+1, k+1). - Mélika Tebni, Jul 05 2022
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EXAMPLE
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a(2) = 1, a(3) = 2 and a(4) = 9 since the possibilities are {BA}, {BCA, CAB} and {BADC, BCDA, BDAC, CADB, CDAB, CDBA, DABC, DCAB, DCBA}. - Henry Bottomley, Jan 17 2001
The Boolean complex of the complete graph K_4 is homotopy equivalent to the wedge of 9 3-spheres.
Necklace problem for n = 6: partitions without part 1 and M2 numbers for n = 6: there are A002865(6) = 4 such partitions, namely (6), (2,4), (3^2) and (2^3) in A-St order with the M2 numbers 5!, 90, 40 and 15, respectively, adding up to 265 = a(6). This corresponds to 1 necklace with 6 beads, two necklaces with 2 and 4 beads respectively, two necklaces with 3 beads each and three necklaces with 2 beads each. - Wolfdieter Lang, Jun 01 2010
G.f. = 1 + x^2 + 9*x^3 + 44*x^4 + 265*x^5 + 1854*x^6 + 14833*x^7 + 133496*x^8 + ...
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MAPLE
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A000166 := proc(n) option remember; if n<=1 then 1-n else (n-1)*(procname(n-1)+procname(n-2)); fi; end;
a:=n->n!*sum((-1)^k/k!, k=0..n): seq(a(n), n=0..21); # Zerinvary Lajos, May 17 2007
ZL1:=[S, {S=Set(Cycle(Z, card>1))}, labeled]: seq(count(ZL1, size=n), n=0..21); # Zerinvary Lajos, Sep 26 2007
with (combstruct):a:=proc(m) [ZL, {ZL=Set(Cycle(Z, card>=m))}, labeled]; end: A000166:=a(2):seq(count(A000166, size=n), n=0..21); # Zerinvary Lajos, Oct 02 2007
Z := (x, m)->m!^2*sum(x^j/((m-j)!^2), j=0..m): R := (x, n, m)->Z(x, m)^n: f := (t, n, m)->sum(coeff(R(x, n, m), x, j)*(t-1)^j*(n*m-j)!, j=0..n*m): seq(f(0, n, 1), n=0..21); # Zerinvary Lajos, Jan 22 2008
a:=proc(n) if `mod`(n, 2)=1 then sum(2*k*factorial(n)/factorial(2*k+1), k=1.. floor((1/2)*n)) else 1+sum(2*k*factorial(n)/factorial(2*k+1), k=1..floor((1/2)*n)-1) end if end proc: seq(a(n), n=0..20); # Emeric Deutsch, Feb 23 2008
G(x):=2*exp(-x)/(1-x): f[0]:=G(x): for n from 1 to 26 do f[n]:=diff(f[n-1], x) od: x:=0: seq(f[n]/2, n=0..21); # Zerinvary Lajos, Apr 03 2009
seq(simplify(KummerU(-n, -n, -1)), n = 0..23); # Peter Luschny, May 10 2022
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MATHEMATICA
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a[0] = 1; a[1] = 0; a[n_] := Round[n!/E] /; n >= 1 (* Michael Taktikos, May 26 2006 *)
Range[0, 20]! CoefficientList[ Series[ Exp[ -x]/(1 - x), {x, 0, 20}], x]
dr[{n_, a1_, a2_}]:={n+1, a2, n(a1+a2)}; Transpose[NestList[dr, {0, 0, 1}, 30]][[3]] (* Harvey P. Dale, Feb 23 2013 *)
a[n_] := (-1)^n HypergeometricPFQ[{- n, 1}, {}, 1]; (* Michael Somos, Jun 01 2013 *)
a[n_] := n! SeriesCoefficient[Exp[-x] /(1 - x), {x, 0, n}]; (* Michael Somos, Jun 01 2013 *)
RecurrenceTable[{a[n] == n*a[n - 1] + (-1)^n, a[0] == 1}, a, {n, 0, 23}] (* Ray Chandler, Jul 30 2015 *)
nxt[{n_, a_}]:={n+1, a(n+1)+(-1)^(n+1)}; NestList[nxt, {0, 1}, 25][[All, 2]] (* Harvey P. Dale, Jun 01 2019 *)
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PROG
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(PARI) {a(n) = if( n<1, 1, n * a(n-1) + (-1)^n)}; /* Michael Somos, Mar 24 2003 */
(PARI) {a(n) = n! * polcoeff( exp(-x + x * O(x^n)) / (1 - x), n)}; /* Michael Somos, Mar 24 2003 */
(PARI) {a(n)=polcoeff(sum(m=0, n, m^m*x^m/(1+(m+1)*x+x*O(x^n))^(m+1)), n)} /* Paul D. Hanna */
(Python) See Hobson link.
(Maxima)
s[0]:1$
s[n]:=n*s[n-1]+(-1)^n$
(Haskell)
a000166 n = a000166_list !! n
a000166_list = 1 : 0 : zipWith (*) [1..]
(zipWith (+) a000166_list $ tail a000166_list)
(Python)
for n in range(10*2):
x, m = x*n + m, -m
(Magma) I:=[0, 1]; [1] cat [n le 2 select I[n] else (n-1)*(Self(n-1)+Self(n-2)): n in [1..30]]; // Vincenzo Librandi, Jan 07 2016
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CROSSREFS
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Cf. A000142, A002467, A003221, A000522, A000240, A000387, A000449, A000475, A129135, A092582, A000255, A002469, A159610, A068985, A068996, A047865, A038205, A008279, A281682.
Cf. A101560, A101559, A000110, A101033, A101032, A000204, A100492, A099731, A000045, A094216, A094638, A000108.
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KEYWORD
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core,nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A038205
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Number of derangements of n where minimal cycle size is at least 3.
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+10
110
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1, 0, 0, 2, 6, 24, 160, 1140, 8988, 80864, 809856, 8907480, 106877320, 1389428832, 19452141696, 291781655984, 4668504894480, 79364592318720, 1428562679845888, 27142690734936864, 542853814536802656, 11399930109077490560, 250798462399300784640
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,4
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COMMENTS
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Permutations with no cycles of length 1 or 2.
Related to (and bounded by) "derangements" (A000166). Minimal cycle size 3 is interesting because of its physical analog. Consider a fully-connected network of n nodes where the objects stored at the nodes must derange but can't do so in such a way that any two objects would collide along the connecting "pipe" between their nodes.
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REFERENCES
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G. Paquin, Dénombrement de multigraphes enrichis, Mémoire, Math. Dept., Univ. Québec à Montréal, 2004.
H. S. Wilf, Generatingfunctionology, Academic Press, NY, 1990, p. 147, Eq. 5.2.9 (q=2).
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LINKS
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FORMULA
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a(n) = Sum_{i=3..n} binomial(n-1,i-1) * (i-1)! * a(n-i).
E.g.f.: exp(-x-x^2/2)/(1-x) = exp( Sum_{k>2} x^k / k ).
a(n) = n! * Sum_{m=1..n} ((Sum_{k=0..m} k!*(-1)^(m-k) *binomial(m,k) *Sum_{i=0..n-m} abs(stirling1(i+k,k)) *binomial(m-k,n-m-i) *2^(-n+m+i) /(i+k)!))/m!; a(0)=1. - Vladimir Kruchinin, Feb 01 2011
a(n) = (n-1)*a(n-1) + (n-1)*(n-2)*a(n-3). - Peter Bala, Apr 18 2012
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EXAMPLE
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a(5) = 24 because, with a minimum cycle size of 3, the only way to derange all 5 elements is to have them move around in one large 5-cycle. The number of possible moves is (5-1)! = 4! = 24.
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MAPLE
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with(combstruct): ZL2:=[S, {S=Set(Cycle(Z, card>2))}, labeled] :seq(count(ZL2, size=n), n=0..21); # Zerinvary Lajos, Sep 26 2007
with(combstruct):a:=proc(m) [ZZ, {ZZ=Set(Cycle(Z, card>m))}, labeled]; end: A038205:=a(2):seq(count(A038205, size=n), n=0..21); # Zerinvary Lajos, Oct 02 2007
G:= exp(-x-x^2/2)/(1-x): Gser:=series(G, x, 26): a:= n-> n!*coeff(Gser, x, n): seq(a(n), n=0..25); # Paul Weisenhorn, May 29 2010
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MATHEMATICA
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PROG
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(PARI) x='x+O('x^30); Vec(serlaplace(exp(-x-x^2/2)/(1-x))) \\ G. C. Greubel, Jun 25 2018
(Magma) m:=30; R<x>:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(Exp(-x-x^2/2)/(1-x))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Jun 25 2018
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A333726
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Triangle read by rows: T(n,k) is the number of permutations in S_n for which all cycles are length k or greater.
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+10
2
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1, 2, 1, 6, 2, 2, 24, 9, 6, 6, 120, 44, 24, 24, 24, 720, 265, 160, 120, 120, 120, 5040, 1854, 1140, 720, 720, 720, 720, 40320, 14833, 8988, 6300, 5040, 5040, 5040, 5040, 362880, 133496, 80864, 58464, 40320, 40320, 40320, 40320, 40320
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OFFSET
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1,2
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LINKS
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FORMULA
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T(n,k) = 0 when n < k,
T(n,k) = (n-1)! when n = k, and otherwise
T(n,k) = (n-1) * T(n-1, k) + A068424(n-1, k-1) * T(n-k, k).
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EXAMPLE
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Triangle begins:
n\k| 1 2 3 4 5 6 7 8 9
---+--------------------------------------------------------
1| 1
2| 2 1
3| 6 2 2
4| 24 9 6 6
5| 120 44 24 24 24
6| 720 265 160 120 120 120
7| 5040 1854 1140 720 720 720 720
8| 40320 14833 8988 6300 5040 5040 5040 5040
9| 362880 133496 80864 58464 40320 40320 40320 40320 40320
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A067736
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Decimal expansion of exp(3/2).
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+10
1
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4, 4, 8, 1, 6, 8, 9, 0, 7, 0, 3, 3, 8, 0, 6, 4, 8, 2, 2, 6, 0, 2, 0, 5, 5, 4, 6, 0, 1, 1, 9, 2, 7, 5, 8, 1, 9, 0, 0, 5, 7, 4, 9, 8, 6, 8, 3, 6, 9, 6, 6, 7, 0, 5, 6, 7, 7, 2, 6, 5, 0, 0, 8, 2, 7, 8, 5, 9, 3, 6, 6, 7, 4, 4, 6, 6, 7, 1, 3, 7, 7, 2, 9, 8, 1, 0, 5, 3, 8, 3, 1, 3, 8, 2, 4, 5, 3, 3, 9, 1, 3, 8, 8, 6, 1
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OFFSET
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1,1
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COMMENTS
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It is well known that derangements, A000166, are related to exp(1) (cf. A001113). It appears that derangements with minimal cycle size 3 relate to exp(1+1/2). for example, 720/160 = 4.5, 5040/1140 = 4.4210, 40320/8988 = 4.4859, 362880/80864 = 4.4875 the pattern continues - derangements with minimal cycle size 4 appear to relate in the same way to exp(1 + 1/2 +1/3).
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LINKS
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FORMULA
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EXAMPLE
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4.4816890703380648226020554601192758190057498683696...
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MATHEMATICA
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A274169
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Decimal expansion of 1/exp(exp(1)-1).
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+10
1
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1, 7, 9, 3, 7, 4, 0, 7, 8, 7, 3, 4, 0, 1, 7, 1, 8, 1, 9, 6, 1, 9, 8, 9, 5, 8, 7, 3, 1, 8, 3, 1, 6, 4, 9, 8, 4, 5, 9, 6, 8, 1, 6, 0, 1, 7, 5, 8, 9, 1, 5, 6, 1, 3, 1, 5, 7, 3, 7, 0, 4, 2, 1, 6, 0, 2, 4, 8, 3, 7, 6, 0, 8, 1, 1, 6, 4, 5, 7, 2, 8, 8, 0, 1, 3, 0, 9, 4, 1, 4, 1, 1, 2, 4, 3, 8, 0, 0, 4, 6, 0, 5, 6, 0
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OFFSET
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0,2
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COMMENTS
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This is the limiting value of the probability that a random n-permutation will have no cycles of length less than k (for any k) as n goes to infinity. For example, the probability (as n goes to infinity) that a random n-permutation has no fixed points is 1/exp(1). The probability that it has no cycles of length 1 or 2 is 1/exp(1+1/2). The probability that it has no cycles of length 1 or 2 or 3 is 1/exp(1+1/2+1/3!)...
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LINKS
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FORMULA
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EXAMPLE
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0.1793740787340171819619895873183164984596816...
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MAPLE
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MATHEMATICA
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RealDigits[1/E^(E - 1), 10, 50][[1]]
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PROG
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A355285
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Expansion of e.g.f. 1 / (1 + x + x^2/2 + x^3/3 + log(1 - x)).
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+10
1
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1, 0, 0, 0, 6, 24, 120, 720, 7560, 76608, 810432, 9141120, 118015920, 1666336320, 25211774016, 404932155264, 6951992261760, 127203705538560, 2467434718218240, 50477473338494976, 1086707769452699904, 24573149993692615680, 582367494447600583680, 14430857455114783119360
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OFFSET
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0,5
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LINKS
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FORMULA
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E.g.f.: 1 / (1 - Sum_{k>=4} x^k/k).
a(0) = 1; a(n) = Sum_{k=4..n} binomial(n,k) * (k-1)! * a(n-k).
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MATHEMATICA
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nmax = 23; CoefficientList[Series[1/(1 + x + x^2/2 + x^3/3 + Log[1 - x]), {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] (k - 1)! a[n - k], {k, 4, n}]; Table[a[n], {n, 0, 23}]
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PROG
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(PARI) my(x='x+O('x^30)); Vec(serlaplace(1/(1 + x + x^2/2 + x^3/3 + log(1 - x)))) \\ Michel Marcus, Jun 27 2022
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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