A000904 -id:A000904 - cp's OEIS frontend

A000179 Ménage numbers: a(0) = 1, a(1) = -1, and for n >= 2, a(n) = number of permutations s of [0, ..., n-1] such that s(i) != i and s(i) != i+1 (mod n) for all i.

1, -1, 0, 1, 2, 13, 80, 579, 4738, 43387, 439792, 4890741, 59216642, 775596313, 10927434464, 164806435783, 2649391469058, 45226435601207, 817056406224416, 15574618910994665, 312400218671253762, 6577618644576902053, 145051250421230224304, 3343382818203784146955, 80399425364623070680706, 2013619745874493923699123

Offset: 0

N. J. A. Sloane

According to rook theory, John Riordan considered a(1) to be -1. - Vladimir Shevelev, Apr 02 2010
This is also the value that the formulas of Touchard and of Wyman and Moser give and is compatible with many recurrences. - William P. Orrick, Aug 31 2020
Or, for n >= 3, the number of 3 X n Latin rectangles the second row of which is full cycle with a fixed order of its elements, e.g., the cycle (x_2,x_3,...,x_n,x_1) with x_1 < x_2 < ... < x_n. - Vladimir Shevelev, Mar 22 2010
Muir (p. 112) gives essentially this recurrence (although without specifying any initial conditions). Compare A186638. - N. J. A. Sloane, Feb 24 2011
Sequence discovered by Touchard in 1934. - L. Edson Jeffery, Nov 13 2013
Although these are also known as Touchard numbers, the problem was formulated by Lucas in 1891, who gave the recurrence formula shown below. See Cerasoli et al., 1988. - Stanislav Sykora, Mar 14 2014
An equivalent problem was formulated by Tait; solutions to Tait's problem were given by Muir (1878) and Cayley (1878). - William P. Orrick, Aug 31 2020
From Vladimir Shevelev, Jun 25 2015: (Start)
According to the ménage problem, 2*n!*a(n) is the number of ways of seating n married couples at 2*n chairs around a circular table, men and women in alternate positions, so that no husband is next to his wife.
It is known [Riordan, ch. 7] that a(n) is the number of arrangements of n non-attacking rooks on the positions of the 1's in an n X n (0,1)-matrix A_n with 0's in positions (i,i), i = 1,...,n, (i,i+1), i = 1,...,n-1, and (n,1). This statement could be written as a(n) = per(A_n). For example, A_5 has the form
    001*11
    1*0011
    11001*                        (1)
    11*100
    0111*0,
where 5 non-attacking rooks are denoted by {1*}.
We can indicate a one-to-one correspondence between arrangements of n non-attacking rooks on the 1's of a matrix A_n and arrangements of n married couples around a circular table by the rules of the ménage problem, after the ladies w_1, w_2, ..., w_n have taken the chairs numbered
    2*n, 2, 4, ..., 2*n-2         (2)
respectively. Suppose we consider an arrangement of rooks: (1,j_1), (2,j_2), ..., (n,j_n). Then the men m_1, m_2, ..., m_n took chairs with numbers
    2*j_i - 3 (mod 2*n),          (3)
where the residues are chosen from the interval[1,2*n]. Indeed {j_i} is a permutation of 1,...,n. So {2*j_i-3}(mod 2*n) is a permutation of odd positive integers <= 2*n-1. Besides, the distance between m_i and w_i cannot be 1. Indeed, the equality |2*(j_i-i)-1| = 1 (mod 2*n) is possible if and only if either j_i=i or j_i=i+1 (mod n) that correspond to positions of 0's in matrix A_n.
For example, in the case of positions of {1*} in(1) we have j_1=3, j_2=1, j_3=5, j_4=2, j_5=4. So, by(2) and (3) the chairs 1,2,...,10 are taken by m_4, w_2, m_1, w_3, m_5, w_4, m_3, w_5, m_2, w_1, respectively. (End)
The first 20 terms of this sequence were calculated in 1891 by E. Lucas (see [Lucas, p. 495]). - Peter J. C. Moses, Jun 26 2015
From Ira M. Gessel, Nov 27 2018: (Start)
If we invert the formula
  Sum_{ n>=0 } u_n z^n = ((1-z)/(1+z)) F(z/(1+z)^2)
that Don Knuth mentions (see link) (i.e., set x=z/(1+z)^2 and solve for z in terms of x), we get a formula for F(z) = Sum_{n >= 0} n! z^n as a sum with all positive coefficients of (almost) powers of the Catalan number generating function.
The exact formula is (5) of the Yiting Li article.
This article also gives a combinatorial proof of this formula (though it is not as simple as one might want). (End)

			a(2) = 0; nothing works. a(3) = 1; (201) works. a(4) = 2; (2301), (3012) work. a(5) = 13; (20413), (23401), (24013), (24103), (30412), (30421), (34012), (34021), (34102), (40123), (43012), (43021), (43102) work.

W. W. R. Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, 13th Ed. Dover, p. 50.
M. Cerasoli, F. Eugeni and M. Protasi, Elementi di Matematica Discreta, Nicola Zanichelli Editore, Bologna 1988, Chapter 3, p. 78.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 185, mu(n).
Kaplansky, Irving and Riordan, John, The probleme des menages, Scripta Math. 12, (1946). 113-124. See u_n.
E. Lucas, Théorie des nombres, Paris, 1891, pp. 491-495.
P. A. MacMahon, Combinatory Analysis. Cambridge Univ. Press, London and New York, Vol. 1, 1915 and Vol. 2, 1916; see vol. 1, p 256.
T. Muir, A Treatise on the Theory of Determinants. Dover, NY, 1960, Sect. 132, p. 112. - N. J. A. Sloane, Feb 24 2011
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 197.
V. S. Shevelev, Reduced Latin rectangles and square matrices with equal row and column sums, Diskr. Mat. (J. of the Akademy of Sciences of Russia) 4(1992), 91-110. - Vladimir Shevelev, Mar 22 2010
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).
H. M. Taylor, A problem on arrangements, Mess. Math., 32 (1902), 60ff.
J. Touchard, Permutations discordant with two given permutations, Scripta Math., 19 (1953), 108-119.
J. H. van Lint, Combinatorial Theory Seminar, Eindhoven University of Technology, Springer Lecture Notes in Mathematics, Vol. 382, 1974. See page 10.

Seiichi Manyama, Table of n, a(n) for n = 0..450 (terms 0..100 from T. D. Noe)
M. A. Alekseyev, Weighted de Bruijn Graphs for the Menage Problem and Its Generalizations. Lecture Notes in Computer Science 9843 (2016), 151-162. doi:10.1007/978-3-319-44543-4_12 arXiv:1510.07926, [math.CO], 2015-2016.
Vladimir Baltic, On the number of certain types of strongly restricted permutations, Applicable Analysis and Discrete Mathematics Vol. 4, No 1 (April, 2010), 119-135.
Kenneth P. Bogart and Peter G. Doyle, Nonsexist solution of the ménage problem, Amer. Math. Monthly 93 (1986), no. 7, 514-519.
A. Cayley, On a problem of arrangements, Proceedings of the Royal Society of Edinburgh 9 (1878) 338-342.
A. Cayley, On a problem of arrangements, Proceedings of the Royal Society of Edinburgh 9 (1878) 338-342.
A. Cayley, On Mr Muir's discussion of Professor Tait's problem of arrangements, Proceedings of the Royal Society of Edinburgh 9 (1878) 388-391.
A. Cayley, On Mr Muir's discussion of Professor Tait's problem of arrangements, Proceedings of the Royal Society of Edinburgh 9 (1878) 388-391.
J. Dutka, On the Probleme des Menages, Mathem. Conversat. (2001) 277-287, reprinted from Math. Intell. 8 (1986) 18-25
A. de Gennaro, How many latin rectangles are there?, arXiv:0711.0527 [math.CO] (2007), see p. 2.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 372
Nick Hobson, Python program for this sequence.
Peter Kagey, Ranking and Unranking Restricted Permutations, arXiv:2210.17021 [math.CO], 2022.
Irving Kaplansky, Solution of the "Problème des ménages", Bull. Amer. Math. Soc. 49, (1943). 784-785.
Irving Kaplansky, Symbolic solution of certain problems in permutations, Bull. Amer. Math. Soc., 50 (1944), 906-914.
I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. [Scan of annotated copy]
S. M. Kerawala, Asymptotic solution of the "Probleme des menages, Bull. Calcutta Math. Soc., 39 (1947), 82-84. [Annotated scanned copy]
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 221.
D. E. Knuth, Comments on A000179, Nov 25 2018 - Nov 27 2018.
D. E. Knuth, Donald Knuth's 24th Annual Christmas Lecture: Dancing Links, Stanfordonline, Video published on YouTube, Dec 12, 2018.
A. R. Kräuter, Über die Permanente gewisser zirkulärer Matrizen..., Séminaire Lotharingien de Combinatoire, B11b (1984), 11 pp.
Yiting Li, Ménage Numbers and Ménage Permutations, J. Int. Seq. 18 (2015) 15.6.8
E. Lucas, Théorie des Nombres, Gauthier-Villars, Paris, 1891, Vol. 1, p. 495.
T. Muir, On Professor Tait's problem of arrangement, Proceedings of the Royal Society of Edinburgh 9 (1878) 382-387.
T. Muir, On Professor Tait's problem of arrangement, Proceedings of the Royal Society of Edinburgh 9 (1878) 382-387.
Vladimir Shevelev and Peter J. C. Moses, The ménage problem with a known mathematician, arXiv:1101.5321 [math.CO], 2011, 2015.
Vladimir Shevelev and Peter J. C. Moses, Alice and Bob go to dinner: A variation on menage, INTEGERS, Vol. 16 (2016), #A72.
R. J. Stones, S. Lin, X. Liu and G. Wang, On Computing the Number of Latin Rectangles, Graphs and Combinatorics (2016) 32:1187-1202; DOI 10.1007/s00373-015-1643-1.
H. M. Taylor, A problem on arrangements, Mess. Math., 32 (1902), 60ff. [Annotated scanned copy]
J. Touchard, Théorie des substitutions. Sur un problème de permutations, C. R. Acad. Sci. Paris 198 (1934), 631-633.
Eric Weisstein's World of Mathematics, Married Couples Problem.
Eric Weisstein's World of Mathematics, Rooks Problem.
Wikipedia, Menage problem.
M. Wyman and L. Moser, On the problème des ménages, Canad. J. Math., 10 (1958), 468-480.
D. Zeilberger, Automatic Enumeration of Generalized Menage Numbers, arXiv preprint arXiv:1401.1089 [math.CO], 2014.

Diagonal of A058087. Also a diagonal of A008305.
Cf. A000904, A059375, A102761, A000186, A094047, A067998, A033999, A258664, A258665, A258666, A258667, A258673, A259212, A213234, A000023.
A000179, A102761, and A335700 are all essentially the same sequence but with different conventions for the initial terms a(0) and a(1). - N. J. A. Sloane, Aug 06 2020

import Data.List (zipWith5)
a000179 n = a000179_list !! n
a000179_list = 1 : -1 : 0 : 1 : zipWith5
   (\v w x y z -> (x * y + (v + 2) * z - w) `div` v) [2..] (cycle [4,-4])
   (drop 4 a067998_list) (drop 3 a000179_list) (drop 2 a000179_list)
-- Reinhard Zumkeller, Aug 26 2013

A000179:= n ->add ((-1)^k*(2*n)*binomial(2*n-k,k)*(n-k)!/(2*n-k), k=0..n); # for n >= 1
U:= proc(n) local k; add( (2*n/(2*n-k))*binomial(2*n-k,k)*(n-k)!*(x-1)^k, k=0..n); end; W := proc(r,s) coeff( U(r),x,s ); end; A000179 := n->W(n,0); # valid for n >= 1

a[n_] := 2*n*Sum[(-1)^k*Binomial[2*n - k, k]*(n - k)!/(2*n - k), {k, 0, n}]; a[0] = 1; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Dec 05 2012, from 2nd formula *)

\\ 3 programs adapted to a(1) = -1 by Hugo Pfoertner, Aug 31 2020

{a(n) = my(A); if( n, A = vector(n,i,i-2); for(k=4, n, A[k] = (k * (k - 2) * A[k-1] + k * A[k-2] - 4 * (-1)^k) / (k-2)); A[n], 1)};/* Michael Somos, Jan 22 2008 */

a(n)=if(n>1, round(2*n*exp(-2)*besselk(n, 2)), 1-2*n) \\ Charles R Greathouse IV, Nov 03 2014

{a(n) = my(A); if( n, A = vector(n,i,i-2); for(k=5, n, A[k] = k * A[k-1] + 2 * A[k-2] + (4-k) * A[k-3] - A[k-4]); A[n], 1)} /* Michael Somos, May 02 2018 */

from math import comb, factorial
def A000179(n): return 1 if n == 0 else sum((-2*n if k & 1 else 2*n)*comb(m:=2*n-k,k)*factorial(n-k)//m for k in range(n+1)) # Chai Wah Wu, May 27 2022

a(n) = ((n^2-2*n)*a(n-1) + n*a(n-2) - 4*(-1)^n)/(n-2) for n >= 3.
a(n) = A059375(n)/(2*n!) for n >= 2.
a(n) = Sum_{k=0..n} (-1)^k*(2*n)*binomial(2*n-k, k)*(n-k)!/(2*n-k) for n >= 1. - Touchard (1934)
G.f.: ((1-x)/(1+x))*Sum_{n>=0} n!*(x/(1+x)^2)^n. - Vladeta Jovovic, Jun 26 2007
a(2^k+2) == 0 (mod 2^k); for k >= 2, a(2^k) == 2(mod 2^k). - Vladimir Shevelev, Jan 14 2011
a(n) = round( 2*n*exp(-2)*BesselK(n,2) ) for n > 1. - Mark van Hoeij, Oct 25 2011
a(n) ~ (n/e)^n * sqrt(2*Pi*n)/e^2. - Charles R Greathouse IV, Jan 21 2016
0 = a(n)*(-a(n+2) +a(n+4)) +a(n+1)*(+a(n+1) +a(n+2) -3*a(n+3) -5*a(n+4) +a(n+5)) +a(n+2)*(+2*a(n+2) +3*a(n+3) -3*a(n+4)) +a(n+3)*(+2*a(n+3) +a(n+4) -a(n+5)) +a(n+4)*(+a(n+4)), for all n>1. If a(-2..1) = (0, -1, 2, -1) then also true for those values of n. - Michael Somos, Apr 29 2018
D-finite with recurrence: 0 = a(n) +n*a(n+1) -2*a(n+2) +(-n-4)*a(n+3) +a(n+4), for all n in Z where a(n) = a(-n) for all n in Z and a(0) = 2, a(1) = -1. - Michael Somos, May 02 2018
a(n) = Sum_{k=0..n} A213234(n,k) * A000023(n-2*k) = Sum_{k=0..n} (-1)^k * n/(n-k) * binomial(n-k, k) * (n-2*k)! Sum_{j=0..n-2*k} (-2)^j/j! for n >= 1. [Wyman and Moser (1958)]. - William P. Orrick, Jun 25 2020
a(k+4*p) - 2*a(k+2*p) + a(k) is divisible by p, for any k > 0 and any prime p. - Mark van Hoeij, Jan 11 2022

More terms from James Sellers, May 02 2000
Additional comments from David W. Wilson, Feb 18 2003
a(1) changed to -1 at the suggestion of Don Knuth. - N. J. A. Sloane, Nov 26 2018

A000271 Sums of ménage numbers.

Original entry on oeis.org

1, 0, 0, 1, 3, 16, 96, 675, 5413, 48800, 488592, 5379333, 64595975, 840192288, 11767626752, 176574062535, 2825965531593, 48052401132800, 865108807357216, 16439727718351881, 328839946389605643, 6906458590966507696

Offset: 0

N. J. A. Sloane

Permanent of the (0,1)-matrix having (i,j)-th entry equal to 0 iff this is on the diagonal or the first upper-diagonal. - Simone Severini, Oct 14 2004
Equivalently, number of permutations p of {1,2,...,n} such that p(i)-i not in {0,1}. - Andrew Howroyd, Sep 19 2017
From Vladimir Shevelev, Jun 21 2015: (Start)
Let 2*n!*V(n)=A137886(n) be the number of ways of seating n married couples at 2*n chairs arranged side-by-side in a straight line, men and women in alternate positions, so that no husband is next to his wife.
It is known [Riordan, Ch. 8, Th. 1, t=0] that, if 2*n!*U(n) is a solution of an analogous problem at a circular table, then U(n) = V(n) - V(n-1), n>=3, where U(n) = A000179(n). Thus V(n) = Sum_{i=3,...,n} A000179(i), n>=1, and comparing the initial conditions, we conclude that a(n) = V(n), n>=1. This gives a combinatorial interpretation for 2*n!*a(n).
(End)

			G.f. = 1 + x^3 + 3*x^4 + 16*x^5 + 96*x^6 + 675*x^7 + 5413*x^8 + ...

W. Ahrens, Mathematische Unterhaltungen und Spiele. Teubner, Leipzig, Vol. 1, 3rd ed., 1921; Vol. 2, 2nd ed., 1918. See Vol. 2, p. 79.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 198.
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).
H. M. Taylor, A problem on arrangements, Mess. Math., 32 (1902), 60ff.

Seiichi Manyama, Table of n, a(n) for n = 0..450 (terms 0..100 from T. D. Noe)
W. Ahrens, Mathematische Unterhaltungen und Spiele, Leipzig: B. G. Teubner, 1901.
J. D. H. Dickson, Discussion of two double series arising from the number of terms in determinants of certain forms, Proc. London Math. Soc., 10 (1879), 120-122.
J. D. H. Dickson, Discussion of two double series arising from the number of terms in determinants of certain forms, Proc. London Math. Soc., 10 (1879), 120-122. [Annotated scanned copy]
Dmitry Efimov, Determinants of generalized binary band matrices, arXiv:1702.05655 [math.RA], 2017.
V. Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 222.
Y. Li, Ménage Numbers and Ménage Permutations, J. Int. Seq. 18 (2015) 15.6.8
H. M. Taylor, A problem on arrangements, Mess. Math., 32 (1902), 60ff. [Annotated scanned copy]
M. Wyman and L. Moser, On the problème des ménages, Canad. J. Math., 10 (1958), 468-480.

Cf. A000179, A000904, A001883, A137886, A292574. A diagonal of A058057.

[ &+[(-1)^(n-k)*Binomial(n+k, 2*k)*Factorial(k): k in [0..n]]: n in [0..21]]; // Bruno Berselli, Apr 11 2011

V := proc(n) local k; add( binomial(2*n-k,k)*(n-k)!*(x-1)^k, k=0..n); end; W := proc(r,s) coeff( V(r),x,s ); end; A000271 := n->W(n-2,0);

Table[Sum[(-1)^(n - k) k! Binomial[n + k, 2 k], {k, 0, n}], {n, 0, 22}] (* Jean-François Alcover, Apr 11 2011, after Paul Barry *)
RecurrenceTable[{a[0] == 1, a[1] == a[2] == 0, a[n] == (n - 1) a[n - 2] + (n - 1) a[n - 1] +  a[n - 3]}, a, {n, 30}] (* Harvey P. Dale, Jun 01 2012 *)
Table[(-1)^n HypergeometricPFQ[{1, -n, n + 1}, {1/2}, 1/4], {n, 20}] (* Michael Somos, May 28 2014 *)

a(n) = if(n, round( 2*exp(-2)*(besselk(n+1,2) + besselk(n,2)) ), 1) \\ Charles R Greathouse IV, May 11 2016

a(n) = (n - 1) a(n - 2) + (n - 1) a(n - 1) + a(n - 3).
From Paul Barry, Feb 08 2009: (Start)
G.f.: 1/(1+x-x/(1+x-x/(1+x-2x/(1+x-2x/(1+x-3x/(1+x-3x/(1+x-4x/(1+... (continued fraction);
a(n) = Sum_{k=0..n} binomial(2n-k,k)*(n-k)!*(-1)^k. (End)
a(n) = (-1)^n*hypergeom([1, -n, n+1],[1/2],1/4). - Mark van Hoeij, Nov 12 2009
a(n) = round( 2*exp(-2)*(BesselK(1+n,2) + BesselK(n,2)) ) for n>0. - Mark van Hoeij, Nov 12 2009
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n+k,2*k)*k!. - Paul Barry, Jun 23 2010
G.f.: Sum_{n>=0} n!*x^n/(1+x)^(2*n+1). - Ira M. Gessel, Jan 15 2013
a(n) ~ exp(-2)*n!. - Vaclav Kotesovec, Mar 10 2014
a(-1 - n) = -a(n) for all n in Z. - Michael Somos, May 28 2014
a(n) = Sum_{i=3..n} A000179(i), n>=1. - Vladimir Shevelev, Jun 21 2015
0 = a(n)*(-a(n+2) - a(n+3)) + a(n+1)*(+a(n+1) + 2*a(n+2) + a(n+3) - a(n+4)) + a(n+2)*(+a(n+2) + 2*a(n+3) - a(n+4)) + a(n+3)*(+a(n+3)) for all n in Z. - Michael Somos, Oct 16 2016

More terms from James Sellers, Aug 21 2000

More terms from Simone Severini, Oct 14 2004

A000905 Hamilton numbers.

Original entry on oeis.org

2, 3, 5, 11, 47, 923, 409619, 83763206255, 3508125906290858798171, 6153473687096578758448522809275077520433167, 18932619208894981833333582059033329370801266249535902023330546944758507753065602135843

Offset: 1

N. J. A. Sloane

a(n) is the minimal degree of an equation from which n successive terms after the first can be removed (by a series of transformation comparable to Tschirnhaus') without requiring the solution of an equation of degree greater than n (and excluding cases where an equation of degree greater than n is needed but is in fact factorizable into several equations of degree all less than n). Hamilton computed the first six terms of this sequence (see reference). That is the reason Sylvester and Hammond named them "Hamilton numbers". - Olivier Gérard, Oct 17 2007

Named after the Irish mathematician William Rowan Hamilton (1805-1865). - Amiram Eldar, Jun 19 2021

			a(1)=2 is the familiar fact than one can always remove the linear term of a quadratic equation.
a(2)=3 because one can put any cubic equation in the form x^3-a=0 by a Tschirnhaus transformation based on the solutions of a quadratic equation.
a(4)=11 because one can remove the 4 terms after the first term in a polynomial of degree 11 without having to solve a quintic.

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).

Amiram Eldar, Table of n, a(n) for n = 1..14
Tigran Ananyan and Melvin Hochster, Universal Lex Ideal Approximations of Extended Hilbert Functions and Hamilton Numbers, arXiv:2003.00589 [math.AC], 2020.
Alexander Chen, Y. H. He and J. McKay, Erland Samuel Bring's "Transformation of Algebraic Equations", arXiv preprint arXiv:1711.09253 [math.HO], 2017. See page 6.
Raymond Garver, The Tschirnhaus transformation, The Annals of Mathematics, 2nd Ser., Vol. 29, No. 1/4. (1927 - 1928), pp. 329.
William Rowan Hamilton, Inquiry Into the Validity of a Method Recently Proposed by George B. Jerrard, Esq. for Transforming and Resolving Equations of Elevated Degrees Undertaken at the Request of the Association. Richard and John E. Taylor, Report of the Sixth Meeting of the British Association for the Advancement of Science, held at Bristol in August 1836 (London: John Murray, Albemarle St., 1837), pp. 295-348.
Edouard Lucas, Théorie des Nombres, Gauthier-Villars, Paris, 1891, Vol. 1, p. 496.
Edouard Lucas, Théorie des Nombres, Gauthier-Villars, Paris, 1891, Vol. 1. [Annotated scan of pages 488-499 only]
Alexander J. Sutherland, Upper Bounds on Resolvent Degree and Its Growth Rate, arXiv:2107.08139 [math.AG], 2021.
James Joseph Sylvester and James Hammond, On Hamilton's numbers, Philosophical Transactions of the Royal Society of London A, Vol. 178 (1887), pp. 285-312, alternative link.
Jesse Wolfson, Tschirnhaus transformations after Hilbert, arXiv:2001.06515 [math.AG], 2020.

Cf. A001660.
Equals A006719(n) - 1.
Cf. A134294.

A000905 := proc(n) option remember; local i; if n=1 then 2 else 2+add((-1)^(i+1)*binomial(A000905(n-i),i+1),i=1..n-1); fi; end;

a[1]=2; a[n_] := a[n] = 2+Sum[(-1)^(i+1)*Product[a[n-i] - k, {k, 0, i}]/(i+1)!, {i, 1, n-1}]; Table[a[n], {n, 1, 11}] (* Jean-François Alcover, May 17 2011, after Maple prog. *)

from sympy import binomial
a = [2]
[a.append(2+sum((-1)**(i)*binomial(a[n-i-1], i+2) for i in range(n))) for n in range(1,11)]
print(a) # Nicholas Stefan Georgescu, Mar 01 2023

The formula given by Lucas on p. 498 is slightly in error - see Maple program given here.

A127548 O.g.f.: Sum_{n>=0} n!*(x/(1+x)^2)^n.

Original entry on oeis.org

1, 1, 0, 1, 4, 19, 112, 771, 6088, 54213, 537392, 5867925, 69975308, 904788263, 12607819040, 188341689287, 3002539594128, 50878366664393, 913161208490016, 17304836525709097, 345279674107957524, 7235298537356113339

Offset: 0

Vladeta Jovovic, Jun 27 2007

a(n+1) = inverse binomial transform of A013999 = Sum_{k=0..n} binomial(n,k)*(-1)^(n-k)*A013999(k). - Emanuele Munarini, Jul 01 2013

Seiichi Manyama, Table of n, a(n) for n = 0..450

Cf. A000179, A078480, A013999, A000271, A000904, A078509.

A127548 := proc(n) if n = 0 then 1 ; else add(factorial(s)*(-1)^(n-s)*binomial(s+n-1,2*s-1),s=1..n) ; fi ; end: for n from 0 to 20 do printf("%d,",A127548(n)) ; od ; # R. J. Mathar, Jul 13 2007

nn = 21; CoefficientList[Series[Sum[n!*(x/(1 + x)^2)^n, {n, 0, nn}], {x, 0, nn}], x] (* Michael De Vlieger, Sep 04 2016 *)

import math
def binomial(n,m):
    a=1
    for k in range(n-m+1,n+1):
        a *= k
    return a//math.factorial(m)
def A127548(n):
    if n == 0:
        return 1
    a=0
    for s in range(1,n+1):
        a += (-1)**(n-s)*binomial(s+n-1,2*s-1)*math.factorial(s)
    return a
for n in range(30):
    print(A127548(n))
# R. J. Mathar, Oct 20 2009

a(n) = Sum_{s=1..n} (-1)^(n-s)*s!*C(s+n-1,2s-1) if n>=1, where C(a,b)=binomial(a,b). - R. J. Mathar, Jul 13 2007
G.f.: Q(0) where Q(k) = 1 + (2*k + 1)*x/( (1+x)^2- 2*x*(1+x)^2*(k+1)/(2*x*(k+1) + (1+x)^2/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Mar 08 2013
a(n) = A000271(n) + A000271(n-1). - Peter Bala, Sep 02 2016
a(n) ~ exp(-2) * n!. - Vaclav Kotesovec, Oct 31 2017

More terms from R. J. Mathar, Jul 13 2007

More terms from R. J. Mathar, Oct 20 2009

A273596 For n >= 2, a(n) is the number of slim rectangular diagrams of length n.

Original entry on oeis.org

1, 3, 9, 32, 139, 729, 4515, 32336, 263205, 2401183, 24275037, 269426592, 3257394143, 42615550453, 599875100487, 9040742057760, 145251748024649, 2478320458476795, 44755020000606961, 852823700470009056, 17101229029400788083, 359978633317886558801, 7936631162022905081707

Offset: 2

Tamas Dekany, May 26 2016

			The initial term is the diagram of the four element diamond shape lattice.

Vaclav Kotesovec, Table of n, a(n) for n = 2..400
P. Bala, Notes on A273596
Gábor Czédli, Tamás Dékány, Gergő Gyenizse, and Júlia Kulin, The number of slim rectangular lattices, Algebra Universalis, 2016, Volume 75, Issue 1, pp 33-50.

Cf. A273988, A000522, A143409, A000179, A000271, A000904, A127548.

Partial sums of A155857.

A273596 := proc (n) option remember; `if`(n = 2, 1, `if`(n = 3, 3, (n-2)*procname(n-1) + procname(n-2) + 2)) end: seq(A273596(n), n = 2..20); # Peter Bala, Jan 08 2017

x = 15;
SRectD = Table[0, {x}];
For[n = 2, n < x, n++,
For[a = 1, a < n, a++,
   For[b = 1, b <= n - a, b++,
    SRectD[[n]] +=
      Binomial[n - a - 1, b - 1]*
       Binomial[n - b - 1, a - 1]*(n - a - b)!;
    ]
   ]
  Print[n, " ", SRectD[[n]]]
]
(* Alternatively: *)
T[n_, k_] := HypergeometricPFQ[{k+1, k-n}, {}, -1];
Table[Sum[T[n,k], {k,0,n}], {n,0,22}] (* Peter Luschny, Oct 05 2017 *)

a(n)= sum(rps=1, n, sum(r=1, n, s = rps-r; binomial(n-r-1, s-1) * binomial(n-s-1, r-1) * (n-r-s)!)); \\ Michel Marcus, Jun 12 2016

a(n) = Sum_{1<=r,s; r+s<=n} binomial(n-r-1, s-1) * binomial(n-s-1, r-1) * (n-r-s)!.
a(n) ~ exp(2) * n! / n^2. - Vaclav Kotesovec, Jun 29 2016
a(n) = Sum_{k=0..n} hypergeom([k+1, k-n], [], -1). - Peter Luschny, Oct 05 2017
From Peter Bala, Jan 08 2018: (Start)
a(n) = Sum_{k = 0..n-2} k!*binomial(n+k-1, 2*k+1).
a(n) = (n - 2)*a(n-1) + a(n-2) + 2, with a(2) = 1, a(3) = 3.
a(n+2) = 1/n!*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)* A000522(n)^2.
Row sums of array A143409 read as a triangle.
O.g.f.: Sum_{n >= 0} n!*x^(n+2)/(1 - x)^(2*n+2). Cf. A000179, A000271, A000904 and A127548.
O.g.f. with offset 0: 1/(1 - x) o 1/(1 - x) = 1 + 3*x + 9*x^2 + 32*x^3 + ..., where o denotes the white diamond multiplication of power series. See the Bala link for details. (End)

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