A335700 -id:A335700 - 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

A000186 Number of 3 X n Latin rectangles in which the first row is in order.

Original entry on oeis.org

1, 0, 0, 2, 24, 552, 21280, 1073760, 70299264, 5792853248, 587159944704, 71822743499520, 10435273503677440, 1776780700509416448, 350461958856515690496, 79284041282622163140608, 20392765404792755583221760, 5917934230798104348783083520, 1924427226324694427836833857536

Offset: 0

N. J. A. Sloane

Or number of n X n matrices with exactly one 1 and one 2 in each row and column which are not in the main diagonal, other entries 0. - Vladimir Shevelev, Mar 22 2010

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 183.
Dulmage, A. L.; McMaster, G. E. A formula for counting three-line Latin rectangles. Proceedings of the Sixth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1975), pp. 279-289. Congressus Numerantium, No. XIV, Utilitas Math., Winnipeg, Man., 1975. MR0392611 (52 #13428). - From N. J. A. Sloane, Apr 06 2012
I. Gessel, Counting three-line Latin rectangles, Lect. Notes Math, 1234(1986), 106-111. [From Vladimir Shevelev, Mar 25 2010]
Goulden and Jackson, Combin. Enum., Wiley, 1983 p. 284.
S. M. Jacob, The enumeration of the Latin rectangle of depth three..., Proc. London Math. Soc., 31 (1928), 329-336.
S. M. Kerawala, The enumeration of the Latin rectangle of depth three by means of a difference equation, Bull. Calcutta Math. Soc., 33 (1941), 119-127.
S. M. Kerawala, The asymptotic number of three-deep Latin rectangles, Bull. Calcutta Math. Soc., 39 (1947), 71-72.
Koichi, Yamamoto, An asymptotic series for the number of three-line Latin rectangles, J. Math. Soc. Japan 1, (1950). 226-241.
W. Moser. A generalization of Riordan's formula for 3Xn Latin rectangles, Discrete Math., 40, 311-313 [From Vladimir Shevelev, Mar 25 2010]
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 210.
V. S. Shevelev, Reduced Latin rectangles and square matrices with identical sums in the rows and columns [Russian], Diskret. Mat., 4 (1992), no. 1, 91-110.
V. S. Shevelev, A generalized Riordan formula for three-rowed Latin rectangles and its applications, DAN of the Ukraine, 2 (1991), 8-12 (in Russian) [From Vladimir Shevelev, Mar 25 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).
D. S. Stones, The many formulas for the number of Latin rectangles, Electron. J. Combin 17 (2010), A1.
RJ Stones, S Lin, X Liu, G Wang, On Computing the Number of Latin Rectangles, Graphs and Combinatorics, Graphs and Combinatorics (2016) 32:1187-1202; DOI 10.1007/s00373-015-1643-1

N. J. A. Sloane, Table of n, a(n) for n = 0..207
F. Alayont and N. Krzywonos, Rook Polynomials in Three and Higher Dimensions, 2012. - From _N. J. A. Sloane_, Jan 02 2013
K. P. Bogart and J. Q. Longyear, Counting 3 by n Latin rectangles, Proc. Amer. Math. Soc., 54 (1976), 463-467.
S. M. Jacob, The enumeration of the Latin rectangle of depth three..., Proc. London Math. Soc., 31 (1928), 329-336.
S. M. Kerawala, The enumeration of the Latin rectangle of depth three by means of a difference equation, Bull. Calcutta Math. Soc., 33 (1941), 119-127. [Annotated scanned copy]
S. M. Kerawala, The asymptotic number of three-deep Latin rectangles, Bull. Calcutta Math. Soc., 39 (1947), 71-72. [Annotated scanned copy]
John Riordan, Table of a(0) - a(25), circa 1968
Vladimir Shevelev, Spectrum of permanent's values and its extremal magnitudes in Lambda_n^3 and Lambda_n(alpha, beta, gamma), arXiv preprint arXiv:1104.4051[math.CO], 2011.
D. S. Stones and I. M. Wanless, Divisors of the number of Latin rectangles, J. Combin. Theory Ser. A 117 (2010), 204-215.
Index entries for sequences related to Latin squares and rectangles

Cf. A000512, A000166, A000179, A335700, A170904.

for n from 1 to 250 do t0:=0; for j from 0 to n do for k from 0 to n-j do t0:=t0 + (2^j/j!)*k!*binomial(-3*(k+1), n-k-j); od: od: t0:=n!*t0; lprint(n,t0); od:
Maple code for A000186 based on Eq. (30) of Riordan, p. 205. Eq. (30a) on p. 206 is wrong. - N. J. A. Sloane, Jan 21 2010. Thanks to Neven Juric for correcting an error in the definition of fU, Mar 01 2010. Additional comment and modifications of code due to changes in underlying sequences from William P. Orrick, Aug 12 2020: Eq. (30) and Eq. (30a) are, in fact, related to each other by a trivial transformation and are both valid. Current code is based on Eq. (30a).
# A000166
unprotect(D);
D := proc(n) option remember; if n<=1 then 1-n else (n-1)*(D(n-1)+D(n-2));fi; end;
[seq(D(n), n=0..30)];
# A000179
U := proc(n) if n=0 then 1 else add ((-1)^k*(2*n)*binomial(2*n-k, k)*(n-k)!/(2*n-k), k=0..n); fi; end;
[seq(U(n), n=0..30)];
# A000186
K:=proc(n) local k; global D, U; add( binomial(n,k)*D(n-k)*D(k)*U(n-2*k), k=0..floor(n/2) ); end;
[seq(K(n), n=0..30)];
# another Maple program:
a:= proc(n) option remember; `if`(n<5, [1, 0, 0, 2, 24][n+1],
     ((n-1)*(n^2-2*n+2)*a(n-1) +(n-1)*(n-2)*(n^2-2*n+2)*a(n-2)
      +(n-1)*(n-2)*(n^2-2*n-2) *a(n-3)
      +2*(n-1)*(n-2)*(n-3)*(n^2-5*n+3) *a(n-4)
      -4*(n-2)*(n-3)*(n-4)*(n-1)^2 *a(n-5)) / (n-2))
    end:
seq(a(n), n=0..25);  # Alois P. Heinz, Nov 02 2013

a[n_] := (t0 = 0; Do[t0 = t0 + (2^j/j!)*k!*Binomial[-3*(k+1), n-k-j], {j, 0, n}, {k, 0, n-j}]; n!*t0); Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Oct 13 2011, after Maple *)

# after Maple code based on Riordan's Eq. (30a)
d = [1,0]
for j in range(2,31):
    d.append((j - 1) * (d[-1] + d[-2]))
def u(n):
    if n == 0:
        return 1
    else:
        return sum((-1)^k * (2 * n) * binomial(2 * n - k, k) * factorial(n - k) / (2 * n - k) for k in range(0, n + 1))
def k(n):
    return sum(binomial(n, k) * d[n - k] * d[k] * u(n - 2 * k) for k in range(0, floor(n / 2) + 1))
[k(n) for n in range(0, 31)] # William P. Orrick, Aug 12 2020

a(n) = n!*Sum_{k+j<=n} (2^j/j!)*k!*binomial(-3*(k+1), n-k-j).
Note that the formula Sum_{k=0..n, k <= n/2} binomial(n, k)*D(n-k)*D(k)*U(n-2*k), where D() = A000166 and U() represents the menage numbers given by Riordan, p. 209 gives the wrong answers unless we set U(1) = -1 (or in other words we must take U() = A000179). With U(1) = 0 (see A335700) it produces A170904. See the Maple code here. - N. J. A. Sloane, Jan 21 2010, Apr 04 2010. Thanks to Vladimir Shevelev for clarifying this comment. Additional changes from William P. Orrick, Aug 12 2020
E.g.f.: exp(2*x) Sum(n>=0; n! x^n /(1+x)^(3*n+3)) from Gessel reference. - Wouter Meeussen, Nov 02 2013
a(n) ~ n!^2/exp(3). - Vaclav Kotesovec, Sep 08 2016
a(n+p)-2*a(n) is divisible by p for any prime p. - Mark van Hoeij, Jun 13 2019

Formula and more terms from Vladeta Jovovic, Mar 31 2001

Edited by N. J. A. Sloane, Jan 21 2010, Mar 04 2010, Apr 04 2010

A102761 Same as A000179, except that a(0) = 2.

Original entry on oeis.org

2, -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, Apr 04 2010, following a suggestion from Vladimir Shevelev

For any integer n>=0, 2 * Integral_{t=-2..2} T_n(t/2)*exp(-t)*dt = 4 * Integral_{z=-1..1} T_n(z)*exp(-2*z)*dz = a(n)*exp(2) - A300484(n)*exp(-2). - Max Alekseyev, Mar 08 2018

J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 197.

Vladimir Shevelev, Spectrum of permanent's values and its extremal magnitudes in Λ_n^3 and Λ_n(α,β,γ), arXiv:1104.4051 [math.CO], 2011.

Row m=2 in A300481.
Cf. A000023, A000179, A000186, A300484.
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

{ A102761(n) = subst( serlaplace( 2*polchebyshev(n, 1, (x-2)/2)), x, 1); } \\ Max Alekseyev, Mar 06 2018

a(n) = Sum_{i=0..n} A127672(n,i) * A000023(i). - Max Alekseyev, Mar 06 2018
a(n) = A300481(2,n) = A300480(-2,n). - Max Alekseyev, Mar 06 2018
a(n) = A335391(0,n) (Touchard). - William P. Orrick, Aug 29 2020

Changed a(0)=2 (making the sequence more consistent with existing formulae) by Max Alekseyev, Mar 06 2018

A170904 Sequence obtained by a formal reading of Riordan's Eq. (30a), p. 206.

Original entry on oeis.org

1, 0, 0, 2, 24, 572, 21280, 1074390, 70299264, 5792903144, 587159944704, 71822748886440, 10435273503677440, 1776780701352504408, 350461958856515690496, 79284041282799128098778, 20392765404792755583221760, 5917934230798152486136427600, 1924427226324694427836833857536

Offset: 0

N. J. A. Sloane, Jan 21 2010

nonn

See the comments in A000186 for further discussion.
Neven Juric alerted me to the fact that Riordan's formula is misleading.
It is not error of Riordan, since, according to the rook theory, he considered U(1) to be -1. [Vladimir Shevelev, Apr 02 2010]
A combinatorial argument, valid for n >= 2, leads to Touchard's formula for the n-th menage number, U(n), a formula which involves the coefficients of Chebyshev polynomials of the first kind. It is combinatorially reasonable to take U(0) = 1 and U(1) = 0, leading to A335700, but taking the connection with Chebyshev polynomials seriously instead gives U(0) = 2 and U(1) = -1, leading to A102761. Riordan derives equation (30) on page 205 for the number of reduced three-line Latin rectangles (A000186) by making use of product identities on Chebyshev polynomials, and therefore requires the second definition; it also requires extending the definition of menage numbers to negative index. Riordan then obtains equation (30a) on page 206 by eliminating the negative indices and redefining U(0) to be 1 (which leads to A000179). A170904 (this sequence) is what is obtained by mistakenly using A335700 instead of A000179 in Riordan's equation (30a). - William P. Orrick, Aug 11 2020

J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 206, 209.

# A000166
unprotect(D);
D := proc(n) option remember; if n<=1 then 1-n else (n-1)*(D(n-1)+D(n-2)); fi; end;
[seq(D(n),n=0..30)];
# A335700 (equals A000179 except that A335700(1) = 0)
U := proc(n) if n<=1 then 1-n else add ((-1)^k*(2*n)*binomial(2*n-k, k)*(n-k)!/(2*n-k), k=0..n); fi; end;
[seq(U(n),n=0..30)];
# bad A000186 (A170904)
Kbad:=proc(n) local k; global D, U; add( binomial(n,k)*D(n-k)*D(k)*U(n-2*k), k=0..floor(n/2) ); end;
[seq(Kbad(n),n=0..30)];

One can enumerate 3 X n Latin rectangles by the formula A000186(2n)=a(2n) and A000186(2n+1)=a(2n+1)-A001700(n)*A000166(n)*A000166(n+1). - Vladimir Shevelev, Apr 04 2010

a(2n)=A000186(2n), a(2n+1)=A000186(2n+1)+A001700(n)*A000166(n)*A000166(n+1). [From Vladimir Shevelev, Apr 02 2010]

Edited by N. J. A. Sloane, Apr 04 2010 following a suggestion from Vladimir Shevelev

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.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

PARI

PARI

PARI

Python

Formula

Extensions

A000186 Number of 3 X n Latin rectangles in which the first row is in order.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

SageMath

Formula

Extensions

A102761 Same as A000179, except that a(0) = 2.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

PARI

Formula

Extensions

A170904 Sequence obtained by a formal reading of Riordan's Eq. (30a), p. 206.

Views

Author

Keywords

Comments

References

Programs

Maple

Formula

Extensions