cp's OEIS Frontend

a(n) counts permutations of [1,...,n+1] having no substring [k,k+1]. - Len Smiley, Oct 13 2001

Also, for n > 0, determinant of the tridiagonal n X n matrix M such that M(i,i)=i and for i=1..n-1, M(i,i+1)=-1, M(i+1,i)=i. - Mario Catalani (mario.catalani(AT)unito.it), Feb 04 2003

Also, for n > 0, maximal permanent of a nonsingular n X n (0,1)-matrix, which is achieved by the matrix with just n-1 0's, all on main diagonal. [For proof, see next entry.] - W. Edwin Clark, Oct 28 2003

Proof from Richard Brualdi and W. Edwin Clark, Nov 15 2003: Let n >= 4. Take an n X n (0,1)-matrix A which is nonsingular. It has t >= n-1, 0's, otherwise there will be two rows of all 1's. Let B be the matrix obtained from A by replacing t-(n-1) of A's 0's with 1's. Let D be the matrix with all 1's except for 0's in the first n-1 positions on the diagonal. This matrix is easily seen to be non-singular. Now we have per(A) < = per(B) < = per (D), where the first inequality follows since replacing 0's by 1's cannot decrease the permanent and the second from Corollary 4.4 in the Brualdi et al. reference, which shows that per(D) is the maximum permanent of ANY n X n matrix with n -1 0's. Corollary 4.4 requires n >= 4. a(n) for n < 4 can be computed directly.

With offset 1, permanent of (0,1)-matrix of size n X (n+d) with d=1 and n zeros not on a line. This is a special case of Theorem 2.3 of Seok-Zun Song et al., Extremes of permanents of (0,1)-matrices, pp. 201-202. - Jaap Spies, Dec 12 2003

Number of fixed-point-free permutations of n+2 that begin with a 2; e.g., for 1234, we have 2143, 2341, 2413, so a(2)=3. Also number of permutations of 2..n+2 that have no agreements with 1..n+1. E.g., for 123 against permutations of 234, we have 234, 342 and 432. Compare A047920. - Jon Perry, Jan 23 2004. [This can be proved by the standard argument establishing that d(n+2) = (n+1)(d(n+1)+d(n)) for derangements A000166 (n+1 choices of where 1 goes, then either 1 is in a transposition, or in a cycle of length at least 3, etc.). - D. G. Rogers, Aug 28 2006]

Stirling transform of A006252(n+1)=[1,1,2,4,14,38,...] is a(n)=[1,3,11,53,309,...]. - Michael Somos, Mar 04 2004

a(n+1) is the sequence of numerators of the self-convergents to 1/(e-2); see A096654. - Clark Kimberling, Jul 01 2004

Euler's interpretation was "fixedpoint-free permutations beginning with 2" and he listed the terms up to 148329 (although he was blind at the time). - Don Knuth, Jan 25 2007

Equals lim_{k->infinity} A153869^k. - Gary W. Adamson, Jan 03 2009

Hankel transform is A059332. - Paul Barry, Apr 22 2009

This sequence appears in the analysis of Euler's divergent series 1 - 1! + 2! - 3! + 4! ... by Lacroix, see Hardy. For information about this and related divergent series see A163940. - Johannes W. Meijer, Oct 16 2009

a(n), n >= 1, enumerates also the ways to distribute n beads, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and one open cord allowed to have any number of beads. Each beadless necklace as well as the beadless cord contributes a factor 1 in the counting, e.g., a(0):=1*1=1. There are k! possibilities for the cord with k>=0 beads, which means that the two ends of the cord should be considered as fixed, in short: a fixed cord. This produces for a(n) the exponential (aka binomial) convolution of the sequences {n!=A000142(n)} and the subfactorials {A000166(n)}.

See the formula below. Alternatively, the e.g.f. for this problem is seen to be (exp(-x)/(1-x))*(1/(1-x)), namely the product of the e.g.f.s for the subfactorials (from the unordered necklace problem, without necklaces with exactly one bead) and the factorials (from the fixed cord problem). Therefore the recurrence with inputs holds also. 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 02 2010

a(n) = (n-1)a(n-1) + (n-2)a(n-2) gives the same sequence offset by a 1. - Jon Perry, Sep 20 2012

Also, number of reduced 2 X (n+2) Latin rectangles. - A.H.M. Smeets, Nov 03 2013

Second column of Euler's difference table (second diagonal in example of A068106). - Enrique Navarrete, Dec 13 2016

If we partition the permutations of [n+2] in A000166 according to their starting digit, we will get (n+1) equinumerous classes each of size a(n) (the class starting with the digit 1 is empty since no derangement starts with 1). Hence, A000166(n+2)=(n+1)*a(n), so a(n) is the size of each nonempty class of permutations of [n+2] in A000166. For example, for n=3 we have 44=4*11 (see link). - Enrique Navarrete, Jan 11 2017

For n >= 1, the number of circular permutations (in cycle notation) on [n+2] that avoid substrings (j,j+2), 1 <= j <= n. For example, for n=2, the 3 circular permutations in S4 that avoid substrings {13,24} are (1234),(1423),(1432). Note that each of these circular permutations represent 4 permutations in one-line notation (see link 2017). - Enrique Navarrete, Feb 15 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) holding for all n and k, which in turn is easily proved by induction making use of the given recurrences. - Peter Bala, Nov 21 2017

Number of permutations of [n] where the k-th fixed points are k-colored and all other points are unicolored. - Alois P. Heinz, Apr 28 2025

The last element of each row is 1, corresponding to the n X n "all 1" matrix with permanent = n!. The first 4 rows were provided by Wouter Meeussen. The 6th row was computed by Gordon F. Royle: 13906734081, 2722682160, 4513642920, 3177532800, 4466769300, 2396826720, 3710999520, 2065521600, 3253760550, 1468314000, 2641593600, 1350475200, 2210277600, 1034061120,... .

The following is Max Alekseyev's proof of the formula: Suppose that we have a (0,1)-matrix M with permanent equal to 1. Then in M there is a unique set of n elements, each equal to 1, whose product makes the permanent equal 1. Permute the columns of M so that these n elements become arranged along the main diagonal, and denote the resulting matrix by M'. It is clear that each M' corresponds to n! different matrices M (this is where the factor n! in the formula comes from).

Let M'' be the same as M' except for zeros on the main diagonal. Then the permanent of M'' is zero. Viewing M'' as an adjacency matrix of a directed graph G, we notice that G cannot have a cycle. Indeed, if there is a cycle x_1 -> x_2 -> ... -> x_k -> x_1, then the set of elements (x_1,x_2), (x_2,x_3), ..., (x_k,x_1) together with (y_1,y_1), ..., (y_{n-k},y_{n-k}), where { y_1, ..., y_{n-k} } is the complement of { x_1, ..., x_k } in the set { 1, 2, ..., n }, form a set of elements of the matrix M' whose product is 1, making the permanent of M' greater than 1.

This works in the reverse direction as well, resulting in the statement: The permanent of M' is 1 if and only if M'' represents the adjacency matrix of some DAG. Therefore there exist A003024(n) distinct matrices M'. - Vladeta Jovovic, Oct 27 2009

This sequence was first provided by Jaap Spies.

a(n) = number of real non-singular (0,1)-matrices of order n having maximal permanent = A000255(n). Proof: [W. Edwin Clark and Richard Brualdi] The maximum permanent is per A where A has all 1's except for n-1 0's on the main diagonal. By Corollary 4.4 in the Brualdi et al. reference for n >= 4 any n X n (0,1)-matrix B with per B = per A can be obtained from A by permuting rows and columns. Since there are n ways to place the single 1 on the main diagonal and then n! ways to permute the distinct rows, a(n) = n*n! if n >=4. Direct computation shows this also holds for n = 1 and 3. - W. Edwin Clark, Nov 15 2003