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

First differs from A334442 for reversed partitions of 9. Namely, this sequence has (1,4,4) before (2,2,5), while A334442 has (2,2,5) before (1,4,4). - Gus Wiseman, May 07 2020

This is the "Abramowitz and Stegun" ordering of the partitions, referenced in numerous other sequences. The partitions are in reverse order of the conjugates of the partitions in Mathematica order (A080577). Each partition is the conjugate of the corresponding partition in Maple order (A080576). - Franklin T. Adams-Watters, Oct 18 2006

The "Abramowitz and Stegun" ordering of the partitions is the graded reflected colexicographic ordering of the partitions. - Daniel Forgues, Jan 19 2011

The "Abramowitz and Stegun" ordering of partitions has been traced back to C. F. Hindenburg, 1779, in the Knuth reference, p. 38. See the Hindenburg link, pp. 77-5 with the listing of the partitions for n=10. This is also mentioned in the P. Luschny link. - Wolfdieter Lang, Apr 04 2011

The "Abramowitz and Stegun" order used here means that the partitions of a given number are listed by increasing number of (nonzero) parts, then by increasing lexicographical order with parts in (weakly) indecreasing order. This differs from n=9 on from A334442 which considers reverse lexicographic order of parts in (weakly) decreasing order. - M. F. Hasler, Jul 12 2015, corrected thanks to Gus Wiseman, May 14 2020

This is the Abramowitz-Stegun ordering of reversed partitions (finite weakly increasing sequences of positive integers). The same ordering of non-reversed partitions is A334301. - Gus Wiseman, May 07 2020

First differs from A334439 for partitions of 9. Namely, this sequence has (4,4,1) before (5,2,2), while A334439 has (5,2,2) before (4,4,1). - Gus Wiseman, May 08 2020

This is also a list of all the possible prime signatures of a number, arranged in graded colexicographic ordering. - N. J. A. Sloane, Feb 09 2014

This is also the Abramowitz-Stegun ordering of reversed partitions (A036036) if the partitions are reversed again after sorting. Partitions sorted first by sum and then colexicographically are A211992. - Gus Wiseman, May 08 2020

Differs from A080576 in a(18): Here, (...,1+3,2+2,4), there (...,2+2,1+3,4).

The representation of the partitions (for fixed n) is as (weakly) increasing lists of parts, the order between individual partitions (for the same n) is lexicographic (see example). - Joerg Arndt, Sep 03 2013

The equivalent sequence for compositions (ordered partitions) is A228369. - Omar E. Pol, Oct 19 2019

The sequence of row lengths of this array is p(n) = A000041(n) (partition numbers).

The sequence of row sums is A006128(n).

The number of times k appears in row n is A008284(n,k). - Franklin T. Adams-Watters, Jan 12 2006

The next level (row) gets created from each node by adding one or two more nodes. If a single node is added, its value is one more than the value of its parent. If two nodes are added, the first is equal in value to the parent and the value of the second is one more than the value of the parent. See A128628. - Alford Arnold, Mar 27 2007

The 1's in the (flattened) sequence mark the start of a new row, the value that precedes the 1 equals the row number minus one. (I.e., the 1 preceded by a 0 is the start of row 1, the 1 preceded by a 6 is the start of row 7, etc.) - M. F. Hasler, Jun 06 2018

Also the maximum part in the n-th partition in graded lexicographic order (sum/lex, A193073). - Gus Wiseman, May 24 2020

First differs from A036037 for partitions of 9. Namely, this sequence has (5,2,2) before (4,4,1), while A036037 has (4,4,1) before (5,2,2).

This is the Abramowitz-Stegun ordering of integer partitions (A334301) except that the finer order is reverse-lexicographic instead of lexicographic. The version for reversed partitions is A334302.

First differs from A334438 (shifted left once) at a(75) = 98, A334438(76) = 99. - Gus Wiseman, May 20 2020

This mapping of the set of all partitions of N >= 0 to {1, 2, 3, ...} (set of natural numbers) is one to one (bijective). The empty partition for N = 0 maps to 1.

A129129 seems to be analogous, except that the partition ordering A080577 is used. This ordering, however, does not care about the number of parts: e.g., 1^2,4 = 4,1^2 comes before 3^2, so a(23)=28 and a(22)=25 are interchanged.

Also Heinz numbers of all reversed integer partitions (finite weakly increasing sequences of positive integers), sorted first by sum, then by length, and finally lexicographically, where the Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). The version for non-reversed partitions is A334433. - Gus Wiseman, May 20 2020

The representation of the partitions (for fixed n) is as (weakly) increasing lists of parts, the order between individual partitions (for the same n) is (list-)reversed lexicographic; see examples. [Joerg Arndt, Sep 03 2013]

Also compositions in the triangle of A066099 that are in nondecreasing order.

The equivalent sequence for compositions (ordered partitions) is A066099.

Row n has length A006128(n).

Row sums give A066186.

First differs from A185974 shifted left once at a(76) = 99, A185974(75) = 98.

A permutation of the positive integers.

This is the Abramowitz-Stegun ordering of integer partitions (A334433) except that the finer order is reverse-lexicographic instead of lexicographic. The version for reversed partitions is A334435.

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). This gives a bijective correspondence between positive integers and integer partitions.

As a triangle with row lengths A000041, the sequence starts {{1},{2},{3,4},{5,6,8},...}, so offset is 0.

Note that a(n) = (a(n-1) + a(n+1))/(n+1). - T. D. Noe, Oct 12 2012; corrected by Gary Detlefs, Oct 26 2018

a(n) = log_2(A073888(n)) = log_3(A073889(n)).

a(n) equals minus the determinant of M(n+2) where M(n) is the n X n symmetric tridiagonal matrix with entries 1 just above and below its diagonal and diagonal entries 0, 1, 2, .., n-1. Example: M(4)=matrix([[0, 1, 0, 0], [1, 1, 1, 0], [0, 1, 2, 1], [0, 0, 1, 3]]). - Roland Bacher, Jun 19 2001

a(n) = A221913(n,-1), n>=1, is the numerator sequence of the n-th approximation of the continued fraction -(0 + K_{k>=1} (-1/k)) = 1/(1-1/(2-1/(3-1/(4-... The corresponding denominator sequence is A058797(n). - Wolfdieter Lang, Mar 08 2013

The recurrence equation a(n+1) = (A*n + B)*a(n) + C*a(n-1) with the initial conditions a(0) = 0, a(1) = 1 has the solution a(n) = Sum_{k = 0..floor((n-1)/2)} C^k*binomial(n-k-1,k)*( Product_{j = 1..n-2k-1} (k+j)*A + B ). This is the case A = 1, B = 1, C = -1. - Peter Bala, Aug 01 2013