cp's OEIS Frontend

T(n,k) is the number of subsets of {1,2,...,n-1} of size k and containing no consecutive integers. Example: T(6,2)=6 because the subsets of size 2 of {1,2,3,4,5} with no consecutive integers are {1,3},{1,4},{1,5},{2,4},{2,5} and {3,5}. Equivalently, T(n,k) is the number of k-matchings of the path graph P_n. - Emeric Deutsch, Dec 10 2003

T(n,k) = number of compositions of n+2 into k+1 parts, all >= 2. Example: T(6,2)=6 because we have (2,2,4),(2,4,2),(4,2,2),(2,3,3),(3,2,3) and (3,3,2). - Emeric Deutsch, Apr 09 2005

Given any recurrence sequence S(k) = x*a(k-1) + a(k-2), starting (1, x, x^2+1, ...); the (k+1)-th term of the series = f(x) in the k-th degree polynomial: (1, (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 + 1), ...). Example: let x = 2, then S(k) = 1, 2, 5, 12, 29, 70, 169, ... such that A000129(7) = 169 = f(x), x^6 + 5x^4 + 6x^2 + 1 = (64 + 80 + 24 + 1). - Gary W. Adamson, Apr 16 2008

Row k gives the nonzero coefficients of U(k,x/2) where U is the Chebyshev polynomial of the second kind. For example, row 6 is 1,5,6,1 and U(6,x/2) = x^6 - 5x^4 + 6x^2 - 1. - David Callan, Jul 22 2008

T(n,k) is the number of nodes at level k in the Fibonacci tree f(k-1). The Fibonacci trees f(k) of order k are defined as follows: 1. f(-1) and f(0) each consist of a single node. 2. For k >= 1, to the root of f(k-1), taken as the root of f(k), we attach with a rightmost edge the tree f(k-2). See the Iyer and Reddy references. These trees are not the same as the Fibonacci trees in A180566. Example: T(3,0)=1 and T(3,1)=2 because in f(2) = /\ we have 1 node at level 0 and 2 nodes at level 1. - Emeric Deutsch, Jun 21 2011

Triangle, with zeros omitted, given by (1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011

Riordan array (1/(1-x),x^2/(1-x). - Philippe Deléham, Dec 12 2011

This sequence is the elements on the rising diagonals of the Pascal triangle, where the sum of the elements in each rising diagonal represents a Fibonacci number. - Mohammad K. Azarian, Mar 08 2012

If we set F(0;x) = 0, F(1;x) = 1, F(n+1;x) = x*F(n;x) + F(n-1;x), then we obtain the sequence of Vieta-Fibonacci polynomials discussed by Gary W. Adamson above. We note that F(n;x) = (-i)^n * U(n;i*x/2), where U denotes the respective Chebyshev polynomial of the second kind (see David Callan's remark above). Let us fix a,b,f(0),f(1) in C, b is not the zero, and set f(n) = a*f(n-1) + b*f(n-2). Then we deduce the relation: f(n) = b^((n-1)/2) * F(n;a/sqrt(b))*f(1) + b^(n/2) * F(n-1;a/sqrt(b))*f(0), where for a given value of the complex root sqrt(b) we set b^(n/2) = (sqrt(b))^n. Moreover, if b=1 then we get f(n+k) + (-1)^k * f(n-k) = L(k;a)*f(n), for every k=0,1,...,n, and where L(0;a)=2, L(1;a)=a, L(n+1;a)=a*L(n;a) + L(n-1;a) are the Vieta-Lucas polynomials. Let us observe that L(n+2;a) = F(n+2;a) + F(n;a), L(m+n;a) = L(m;a)*F(n;a) + L(m-1;a)*F(n-1;a), which implies also L(n+1;a) = a*F(n;a) + 2*F(n-1;a). Further we have L(n;a) = 2*(-i)^n * T(n;i*x/2), where T(n;x) denotes the n-th Chebyshev polynomial of the first kind. For the proofs, other relations and facts - see Witula-Slota's papers. - Roman Witula, Oct 12 2012

The diagonal sums of this triangle are A000930. - John Molokach, Jul 04 2013

Aside from signs and index shift, the coefficients of the characteristic polynomial of the Coxeter adjacency matrix for the Coxeter group A_n related to the Chebyshev polynomial of the second kind (cf. Damianou link p. 19). - Tom Copeland, Oct 11 2014

For a mirrored, shifted version showing the relation of these coefficients to the Pascal triangle, Fibonacci, and other number triangles, see A030528. See also A053122 for a relation to Cartan matrices. - Tom Copeland, Nov 04 2014

For a relation to a formulation for a universal Lie Weyl algebra for su(1,1), see page 16 of Durov et al. - Tom Copeland, Nov 29 2014

A reversed, signed and aerated version is given by A049310, related to Chebyshev polynomials. - Tom Copeland, Dec 06 2015

For n >= 3, the n-th row gives the coefficients of the independence polynomial of the (n-2)-path graph P_{n-2}. - Eric W. Weisstein, Apr 07 2017

For n >= 2, the n-th row gives the coefficients of the matching-generating polynomial of the (n-1)-path graph P_{n-1}. - Eric W. Weisstein, Apr 10 2017

Antidiagonals of the Pascal matrix A007318 read bottom to top. These are also the antidiagonals read from top to bottom of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper), which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Reverse is A102426. - Tom Copeland, Jul 02 2018

T(n,k) is the number of Markov equivalence classes with skeleton the path on n+1 nodes having exactly k immoralities. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018

T(n, k) = number of compositions of n+1 into n+1-2*k odd parts. For example, T(6,2) = 6 because 7 = 5+1+1 = 3+3+1 = 3+1+3 = 1+1+5 = 1+3+3 = 1+1+5. - Michael Somos, Sep 19 2019

From Gary W. Adamson, Apr 25 2022: (Start)

Alternate rows can be parsed into those with odd integer coefficients to the right of the leftmost 1, and those with even integer coefficients to the right of the leftmost 1. The first set is shown in A054142 and are characteristic polynomials of submatrices of an infinite tridiagonal matrix (A332602) with all -1's in the super and subdiagonals and (1,2,2,2,...) as the main diagonal. For example, the characteristic equation of the 3 X 3 submatrix (1,-1,0; -1,2,-1; 0,-1,2) is x^3 - 5x^2 + 6x - 1. The roots are the Beraha constants B(7,1) = 3.24697...; B(7,2) = 1.55495...; and B(7,3) = 0.198062.... For n X n matrices of this form, the largest eigenvalue is B(2n+1, 1). The 3 X 3 matrix has an eigenvalue of 3.24697... = B(7,1).

Polynomials with even integer coefficients to the right of the leftmost 1 are in A053123 with roots being the even-indexed Beraha constants. The generating Cartan matrices are those with (2,2,2,...) as the main diagonal and -1's as the sub- and superdiagonals. The largest eigenvalue of n X n matrices of this form are B(2n+2,1). For example, the largest eigenvalue of (2,-1,0; -1,2,-1; 0,-1,2) is 3.414... = B(8,1) = a root to x^3 - 6x^2 + 10x - 4. (End)

T(n,k) is the number of edge covers of P_(n+2) with (n-k) edges. For example, T(6,2)=6 because among edges 1, 2, ..., 7 of P_8, we can eliminate any two non-consecutive edges among 2-6. These numbers can be found using the recurrence relation for the edge cover polynomial of P_n, which is E(P_n,x) = xE(P_(n-1),x)+xE(P_(n-2),x) and E(P_1,x)=0, E(P_2,x)=x (ref. Akbari and Oboudi). - Feryal Alayont, Jun 03 2022

T(n,k) is the number of ways to tile an n-board (an n X 1 array of 1 X 1 cells) using k dominoes and n-2*k squares. - Michael A. Allen, Dec 28 2022

T(n,k) is the number of positive integer sequences (s(1),s(2),...,s(n-2k)) such that s(i) < s(i+1), s(1) is odd, s(n-2k) <= n, and s(i) and s(i+1) have opposite parity (ref. Donnelly, Dunkum, and McCoy). Example: T(6,0)=1 corresponds to 123456; T(6,1)=5 corresponds to 1234, 1236, 1256, 1456, 3456; T(6,2)=6 corresponds to 12, 14, 16, 34, 36; and T(6,3)=1 corresponds to the empty sequence () with length 0. - Molly W. Dunkum, Jun 27 2023

a(n) is the number of permutations in the symmetric group S_n that have a fixed point, i.e., they are not derangements (A000166). - Ahmed Fares (ahmedfares(AT)my-deja.com), May 08 2001

a(n+1)=p(n+1) where p(x) is the unique degree-n polynomial such that p(k)=k! for k=0,1,...,n. - Michael Somos, Oct 07 2003

The termwise sum of this sequence and A000166 gives the factorial numbers. - D. G. Rogers, Aug 26 2006, Jan 06 2008

a(n) is the number of deco polyominoes of height n and having in the last column an odd 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. Example: a(2)=1 because the horizontal domino is the only deco polyomino of height 2 having an odd number of cells in the last column. - Emeric Deutsch, May 08 2008

Starting (1, 4, 15, 76, 455, ...) = eigensequence of triangle A127899 (unsigned). - Gary W. Adamson, Dec 29 2008

(n-1) | a(n), hence a(n) is never prime. - Jonathan Vos Post, Mar 25 2009

a(n) is the number of permutations of [n] that have at least one fixed point = number of positive terms in n-th row of the triangle in A170942, n > 0. - Reinhard Zumkeller, Mar 29 2012

Numerator of partial sum of alternating harmonic series, provided that the denominator is n!. - Richard Locke Peterson, May 11 2020

a(n) is the number of terms in the polynomial expansion of the determinant of a n X n matrix that contains at least one diagonal element. - Adam Wang, May 28 2025

Number of edges of the complete bipartite graph of order n+n^n, K_n,n^n. - Roberto E. Martinez II, Jan 07 2002

All rational solutions to the equation x^y = y^x, with x < y, are given by x = A000169(n+1)/A000312(n), y = A000312(n+1)/A007778(n), where n >= 1. - Nick Hobson, Nov 30 2006

a(n) is also the number of ways of writing an n-cycle as the product of n+1 transpositions. - Nikos Apostolakis, Nov 22 2008

a(n) is the total number of elements whose preimage is the empty set summed over all partial functions from [n] into [n]. - Geoffrey Critzer, Jan 12 2022

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)

Also number of partitions of n with positive crank (n>=2), cf. A064391. - Vladeta Jovovic, Sep 30 2001

Number of smooth weakly unimodal compositions of n into positive parts such that the first and last part are 1 (smooth means that successive parts differ by at most one), see example. Dropping the requirement for unimodality gives A186085. - Joerg Arndt, Dec 09 2012

Number of weakly unimodal compositions of n where the maximal part m appears at least m times, see example. - Joerg Arndt, Jun 11 2013

Also weakly unimodal compositions of n with first part 1, maximal up-step 1, and no consecutive up-steps; see example. The smooth weakly unimodal compositions are recovered by shifting all rows above the bottom row to the left by one position with respect to the next lower row. - Joerg Arndt, Mar 30 2014

It would seem from Stanley that he regards a(0)=0 for this sequence and A001523. - Michael Somos, Feb 22 2015

From Gus Wiseman, Mar 30 2021: (Start)

Also the number of odd-length compositions of n with alternating parts strictly decreasing. These are finite odd-length sequences q of positive integers summing to n such that q(i) > q(i+2) for all possible i. The even-length version is A064428. For example, the a(1) = 1 through a(9) = 14 compositions are:

(1) (2) (3) (4) (5) (6) (7) (8) (9)

(211) (221) (231) (241) (251) (261)

(311) (312) (322) (332) (342)

(321) (331) (341) (351)

(411) (412) (413) (423)

(421) (422) (432)

(511) (431) (441)

(512) (513)

(521) (522)

(611) (531)

(612)

(621)

(711)

(32211)

(End)

In the Ferrers diagram of a partition x of n, count the dots in each diagonal parallel to the main diagonal (starting at the top-right, say). The result diag(x) is a smooth weakly unimodal composition of n into positive parts such that the first and last part are 1. For example, diag(5541) = 11233221. The function diag is many-to-one; the size of its codomain as a set is a(n). If diag(x) = diag(y), each hook of x can be slid by the same amount past the main diagonal to get y. For example, diag(5541) = diag(44331). - George Beck, Sep 26 2021

From Gus Wiseman, May 23 2022: (Start)

Conjecture: Also the number of integer partitions y of n with a fixed point y(i) = i. These partitions are ranked by A352827. The conjecture is stated at A238395, but Resta tells me he may not have had a proof. The a(1) = 1 through a(8) = 10 partitions are:

(1) (11) (111) (22) (32) (42) (52) (62)

(1111) (221) (222) (322) (422)

(11111) (321) (421) (521)

(2211) (2221) (2222)

(111111) (3211) (3221)

(22111) (4211)

(1111111) (22211)

(32111)

(221111)

(11111111)

Note that these are not the same partitions (compare A352827 to A352874), only the same count (apparently).

(End)

The above conjecture is true. See Section 4 of the Blecher-Knopfmacher paper in the Links section. - Jeremy Lovejoy, Sep 26 2022

Also, table of Pochhammer sequences read by antidiagonals (see Rudolph-Lilith, 2015). - N. J. A. Sloane, Mar 31 2016

Reverse of A008279. Row sums are A000522. Diagonal sums are A003470. Rows of inverse matrix begin {1}, {-1,1}, {0,-2,1}, {0,0,-3,1}, {0,0,0,-4,1} ... The signed lower triangular matrix (-1)^(n+k)n!/k! has as row sums the signed rencontres numbers Sum_{k=0..n} (-1)^(n+k)n!/k!. (See A000166). It has matrix inverse 1 1,1 0,2,1 0,0,3,1 0,0,0,4,1,...

Exponential Riordan array [1/(1-x),x]; column k has e.g.f. x^k/(1-x). - Paul Barry, Mar 27 2007

From Tom Copeland, Nov 01 2007: (Start)

T is the umbral extension of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! * Lag(n,x,-1-n) = Sum_{j=0..n} binomial(n,j) * j! * x^(n-j) = Sum_{j=0..n} (n!/j!) x^j. The inverse operator is A132013 with generalizations discussed in A132014.

b = T*a can be characterized several ways in terms of a(n) and b(n) or their o.g.f.'s A(x) and B(x).

1) b(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0], umbrally,

2) b(n) = (-1)^n n! Lag(n,a(.),-1-n)

3) b(n) = Sum_{j=0..n} (n!/j!) a(j)

4) B(x) = (1-xDx)^(-1) A(x), formally

5) B(x) = Sum_{j=0,1,...} (xDx)^j A(x)

6) B(x) = Sum_{j=0,1,...} x^j * D^j * x^j A(x)

7) B(x) = Sum_{j=0,1,...} j! * x^j * L(j,-:xD:,0) A(x) where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the operator series at the j = n term gives an o.g.f. for b(0) through b(n).

c = (0!,1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314 so T(n,k) = binomial(n,k)*c(n-k). The reciprocal sequence is d = (1,-1,0,0,0,...). (End)

From Peter Bala, Jul 10 2008: (Start)

This array is the particular case P(1,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:

n\k|0.....................1...............2.......3......4

----------------------------------------------------------

0..|1.....................................................

1..|a....................1................................

2..|a(a+b)...............2a..............1................

3..|a(a+b)(a+2b).........3a(a+b).........3a........1......

4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1

...

The entries A(n,k) of this array satisfy the recursion A(n,k) = (a+b*(n-k-1))*A(n-1,k) + A(n-1,k-1), which reduces to the Pascal formula when a = 1, b = 0.

Various cases are recorded in the database, including: P(1,0) = Pascal's triangle A007318, P(2,0) = A038207, P(3,0) = A027465, P(2,1) = A132159, P(1,3) = A136215 and P(2,3) = A136216.

When b <> 0 the array P(a,b) has e.g.f. exp(x*y)/(1-b*y)^(a/b) = 1 + (a+x)*y + (a*(a+b)+2a*x+x^2)*y^2/2! + (a*(a+b)*(a+2b) + 3a*(a+b)*x + 3a*x^2+x^3)*y^3/3! + ...; the array P(a,0) has e.g.f. exp((x+a)*y).

We have the matrix identities P(a,b)*P(a',b) = P(a+a',b); P(a,b)^-1 = P(-a,b).

An analog of the binomial expansion for the row entries of P(a,b) has been proved by [Echi]. Introduce a (generally noncommutative and nonassociative) product ** on the ring of polynomials in two variables by defining F(x,y)**G(x,y) = F(x,y)G(x,y) + by^2*d/dy(G(x,y)).

Define the iterated product F^(n)(x,y) of a polynomial F(x,y) by setting F^(1) = F(x,y) and F^(n)(x,y) = F(x,y)**F^(n-1)(x,y) for n >= 2. Then (x+a*y)^(n) = x^n + C(n,1)*a*x^(n-1)*y + C(n,2)*a*(a+b)*x^(n-2)*y^2 + ... + C(n,n)*a*(a+b)*(a+2b)*...*(a+(n-1)b)*y^n. (End)

(n+1) * n-th row = reversal of triangle A068424: (1; 2,2; 6,6,3; ...) - Gary W. Adamson, May 03 2009

Let G(m, k, p) = (-p)^k*Product_{j=0..k-1}(j - m - 1/p) and T(n,k,p) = G(n-1,n-k,p) then T(n, k, 1) is this sequence, T(n, k, 2) = A112292(n, k) and T(n, k, 3) = A136214. - Peter Luschny, Jun 01 2009, revised Jun 18 2019

The higher order exponential integrals E(x,m,n) are defined in A163931. For a discussion of the asymptotic expansions of the E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - ...) see A130534. The asymptotic expansion of E(x,m=1,n) leads for n >= 1 to the left hand columns of the triangle given above. Triangle A165674 is generated by the asymptotic expansions of E(x,m=2,n). - Johannes W. Meijer, Oct 07 2009

T(n,k) = n!/k! = number of permutations of [n+1] with exactly k+1 cycles and with elements 1,2,...,k+1 in separate cycles. See link and example below. - Dennis P. Walsh, Jan 24 2011

T(n,k) is the number of n permutations that leave some size k subset of {1,2,...,n} fixed. Sum_{k=0..n}(-1)^k*T(n,k) = A000166(n) (the derangements). - Geoffrey Critzer, Dec 11 2011

T(n,k) = A162995(n-1,k-1), 2 <= k <= n; T(n,k) = A173333(n,k), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012

The row polynomials form an Appell sequence. The matrix is a special case of a group of general matrices sketched in A132382. - Tom Copeland, Dec 03 2013

For interpretations in terms of colored necklaces, see A213936 and A173333. - Tom Copeland, Aug 18 2016

See A008279 for a relation of this entry to the e.g.f.s enumerating the faces of permutahedra and stellahedra. - Tom Copeland, Nov 14 2016

Also, T(n,k) is the number of ways to arrange n-k nonattacking rooks on the n X (n-k) chessboard. - Andrey Zabolotskiy, Dec 16 2016

The infinitesimal generator of this triangle is the generalized exponential Riordan array [-log(1-x), x] and equals the unsigned version of A238363. - Peter Bala, Feb 13 2017

Formulas for exponential and power series infinitesimal generators for this triangle T are given in Copeland's 2012 and 2014 formulas as T = unsigned exp[(I-A238385)] = 1/(I - A132440), where I is the identity matrix. - Tom Copeland, Jul 03 2017

If A(0) = 1/(1-x), and A(n) = d/dx(A(n-1)), then A(n) = n!/(1-x)^(n+1) = Sum_{k>=0} (n+k)!/k!*x^k = Sum_{k>=0} T(n+k, k)*x^k. - Michael Somos, Sep 19 2021

For a partition p, let l(p) = largest part of p, w(p) = number of 1's in p, m(p) = number of parts of p larger than w(p). The crank of p is given by l(p) if w(p) = 0, otherwise m(p)-w(p).

From Gus Wiseman, Mar 30 2021 and May 21 2022: (Start)

Also the number of even-length compositions of n with alternating parts strictly decreasing, or properly 2-colored partitions (proper = no equal parts of the same color) with the same number of parts of each color, or ordered pairs of strict partitions of the same length with total n. The odd-length case is A001522, and there are a total of A000041 compositions with alternating parts strictly decreasing (see A342528 for a bijective proof). The a(2) = 1 through a(7) = 8 ordered pairs of strict partitions of the same length are:

(1)(1) (1)(2) (1)(3) (1)(4) (1)(5) (1)(6)

(2)(1) (2)(2) (2)(3) (2)(4) (2)(5)

(3)(1) (3)(2) (3)(3) (3)(4)

(4)(1) (4)(2) (4)(3)

(5)(1) (5)(2)

(21)(21) (6)(1)

(21)(31)

(31)(21)

Conjecture: Also the number of integer partitions y of n without a fixed point y(i) = i, ranked by A352826. This is stated at A238394, but Resta tells me he may not have had a proof. The a(2) = 1 through a(7) = 8 partitions without a fixed point are:

(2) (3) (4) (5) (6) (7)

(21) (31) (41) (33) (43)

(211) (311) (51) (61)

(2111) (411) (331)

(3111) (511)

(21111) (4111)

(31111)

(211111)

The version for permutations is A000166, complement A002467.

The version for compositions is A238351.

This is column k = 0 of A352833.

A238352 counts reversed partitions by fixed points, rank statistic A352822.

A238394 counts reversed partitions without a fixed point, ranked by A352830.

A238395 counts reversed partitions with a fixed point, ranked by A352872. (End)

The above conjecture is true. See Section 4 of the Blecher-Knopfmacher paper in the Links section. - Jeremy Lovejoy, Sep 26 2022

In other words, T(n,k) is the number of permutations of n letters that are at Hammimg distance k from the identity permutation (cf. Diaconis, p. 112). - N. J. A. Sloane, Mar 02 2007

The sequence d(n) = 1, 0, 1, 2, 9, 44, 265, 1854, 14833, ... (A000166) is the number of derangements, that is, the number of permutations of n distinct, ordered items in which none of the items is in its natural ordered position.

a(n) is also the number of permutations of [n] having no circular succession. A circular succession in a permutation p of [n] is either a pair p(i), p(i+1), where p(i+1)=p(i)+1 or the pair p(n), p(1) if p(1)=p(n)+1. a(4)=8 because we have 1324, 1432, 4132, 2143, 2413, 3214, 3241, and 4321. - Emeric Deutsch, Sep 06 2010

a(n) is also the number of permutations of [n] having no substring in {12, 23, ..., (n-1)n, n1}. For example, a(4) = 8 since we have 1324, 1432, 4213, 2143, 2431, 3214, 3142, 4321 (different from permutations having no circular succession). - Enrique Navarrete, Oct 07 2016

a(n-1) is also the number of permutations of [n] that allow the substring n1 in the set of permutations of [n] having no substring in {12, 23, ..., (n-1)n}. For example, for n=5 the 8 permutations in S5 having no substring in {12,23,34,45} that allow the substring 51 are {51324,51432,25143,24351,35142,32514,42513,43251} (see link). - Enrique Navarrete, Jan 11 2017

From Enrique Navarrete, Mar 25 2017: (Start)

Let D(n,k) be the set of permutations on [n] that for fixed k, 0 < k < n, avoid substrings j(j+k) for 1 <= j <= n - k, and avoid substrings j(j+k) (mod n) for n-k < j <= n. Then the number of permutations in D(n,k) with k relative prime to n, n>=2, is given by a(n). For example, the forbidden substrings in D(4,3) are {14;21,32,43} (the forbidden substrings (mod 4) are written after the semicolon and lie below the diagonal in the chessboard below):

1 2 3 4

1 |||_|x|

2 |x|||_|

3 ||x||_|

4 |||x|_|

_

Then since 4 and 3 are relatively prime, a(4)=8, and the permutations in D(4,3) are 1234, 1342, 2341, 2413, 3124, 3412, 4123, 4231.

For another example, the forbidden substrings in D(8,5) are {16,27,38;41, 52,63,74,85} and the number of permutations in D(8,5) is a(8)=14832 (see the "K-Shift Forbidden Substrings" link).

(End)