cp's OEIS Frontend

T(n,k) is the number of ways to distribute k identical objects in n distinct containers; containers may be left empty.

T(n,k) is the number of nondecreasing functions f from {1,...,k} to {1,...,n}. - Dennis P. Walsh, Apr 07 2011

Coefficients of Faber polynomials for function x^2/(x-1). - Michael Somos, Sep 09 2003

Consider k-fold Cartesian products CP(n,k) of the vector A(n)=[1,2,3,...,n].

An element of CP(n,k) is a n-tuple T_t of the form T_t=[i_1,i_2,i_3,...,i_k] with t=1,...,n^k.

We count members T of CP(n,k) which satisfy some condition delta(T_t), so delta(.) is an indicator function which attains values of 1 or 0 depending on whether T_t is to be counted or not; the summation sum_{CP(n,k)} delta(T_t) over all elements T_t of CP produces the count.

For the triangle here we have delta(T_t) = 0 if for any two i_j, i_(j+1) in T_t one has i_j > i_(j+1), T(n,k) = Sum_{CP(n,k)} delta(T_t) = Sum_{CP(n,k)} delta(i_j > i_(j+1)).

The indicator function which tests on i_j = i_(j+1) generates A158497, which contains further examples of this type of counting.

Triangle of the numbers of combinations of k elements with repetitions from n elements {1,2,...,n} (when every element i, i=1,...,n, appears in a k-combination either 0, or 1, or 2, ..., or k times). - Vladimir Shevelev, Jun 19 2012

G.f. for Faber polynomials is -log(-t*x-(1-sqrt(1-4*t))/2+1)=sum(n>0, T(n,k)*t^k/n). - Vladimir Kruchinin, Jul 04 2013

Values of complete homogeneous symmetric polynomials with all arguments equal to 1, or, equivalently, the number of monomials of degree k in n variables. - Tom Copeland, Apr 07 2014

Row k >= 0 of the infinite square array A[k,n] = C(n,k), n >= 0, would start with k zeros in front of the first nonzero element C(k,k) = 1; this here is the triangle obtained by taking the first k+1 nonzero terms C(k .. 2k, k) of rows k = 0, 1, 2, ... of that array. - M. F. Hasler, Mar 05 2017

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Mirror image of triangle A133336. - Philippe Deléham, Dec 10 2008

From Tom Copeland, Oct 09 2011: (Start)

With polynomials

P(0,t) = 0

P(1,t) = 1

P(2,t) = t

P(3,t) = t + 2 t^2

P(4,t) = t + 5 t^2 + 5 t^3

P(5,t) = t + 9 t^2 + 21 t^3 + 14 t^4

The o.g.f. A(x,t) = {1+x-sqrt[(1-x)^2-4xt]}/[2(1+t)] (see Drake et al.).

B(x,t)= x-t x^2/(1-x)= x-t(x^2+x^3+x^4+...) is the comp. inverse in x.

Let h(x,t) = 1/(dB/dx) = (1-x)^2/(1+(1+t)*x*(x-2)) = 1/(1-t(2x+3x^2+4x^3+...)), an o.g.f. in x for row polynomials in t of A181289. Then P(n,t) is given by (1/n!)(h(x,t)*d/dx)^n x, evaluated at x=0, A = exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t). These results are a special case of A133437 with u(x,t) = B(x,t), i.e., u_1=1 and (u_n)=-t for n > 1. See A001003 for t = 1. (End)

Let U(x,t) = [A(x,t)-x]/t, then U(x,0) = -dB(x,t)/dt and U satisfies dU/dt = UdU/dx, the inviscid Burgers' equation (Wikipedia), also called the Hopf equation (see Buchstaber et al.). Also U(x,t) = U(A(x,t),0) = U(x+tU,0) since U(x,0) = [x-B(x,t)]/t. - Tom Copeland, Mar 12 2012

Diagonals of A132081 are essentially rows of this sequence. - Tom Copeland, May 08 2012

T(r, s) is the number of [0,r]-covering hierarchies with s segments (see Kreweras). - Michel Marcus, Nov 22 2014

From Yu Hin Au, Dec 07 2019: (Start)

T(n,k) is the number of small Schröder n-paths (lattice paths from (0,0) to (2n,0) using steps U=(1,1), F=(2,0), D=(1,-1) with no F step on the x-axis) that has exactly k U steps.

T(n,k) is the number of Schröder trees (plane rooted tree where each internal node has at least two children) with exactly n+1 leaves and k internal nodes. (End)

The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Also numbers k such that the k-th composition in standard order has reverse-alternating sum != 0.