cp's OEIS Frontend

a(n) is the sum of pyramid weights of all Dyck paths of length 2n (for pyramid weight see Denise and Simion). Equivalently, a(n) is the sum of the total lengths of end branches of an ordered tree, summation being over all ordered trees with n edges. For example, the five ordered trees with 3 edges have total lengths of endbranches 3,2,3,3 and 3. - Emeric Deutsch, May 30 2003

a(n) is the number of Motzkin paths of length 2n with exactly one level segment. (A level segment is a maximal sequence of contiguous flatsteps.) Example: for n=2, the paths counted are FFFF, FFUD, UDFF, UFFD. The formula for a(n) below counts these paths by length of the level segment. - David Callan, Jul 15 2004

The inverse Catalan transform yields A024495, shifted once left. - R. J. Mathar, Jul 07 2009

From Paul Barry, Mar 29 2010: (Start)

Hankel transform is A138341.

The aerated sequence 0, 0, 1, 0, 4, 0, 14, 0, 49, ... has e.g.f. int(cosh(x-t)*Bessel_I(1,2t), t = 0..x). (End)

a(n) is the number of terms of A031443 not exceeding 4^n. - Vladimir Shevelev, Oct 01 2010

Also the number of nonempty subsets of {1..2n} with median n, bisection of A361801. The version containing n is A001700 (bisected). Replacing 2n with 2n+1 and n with n+1 gives A006134. For mean instead of median we have A212352. - Gus Wiseman, Apr 16 2023

Total number of leaves in all rooted ordered trees with at most n edges. - Michael Somos, Feb 14 2006

Also: Number of UH-free Schroeder paths of semilength n with horizontal steps only at level less than two [see Yan]. - R. J. Mathar, May 24 2008

Hankel transform is A010892. - Paul Barry, Apr 28 2009

Binomial transform of A005773. - Philippe Deléham, Dec 13 2009

Number of vertices all of whose children are leaves in all ordered trees with n+1 edges. Example: a(3) = 15; for an explanation see David Callan's comment in A001519. - Emeric Deutsch, Feb 12 2015

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

Comments from Alexander Adamchuk, Jul 02 2006: (Start)

Every a(n) is divisible by prime 2, a(n)/2 = A079309(n).

a(n) is divisible by prime 3 only for n=12,30,36,84,90,108,120,... A083096.

a(p) is divisible by p^2 for primes p=5,11,17,23,29,41,47,... Primes of form 6n-1. A007528.

a(p-1) is divisible by p^2 for primes p=7,13,19,31,37,43,... Primes of form 6n+1. A002476.

Every a(n) from a((p-1)/2) to a(p-1) is divisible by prime p for p=7,13,19,31,37,43,... Primes of form 6n+1. A002476.

Every a(n) from a((p^2-1)/2) to a(p^2-1) is divisible by prime p>3.

a(p^2-1), a(p^2-2) and a(p^2-3) are divisible by p^2 for prime p>3.

a(p^2-4) is divisible by p^2 for prime p>5.

a(p^2-5) is divisible by p^2 for prime p>7.

a(p^2-6) is divisible by p^2 for prime p>7.

a(p^2-7) is divisible by p^2 for prime p>11.

a(p^2-8) is divisible by p^2 for prime p>13.

a(p^3) is divisible by p^2 for prime 2 and prime p=5,11,... Primes of form 6n-1. A007528.

a(p^3-1) is divisible by p^2 for prime p=7,13,... Primes of form 6n+1. A002476.

a(p^4-1) is divisible by p^2 for prime p>3. (End)

Mod[ a(3^k), 9 ] = 1 for integer k>0. Smallest number k such that 2^n divides a(k) is k(n) = {1,2,2,11,11,46,46,707,707,707,...}. Smallest number k such that 3^n divides a(k) is k(n) = {12,822,2466,...}. a(2(p-1)/3) is divisible by p^2 for prime p = {7,13,19,31,37,43,61,...} = A002476 Primes of form 6n+1. Every a(n) from a(p^2-(p+1)/2) to a(p^2-1) is divisible by p^2 for prime p>3. Every a(n) from a((4p+3)(p-1)/6) to a((2p+3)(p-1)/3) is divisible by p^2 for prime p = {7,13,19,31,37,43,61,...} = A002476 Primes of form 6n+1. - Alexander Adamchuk, Jan 04 2007

From Gus Wiseman, Apr 19 2023: (Start)

Appears to be the number of nonempty subsets of {1,...,n} with median k, where the median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length). For example, row n = 5 counts the following subsets:

{1} {2} {3} {4} {5}

{1,3} {1,5} {3,5}

{1,2,3} {2,4} {1,4,5}

{1,2,4} {1,3,4} {2,4,5}

{1,2,5} {1,3,5} {3,4,5}

{2,3,4}

{2,3,5}

{1,2,4,5}

{1,2,3,4,5}

Including half-steps gives A231147.

For mean instead of median we have A327481.

(End)

From Gus Wiseman, Mar 19 2023: (Start)

Also appears to be the number of nonempty subsets of {1,...,n} with median k, where k ranges from 1 to n in steps of 1/2, and the median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length). For example, row n = 5 counts the following subsets:

{1} {1,2} {2} {1,4} {3} {2,5} {4} {4,5} {5}

{1,3} {2,3} {1,5} {3,4} {3,5}

{1,2,3} {1,2,3,4} {2,4} {1,3,4,5} {1,4,5}

{1,2,4} {1,2,3,5} {1,3,4} {2,3,4,5} {2,4,5}

{1,2,5} {1,3,5} {3,4,5}

{2,3,4}

{2,3,5}

{1,2,4,5}

{1,2,3,4,5}

Central diagonals T(n,(n+1)/2) appear to be A100066 (bisection A006134).

For mean instead of median we have A327481.

For partitions instead of subsets we have A359893, full steps A359901.

Central diagonals T(n,n/2) are A361801 (bisection A079309).

(End)

a(n+1) is the number of UDU-free paths of n upsteps (U) and n downsteps (D), n>=0. - David Callan, Aug 19 2004

Hankel transform is A120580. - Paul Barry, Mar 26 2010

If interpreted with offset 0, the inverse binomial transform of A006134 - Gary W. Adamson, Nov 10 2007

Also the number of different integer sets { k_1, k_2, ..., k_(i+1) } with Sum_{j=1..i+1} k_j = i and k_j >= 0, see the "central binomial coefficients" (A000984), without all sets in which any two successive k_j and k_(j+1) are zero. See the partition problem eq. 3.12 on p. 19 in my dissertation below. - Eva Kalinowski, Oct 18 2012

p divides a((p+1)/2) for prime p = 3, 11, 17, 19, 41, 43, 59, 67, 73, 83, 89, 97, 107, 113, ... (A033200: primes congruent to {1, 3} mod 8; or, odd primes of the form x^2 + 2*y^2).

p divides a((p-3)/2) for prime p = 17, 41, 73, 89, 97, 113, 137, ... (A007519: primes of the form 8n+1).

Essentially the same as partial sums of A072547. - Seiichi Manyama, Jan 30 2023

The median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length).