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The FP2 polynomials appear in the numerators of the GF2 o.g.f.s. of the left hand columns of A156920. The FP2 can be calculated with the formula of the LHC sequence, see A156920, and the formula for the general structure of the generating function GF2, see below.

An appropriate name for the FP2 polynomials seems to be the flower polynomials of the second kind because the zero patterns of these polynomials look like flowers. The zero patterns of the FP2 and the FP1, see A156921, resemble each other closely.

A Maple program that generates for a left hand column with a certain LHCnr its GF2 and FP2 can be found below. LHCnr stands for left hand column number and starts from 1.

The FP1 polynomials appear in the numerators of the GF1 o.g.f.s. of the right hand columns of A156920. The FP1 can be calculated with the formula for the RHC sequence, see A156920, and the formula for the general structure of the generating function GF1, see below.

An appropriate name for the FP1 polynomials seems to be the flower polynomials of the first kind because the zero patterns of these polynomials look like flowers. The zero patterns of the FP2, see A156925, and the FP1 resemble each other closely.

A Maple program that generates for a right hand column with a certain RHCnr its GF1 and FP1 can be found below. RHCnr stands for right hand column number and starts from 1.

From Johannes W. Meijer, Feb 20 2009: (Start)

Second right hand column (n-m=1) of the A156920 triangle.

The generating function of this sequence enabled the analysis of the polynomials A156921 and A156925.

(End)

Partial sums of A003462, and thus the second partial sums of A000244 (3^n). Also column k=2 of A106516. - John Keith, Jan 04 2022

The Hankel transform of this sequence is A121835. - Philippe Deléham, Aug 31 2006

a(n) is the moment of order (n-1) for the discrete measure associated to the weight rho(j + 1/2) = 2^(j + 1/2)/(Pi*binomial(2*j + 1, j + 1/2)), with j integral. So we have a(n) = Sum_{j >= 0} (j + 1/2)^(n-1)*rho(j + 1/2). - Groux Roland, Jan 05 2009

Let f(n) = Sum_{j >= 1} j^n*2^j/binomial(2*j, j) = r_n*Pi/2 + s_n; sequence gives r_{n-1}. For example, f(0) through f(5) are [1 + (1/2)*Pi, 3 + Pi, 11 + (7/2)*Pi, 55 + (35/2)*Pi, 355 + 113*Pi, 2807 + (1787/2)*Pi]. For s_n, see A180875. - N. J. A. Sloane, following a suggestion from Herb Conn, Feb 08 2011

Ren gives seven combinatorial interpretations for this sequence. - Peter Bala, Feb 01 2013

Number of left-right arrangements of [n] [Crane, 2015]. - N. J. A. Sloane, Nov 21 2014

In Dyson et al. (2010-2011, 2013), we have S_n(2) = Sum_{j>=1} j^n*2^j/binomial(2*j, j) = A014307(n+1)*Pi/2 + A180875(n) for n >= 1 (and S_0(2) is not defined). This series was originally defined by Lehmer (1985). - Petros Hadjicostas, May 14 2020

Number of words of length n+1 where first element is from {0,1,2}, other elements are from {0,1} and sequence does not decrease (for n=2 there are 3*2^2 sequences, but 000, 100, 110, 111, 200, 210, 211 decrease, so a(2) = 12-7 = 5).

Number of subgroups of C_(2^n) X C_(2^n) (see A060724).

Starting with 1 = row sums of triangle A054582. - Gary W. Adamson, Jun 23 2008

Starting with "1" equals the eigensequence of a triangle with integer squares (1, 4, 9, 16, ...) as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010

(1 + 2x + 2x^2 + 2x^3 + ...)*(1 + 3x + 7x^2 + 15x^3 + ...) = (1 + 5x + 15x^2 + 37x^3 + ...). - Gary W. Adamson, Mar 14 2012

The partial sums of A033484. - J. M. Bergot, Oct 03 2012

Binomial transform is 0, 1, 7, 33, ... (shifted A066810); inverse binomial transform is 0, 1, 3, 3, ... (3 repeated). - R. J. Mathar, Oct 05 2012

Define a triangle by T(n,0) = n*(n+1) + 1, T(n,n) = n + 1, and T(r,c) = T(r-1,c-1) + T(r-1,c) otherwise; then a(n+1) is the sum of the terms of row n. - J. M. Bergot, Mar 30 2013

Starting with "1" are also the antidiagonal sums of the array formed by partial sums of integer squares (1, 4, 9, 16, ...). - Luciano Ancora, Apr 24 2015

Sums of 2 adjacent terms in diagonal k=2 of Eulerian triangle A008292. I.e., T(n,2)+T(n-1,2) for n > 0. Also, 4th NW-SE diagonal of A126277. In other words, a(n) = A000295(n) + A000295(n+1). - Gregory Gerard Wojnar, Sep 30 2018

Essentially the same as A185411. Row reverse of A185410. - Peter Bala, Jul 24 2012

The SF(z; n) formulas, see below, were discovered while studying certain properties of the Dirichlet eta function.

From Peter Bala, Apr 03 2011: (Start)

Let D be the differential operator 2*x*d/dx. The row polynomials of this table come from repeated application of the operator D to the function g(x) = 1/sqrt(1 - x). For example,

D(g) = x*g^3

D^2(g) = x*(2 + x)*g^5

D^3(g) = x*(4 + 10*x + x^2)*g^7

D^4(g) = x*(8 + 60*x + 36*x^2 + x^3)*g^9.

Thus this triangle is analogous to the triangle of Eulerian numbers A008292, whose row polynomials come from the repeated application of the operator x*d/dx to the function 1/(1 - x). (End)

Previous name: Table of coefficients of row polynomials of certain o.g.f.s.

The o.g.f.s G(k, x) for the k-family of sequences S(k, n):= Sum_{p=0..n} p^k*binomial(2*p, p)*binomial(2*(n-p), n-p), k=0,1,... (convolution of two sequences involving the central binomial coefficients) are 1/(1-4*x) for k=0 and 2*x*P(k, x)/(1-4*x)^(k+1) for k=1,2,..., with the row polynomials P(k, x) = Sum_{m=0..k-1} a(n,m)*x^m).

The author was led to compute the sums S(k, n) by a question asked by M. Greiter, Jun 27 2008.

In order to keep the index k>=1 of Sigma(k, n) also for the polynomials P(k, x), their degree is then k-1.