cp's OEIS Frontend

The general formula for alternating sums of powers of odd integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,0)-(-1)^k*P(n,2*k))/2. Here n=3, thus a(k) = |(P(3,0)-(-1)^k*P(3,2*k))/2|. - Peter Luschny, Jul 12 2009

Partial sums of A069190. - J. M. Bergot, Jul 13 2013

From the formula a(n) = n^3 - a(n-1) it follows that a(n-1) + a(n) = n^3. Thus the sum of two consecutive terms (call them the "former" and "latter" terms) is a cube of the index of the "latter" term. - Alexander R. Povolotsky, Jan 09 2008

The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) (A153641) 2^(-n-1)*(P(n,1)-(-1)^k P(n,2*k+1)). Thus we get expression a(k) = |2^(-4)*(P(3,1)-(-1)^k P(3,2*k+1))|. - Peter Luschny, Jul 12 2009

a(n) is the number of (w,x,y) having all terms in {0,...,n} and w < floor((x+y)/2). Also, the number of (w,x,y) having all terms in {0,...,n} and w >= floor((x+y)/2). - Clark Kimberling, Jun 02 2012

a(n) = E(4)*binomial(n+4,4) where E(n) are the Euler number in the enumeration A122045.

a(n) is the special case k=4 in the sequence of diagonals in the triangle A153641.

a(n) is the 5th row in A093375.

a(n) is the 5th column in A103406.

a(n) is the 5th antidiagonal in A103283.

(a(n+1) - a(n))/5 are the pyramidal numbers A000292 (n>1).

(a(n+2) - 2a(n+1) + a(n))/5 are the triangular numbers A000217 (n>2).

(a(n+3) - 3a(n+2) + 3a(n+1) - a(n))/5 are the natural numbers A000027 (n > 3).

Number of orbits of Aut(Z^7) as function of the infinity norm (n+4) of the representative integer lattice point of the orbit, when the cardinality of the orbit is equal to 107520. - Philippe A.J.G. Chevalier, Dec 28 2015

Row sums are A011782. Diagonal sums are F(n+1)*(1+(-1)^n)/2 (aerated version of A001519). Product by Pascal's triangle A007318 is A119468. Schur product of (1/(1-x),x/(1-x)) and (1/(1-x^2),x).

Exponential Riordan array (cosh(x),x). Inverse is (sech(x),x) or A119879. - Paul Barry, May 26 2006

Rows give coefficients of polynomials p_n(x) = Sum_{k=0..n} (k+1 mod 2)*binomial(n,k)*x^(n-k) having e.g.f. exp(x*t)*cosh(t)= 1*(t^0/0!) + x*(t^1/1!) + (1+x^2)*(t^2/2!) + ... - Peter Luschny, Jul 14 2009

Inverse of the coefficient matrix of the Swiss-Knife polynomials in ascending order of x^i (reversed and aerated rows of A153641). - Peter Luschny, Jul 16 2012

Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array

/I_k 0\

\ 0 M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*... is equal to A136630 but with the first row and column omitted. - Peter Bala, Jul 28 2014

The row polynomials SKv(n,x) = [(x+1)^n + (x-1)^n]/2 , with e.g.f. cosh(t)*exp(xt), are the umbral compositional inverses of the row polynomials of A119879 (basically the Swiss Knife polynomials SK(n,x) of A153641); i.e., umbrally SKv(n,SK(.,x)) = x^n = SK(n,SKv(.,x)). Therefore, this entry's matrix and A119879 are an inverse pair. Both sequences of polynomials are Appell sequences, i.e., d/dx P(n,x) = n * P(n-1,x) and (P(.,x)+y)^n = P(n,x+y). In particular, (SKv(.,0)+x)^n = SKv(n,x), reflecting that the first column has the e.g.f. cosh(t). The raising operator is R = x + tanh(d/dx); i.e., R SKv(n,x) = SKv(n+1,x). The coefficients of this operator are basically the signed and aerated zag numbers A000182, which can be expressed as normalized Bernoulli numbers. The triangle is formed by multiplying the n-th diagonal of the lower triangular Pascal matrix by the Taylor series coefficient a(n) of cosh(x). More relations for this type of triangle and its inverse are given by the formalism of A133314. - Tom Copeland, Sep 05 2015

The signed version of this matrix has the e.g.f. cos(t) e^{xt}, generating Appell polynomials that have only real, simple zeros and whose extrema are maxima above the x-axis and minima below and situated above and below the zeros of the next lower degree polynomial. The bivariate versions appear on p. 27 of Dimitrov and Rusev in conditions for entire functions that are cosine transforms of a class of functions to have only real zeros. - Tom Copeland, May 21 2020

The n-th row of the triangle is obtained by multiplying by 2^(n-1) the elements of the first row of the limit as k approaches infinity of the stochastic matrix P^(2k-1) where P is the stochastic matrix associated with the Ehrenfest model with n balls. The elements of a stochastic matrix P give the probabilities of arriving in a state j given the previous state i. In particular the sum of every row of the matrix must be 1, and so the sum of the terms of the n-th row of this triangle is 2^(n-1). Furthermore, by the properties of Markov chains, we can interpret P^(2k-1) as the (2k-1)-step transition matrix of the Ehrenfest model and its limit exists and it is again a stochastic matrix. The rows of the triangle divided by 2^(n-1) are the even rows (second, fourth, ...) and the odd rows (first, third, ...) of the limit matrix P^(2k-1). - Luca Onnis, Oct 29 2023

A signed version of A001586 (Springer numbers).

Equals the logarithmic derivative of A188514 (ignoring the initial term of this sequence); note that the unsigned version (A001586) does not form a logarithmic derivative of an integer sequence.

Transform of 2^n under the matrix A119879.

Also the Swiss-Knife polynomials A153641 evaluated at x=2. - Peter Luschny, Nov 23 2012

These are the coefficients of the Swiss-Knife polynomials A153641. - Peter Luschny, Jul 21 2012

Nonzero diagonals of the triangle are of the form A000364(k)*binomial(n+2k,2k)*(-1)^k.

A363393 is the dual triangle ('dual' in the sense of Euler-tangent versus Euler-secant numbers). - Peter Luschny, Jun 05 2023

The general formula for alternating sums of powers of even integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,1)-(-1)^k P(n,2k+1))/2. Here n=1 and k shifted one place, thus a(k) = (P(1,1)-(-1)^(k-1) P(1,2(k-1)+1))/2. - Peter Luschny, Jul 12 2009

With just one 0 at the beginning, this is a permutation of all the even integers. - Alonso del Arte, Jun 24 2012

Definition. V_n(x) = (skp(n, x+1) - skp(n, x-1))/2 where skp(n,x) are the Swiss-Knife polynomials A153641. - Peter Luschny, Jul 23 2012

Equivalently, let the polynomials V_n(x) (n>=0) defined by V_n(x) = Sum_{k=0..n} Sum_{v=0..k} (-1)^v*C(k,v)*L(k)*(x+v+1)^n; the sequence L(k) = -1 - H(k-1)*(-1)^floor((k-1)/4) / 2^floor(k/2) if k > 0 and L(0)=0; H(k) = 1 if k mod 4 <> 0, otherwise 0.

(1) V_n(0) = 2^n * Euler(n,1) for n > 0, A155585.

(2) V_n(1) = 1 - Euler(n).

(3) V_{n-1}(0) n / (4^n - 2^n) = B_n gives for n > 1 the Bernoulli numbers A027641/A027642.

(4) V_{n-1}(0) n (2/2^n-2)/(2^n-1) = G_n the Genocchi number A036968 for n > 1.

(5) V_n(1/2)2^{n} - 1 is a signed version of the generalized Euler (Springer) numbers, see A001586.

The Swiss-Knife polynomials (A153641) are complementary to the polynomials defined here. Adding both gives polynomials with e.g.f. exp(x*t)*(sech(t)+tanh(t)), the coefficients of which are a signed variant of A109449.

The Swiss-Knife polynomials as well as the complementary Swiss-Knife polynomials are closely related to the Bernoulli and Euler polynomials. Let F be a sequence and

P_{F}[n](x) = Sum_{k=0..n} Sum_{v=0..k} (-1)^v*C(k,v)*F(k)*(x+v+1)^n.

V_n(x) = P_{F}[n](x) with F(k)=L(k) defined above, are the Co-Swiss-Knife polynomials,

W_n(x) = P_{F}[n](x) with F(k)=c(k) the Chen sequence defined in A153641 are the Swiss-Knife polynomials.

B_n(x) = P_{F}[n](x-1) with F(k)=1/(k+1) are the Bernoulli polynomials,

E_n(x) = P_{F}[n](x-1) with F(k)=2^(-k) are the Euler polynomials.

The most striking formal difference between the Swiss-Knife-type polynomials and the Bernoulli-Euler type polynomials is: The SK-type polynomials have integer coefficients whereas the BE-type polynomials have rational coefficients.

Let R be the exponential Riordan array (exp(x)*sech(x), x) = P * A119879 = 2*P(I + P^2)^(-1) where P denotes Pascal's triangle A007318. Then T = R - I. - Peter Bala, Mar 07 2024

Sum of row n equals Euler(n) (in the sense of the non-official version A122045; R. P. Stanley calls A000111 Euler numbers.)