cp's OEIS Frontend

Convolution of the cubes (A000578) with the positive integers a(n)=n+1, where all sequences have offset zero. - Graeme McRae, Jun 06 2006

a(n) gives the n-th antidiagonal sum of the convolution array A212891. - Clark Kimberling, Jun 16 2012

In general, the r-th successive summation of the cubes from 1 to n is (6*n^2 + 6*n*r + r^2 - r)*(n+r)!/((r+3)!*(n-1)!), n>0. Here r = 2. - Gary Detlefs, Mar 01 2013

The inverse binomial transform is (essentially) row n=2 of A087127. - R. J. Mathar, Aug 31 2022

The array is a generalization of the number of involutions of permutations on n letters, A000085, also known as 'telephone numbers'. According to Bednarz et al. the telephone number interpretation "is due to John Riordan, who noticed that T(n, 1) is the number of connection patterns in a telephone system with n subscribers."

In graph theory, the n-th telephone number is the total number of matchings of a complete graph K_n (see the Wikipedia entry). Assuming a network with k possibilities of connections leads to a network that can be modeled by a complete multigraph K(n, k). The total number of connection patterns in such a network is given by T(n, k).

The n+1 consecutive triangular numbers start with the A028387(n-2)-th triangular number A000217(n^2-n-1), while the n-1 consecutive triangular numbers start with the A000290(n)-th triangular number A000217(n^2).

Similar sums of consecutive integers are A059270.

Similar sums of consecutive squares are A059255.

Berselli points out that a(n) = 10*A024166(n-1) = A000292(n-1)*(3*n^2 - 2). Since a(n) is a sum of triangular numbers, 10=1+2+3+4 is the 4th triangular number, A024166 is a sum of cubes, and A000292 is a tetrahedral number, is there a geometric proof of Berselli's formula? (Compare Nelsen and Unal's "Proof Without Words: Runs of Triangular Numbers.") [Jonathan Sondow, Mar 04 2013]

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

Primes p such that 3*(p+2) or (p+2)/3 is a square. - Vincenzo Librandi, Feb 16 2016

m is a member of A063213. - Bruno Berselli, Feb 16 2016

Quasipolynomial of order 2 and degree 2. - Charles R Greathouse IV, Apr 12 2016

From Ray Chandler, Apr 13 2016: (Start)

Square roots of resulting squares gives A001651.

Sequence is the union of A141631 and A271740. (End)

7 is the unique prime in this sequence. If m is in this sequence, then 3*m + 4 = k^2 for k is nonzero integer, that is, m = (k^2 - 4)/3 = (k-2)*(k+2)/3. So m can be only prime if one of divisors is prime and another one is 1. Otherwise there should be more than 1 prime divisors, that is n must be composite. - Altug Alkan, Apr 12 2016

From Ray Chandler, Apr 12 2016: (Start)

Square roots of resulting squares gives A001651 (with a different starting point).

Sequence is the union of (positive terms) in A140676 and A270710. (End)

The sequence terms are the exponents in the expansion of Sum_{n >= 0} q^n*(1 - q)*(1 - q^3)*...*(1 - q^(2*n+1)) = 1 - q^4 - q^7 + q^15 + q^20 - q^32 - q^50 + + - - .... - Peter Bala, Dec 19 2024

A combination of a multiset M is an unordered selection of k objects of M, where every object can appear at most as many times as it appears in M.

A(n, k) = Cardinality(Union_{j=0..k} Combination(MultiSet(1^[j*n], 0^[(k-j)*n]))), where MultiSet(r^[s], u^[v]) denotes a set that contains the element r with multiplicity s and the element u with multiplicity v; thus the multisets under consideration have n*k elements. Since the base set is {1, 0} the elements can be represented as binary strings. Applying the combination operator to the multisets results in a set of binary strings where '0' resp. '1' can appear at most j*n resp. (k-j)*n times. 'At most' means that they do not have to appear; in other words, the resulting set always includes the empty string ''.

In contrast to the procedure in A361045 we consider here the cardinality of the set union and not the sum of the individual cardinalities. If you want to exclude the empty string, you will find the sequences listed in A361521. The same construction with multiset permutations instead of multiset combinations results in A361043.

A different view can be taken if one considers the hypergeometric representation, hypergeom([-k, -m], [1], n). This is a family of arrays that includes the 'rascal' triangle: the all 1's array A000012 (m = 0), the rascal array A077028 (m = 1), this array (m = 2), and A361731 (m = 3).

Right border of the triangle = A100536: (1, 10, 25, 46, 73,...).

Left border = A056107: (1, 4, 13, 28, 49,...).

Row sums = A005915: (1, 14, 57, 148, 305,...).

n-th row = (right border then going to the left): (n-th term of A100536 followed by (n-1) operations of (-6), (-6), (-6),... As a Connell-like triangle, odd row terms are in the subset 6n-5; even row terms are in the set 6n-2.

Row 3 = (13, 19, 25) beginning with A100536(3) = 25 at the right then following the trajectory (-6), (-6).

Using the modular rules, the triangle begins (1; 4, 10; 13, 19, 25;...) since 1 == 6n-5, while 4 is the next higher term in the set 6n-2, then 10 also in the set 6n-2, being an even row.

Binomial transform of 1, 14, 12, 0, 0, 0 (0 continued).

Sum_{n>=0} 1/a(n) = (Psi(0,(4+sqrt(10))/6) - Psi(0,(4-sqrt(10))/6))/(2*sqrt(10)) = 1.14373625509612753878..., where Psi(n,k) is the n^th derivative of the digamma function. - Bruno Berselli, Mar 16 2014