cp's OEIS Frontend

This number triangle results from the array A(n, m) = T(n+m+1) - T(n-1), with T = A000217, for n, m >= 1. For this array see the example by Bob Selcoe, in A111774 (but with rows continued). The present triangle is obtained by reading the array by upwards antidiagonals: T(n, k) = A(n+1-k, k). See also the Jul 09 2019 comment by Ralf Steiner with the formula c_k(n) (rows k >= 1, columns n >= 3), rewritten for A(n, m) = (m+2)*(2*n+m+1)/2, leading to T(n, k) = (k+2)*(2*n-k+3)/2.

Therefore this triangle is related to the problem of giving the numbers which are sums of at least three consecutive positive integers given as sequence A111774. It allows us to find the multiplicities for the numbers of A111774. They are given in A338428(n).

To obtain the multiplicity for number N (>= 6) from A111774 one has to consider only the triangle rows n = 1, 2, ..., floor((N-3)/3).

The row reversed triangle, considered by Bob Selcoe in A111774, is T(n, n-k+1) = T(n+2) - T(k-1), for n >= 1, and k=1, 2, ..., n.

This triangle contains no odd prime numbers and no exact powers 2^m, for m >= 0. This can be seen by considering the diagonal sequences D(d, k), for d >= 1, k >= 1 or the row sequences of the array A(n, m), for n >= 1 and m >= 1. The result is A(r+1, s-2) = s*(s + 2*r + 1)/2, for r >= 0 and s >= 3 (from the g.f. of the diagonals of T given below). This is also given in the Jul 09 2019 comment by Ralf Steiner in A111774. Therefore A(r+1, s-2) is a product of two numbers >= 2, hence not a prime. And in both cases (i) s/2 integer or (ii) (s + 2*r + 1)/2 integer not both numbers can be powers of 2 by simple parity arguments.

The previous comment means that each T(n, k) has at least one odd prime as a proper divisor.

A number N appears in this triangle, or in A111774, if and only if floor(N/2) - delta(N) >= 1, where delta(N) = A055034(N). For the sequence b(n) := floor(n/2) - delta(n), for n >= 2, see A219839(n), b(1) = -1. See a W. Lang comment in A111774 for the proof.

The length of row n is given in A338431(n), for n >= 1.

The length of the sublists t(n, k) of the power basis coefficients of DSR(n, k), for k = 1, 2, ..., floor(n/2), is 1 if n = 1, for n >= 2 it is k except for n = n(j) = A111774(j) for which the final A219839(n) sublists have fewer than k members.

Trailing vanishing coefficients of the delta(n) = A055034(n) power base elements <1 = rho(n)^0, rho(n)^1, ..., rho(n)^{delta(n)-1}> are not recorded. The coefficients of the minimal polynomial C(n, x) of rho(n) = 2*cos(Pi/n) of degree delta(n) are given in A187360. C(n, rho(n)) = 0 is used to eliminate all powers of rho with exponent >= delta(n).

The length ratios DSR(n, k) := diagonal(n, k)/side(n) of regular n-gons, for n >= 2, and k = 1, 2, ..., floor(n/2) (distinct diagonals, starting with the side for k = 1, in increasing order) are given by DSR(n, k) = S(k-1, rho(n)), with the Chebyshev S polynomials (A049310). See the W. Lang link.

For n = 2, the degenerate case, diagonal/side = side/side = 1 for k = 1. For n = 1 (a point) diagonal/side is undetermined, and T(1, 1) is set to 1.

For the power basis sublists t(n, k), for k = 1, 2, ..., delta(n), only the k coefficients of S(k-1, x) are present (trailing vanishing coefficients are not recorded). For k = delta(n)+1, ..., floor(n/2) less than k coefficients appear due to elimination via C(n, rho(n)) = 0. E.g., for n = 6 with delta(6) = 2 the only coefficient for k = 3 is 2 (coefficient of rho^0). This appears for n = n(j) = A111774(j), because then floor(n/2) - delta(n) = A219839(n) > 0.

Because A219839(n) = 0 means that n is from A174090, i.e., a prime or a power of 2 (complement of A111774), these rows n have all the sublists t(n, k) with the k coefficients of S(k-1, x), hence they are identical (but the basis differs). See especially the table for the pairs of consecutive numbers n with identical coefficients, like (2, 3), (4, 5), (16, 17), (256, 267), (65536, 65537), ?... (cf. Fermat primes A019434).