cp's OEIS Frontend

In this sequence there are no (odd) primes and there are no powers of 2.

So we have only three kinds of natural numbers: the odd primes, the powers of 2 and the numbers that can be represented as a sum of at least three consecutive integers.

Odd primes can only be written as a sum of two consecutive integers. Powers of 2 do not have a representation as a sum of k consecutive integers (other than the trivial n=n, for k=1).

Numbers of the form (x*(x+1)-y*(y+1))/2 for nonnegative integers x,y with x-y >= 3. - Bob Selcoe, Feb 21 2014

Numbers of the form (x + 1)*(x + 2*y)/2 for integers x,y with x >= 2 and y >= 1. For y = 1 only triangular numbers (A000217) >= 6 occur. - Ralf Steiner, Jun 27 2019

From Ralf Steiner, Jul 09 2019: (Start)

If k >= 1 sequences are c_k(n) = c_k(n - 1) + n + k - 1, c_k(0) = 0, means c_k(n) = n*(n + 2*k - 1)/2: A000217, A000096, A055998, A055999, A056000, ... then this sequence is the union of c_k(n), n >= 3. (End)

From Wolfdieter Lang, Oct 28 2020: (Start)

This sequence gives all positive integers that have at least one odd prime as proper divisor. The proof follows from the first two comments.

The set {a(n)}_{n>=1} equals the set {k positive integer : floor(k/2) - delta(k) >= 1}, where delta(k) = A055034(k). Proof: floor(k/2) gives the number of positive odd numbers < k, and delta(k), gives the number of positive odd numbers coprime to k. delta(1) = 1 but 1 is not < 1, therefore k = 1 is not a member of this set. Hence a member >= 2 of this set has at least one odd number > 1 and < k missing in the set of odd numbers relative prime to k. Therefore there exists at least one odd prime < k dividing k. (End)

For the multiplicity of a(n) see A338428, obtained from triangle A337940 (the array is given Bob Selcoe as example below, and in the Ralf Steiner comment above). - Wolfdieter Lang, Dec 09 2020

a(n) is also the number of linearly dependent diagonal/side length ratios R(n,k), in the regular n-gon. The following will explain this. In the regular n-gon inscribed in a circle the number of distinct diagonals including the side is floor(n/2). Not all of the corresponding length ratios R(n,k) = d(n,k)/d(n,1), k = 1..floor(n/2), with d(n,1) = s(n) (the length of the side), d(n,2) the length of the smallest diagonal, etc., are linearly independent because C(n,R(n,2)) = 0, where C is the minimal polynomial of R(n,2) = 2*cos(Pi/n) (see A187360) with degree delta(n) = A055034(n). Thus every ratio R(n,j), with j = delta(n)+1, ..., floor(n/2) can be expressed as a linear combination of the independent R(n,k), k=1, ..., delta(n). See the comment from Sep 21 2013 on A053121 for powers of R(n,2) (called there rho(N)). Therefore, a(n) = floor(n/2) - delta(n) is, for n>=2, the number of linearly dependent ratios R(n,k) in the regular n-gon. - Wolfdieter Lang, Sep 23 2013

From Wolfdieter Lang, Nov 23 2020: (Start)

This sequence gives the difference between the number of odd numbers in the smallest nonnegative residue system modulo n (called here RS(n)) and the smallest nonnegative restricted residue system (called here RRS(n), but RRS(1) = {1}, not {0}).

This sequence can be used to find sequence A111774 by recording the positions of the entries >= 1. See a W. Lang comment there, and also A337940, for the proof. Hence the complement of A111774, given in A174090, is given by the numbers m with a(m) = 0. (End)

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.