cp's OEIS Frontend

a(n)=0 if n is an odd prime or a power of 2. For numbers of the third kind we proceed as follows: suppose n is to be written as sum of k consecutive integers starting with m, then 2n = k(2m + k - 1). Let p be the smallest odd prime divisor of n then a(n) = min(p,2n/p).

Powers of 2 and (odd) primes cannot be written as a sum of at least three consecutive integers. a(n) strongly depends on the number of odd divisors of n (A001227): Suppose n is to be written as sum of k consecutive integers starting with m, then 2n = k(2m + k - 1). Only one of the factors is odd. For each odd divisor of n there is a unique corresponding k, k=1 and k=2 must be excluded.

When the initial 0 term is a(1), a(n) is the number of times n occurs after the second column in the square array of A049777. - Bob Selcoe, Feb 14 2014

For nonnegative integers x,y where x-y>=3: a(n) equals the number of ways n can be expressed as a function of (x*(x+1)-y*(y+1))/2 when the initial 0 term is a(1). - Bob Selcoe, Feb 14 2014

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

The Dancing School Problem: a line of g girls (g>0) with integer heights ranging from m to m+g-1 cm and a line of g+h boys (h>=0) ranging in height from m to m+g+h-1 cm are facing each other in a dancing school (m is the minimal height of both girls and boys).

A girl of height l can choose a boy of her own height or taller with a maximum of l+h cm. We call the number of possible matchings f(g,h).

This problem is equivalent to a rooks problem: The number of possible placings of g non-attacking rooks on a g X g+h chessboard with the restriction i <= j <= i+h for the placement of a rook on square (i,j): f(g,h) = per(B), the permanent of the corresponding (0,1)-matrix B, b(i, j)=1 if and only if i <= j <= i+h

f(g,h) = per(B), the permanent of the (0,1)-matrix B of size g X g+h with b(i,j)=1 if and only if i <= j <= i+h.

For fixed g, f(g,n) is polynomial in n for n >= g-2. See reference.