A243224 -id:A243224

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A243223

Number of partitions of n into positive summands in arithmetic progression with common difference 3.

+10
3

0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 2, 0, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 2, 0, 1, 2, 1, 0, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 0, 3, 1, 1, 1, 1, 2, 3, 0, 1, 2, 2, 0, 3, 1, 1, 2, 1, 1, 3, 0, 2, 2, 1, 0, 3, 3, 1, 1, 1, 1, 4, 0, 2, 2, 1, 1, 3, 1, 1, 2, 2, 1, 3, 0, 1, 3, 2, 1, 3, 1, 2, 1, 1

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OFFSET

1,15

COMMENTS

This sequence gives the number of ways to write n as n = a + a+3 + ... + a+3r = (r+1)(2a+3r)/2, with a and r integers > 0.

LINKS

Jean-Christophe Hervé, Table of n, a(n) for n = 1..10045

J. W. Andrushkiw, R. I. Andrushkiw and C. E. Corzatt, Representations of Positive Integers as Sums of Arithmetic Progressions, Mathematics Magazine, Vol. 49, No. 5 (Nov., 1976), pp. 245-248.

M. A. Nyblom and C. Evans, On the enumeration of partitions with summands in arithmetic progression, Australasian Journal of Combinatorics, Vol. 28 (2003), pp. 149-159.

FORMULA

a(n) = d1(n) - 1 - f(n) with d1(n) = number of odd divisors of n (A001227) and f(n) = the number of those odd divisors d of n such that d > 1 and d(1+d/3)/2 <= n <= 3d(d-1)/2. f(n) is in A243224.

EXAMPLE

a(15) = 2 because 15 = 6 + 9 = 2 + 5 + 8.

CROSSREFS

Cf. A072670 (same with common differences = 2).

A243225 gives the integers n that are not such sums for which a(n) = 0.

KEYWORD

nonn

AUTHOR

Jean-Christophe Hervé, Jun 01 2014

STATUS

approved

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