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%I #11 Jun 02 2014 04:21:18
%S 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,
%T 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,
%U 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
%N Number of partitions of n into positive summands in arithmetic progression with common difference 3.
%C 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.
%H Jean-Christophe Hervé, <a href="/A243223/b243223.txt">Table of n, a(n) for n = 1..10045</a>
%H J. W. Andrushkiw, R. I. Andrushkiw and C. E. Corzatt, <a href="http://www.jstor.org/stable/2689456">Representations of Positive Integers as Sums of Arithmetic Progressions</a>, Mathematics Magazine, Vol. 49, No. 5 (Nov., 1976), pp. 245-248.
%H M. A. Nyblom and C. Evans, <a href="http://ajc.maths.uq.edu.au/pdf/28/ajc_v28_p149.pdf">On the enumeration of partitions with summands in arithmetic progression</a>, Australasian Journal of Combinatorics, Vol. 28 (2003), pp. 149-159.
%F 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.
%e a(15) = 2 because 15 = 6 + 9 = 2 + 5 + 8.
%Y Cf. A072670 (same with common differences = 2).
%Y A243225 gives the integers n that are not such sums for which a(n) = 0.
%K nonn
%O 1,15
%A _Jean-Christophe Hervé_, Jun 01 2014
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