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A068322
Number of arithmetic progressions of positive odd integers, strictly increasing with sum n.
6
1, 0, 1, 1, 1, 1, 1, 2, 2, 2, 1, 3, 1, 3, 3, 5, 1, 4, 1, 5, 4, 5, 1, 7, 2, 6, 5, 8, 1, 7, 1, 9, 6, 8, 2, 11, 1, 9, 7, 12, 1, 10, 1, 12, 10, 11, 1, 15, 2, 12, 9, 15, 1, 13, 3, 16, 10, 14, 1, 18, 1, 15, 12, 20, 4, 17, 1, 19, 12, 17, 1, 22, 1, 18, 16, 22, 2, 20, 1, 24, 15, 20, 1, 25, 5, 21, 15, 26
OFFSET
1,8
LINKS
Sadek Bouroubi and Nesrine Benyahia Tani, Integer partitions into arithmetic progressions, Rostok. Math. Kolloq. 64 (2009), 11-16.
Sadek Bouroubi and Nesrine Benyahia Tani, Integer partitions into arithmetic progressions with an odd common difference, Integers 9(1) (2009), 77-81.
Augustine O. Munagi, Combinatorics of integer partitions in arithmetic progression, Integers 10(1) (2010), 73-82.
Augustine O. Munagi and Temba Shonhiwa, On the partitions of a number into arithmetic progressions, Journal of Integer Sequences 11 (2008), Article 08.5.4.
A. N. Pacheco Pulido, Extensiones lineales de un poset y composiciones de números multipartitos, Maestría thesis, Universidad Nacional de Colombia, 2012.
FORMULA
From Petros Hadjicostas, Oct 01 2019: (Start)
a(n) = A068324(n) - A001227(n) + (1/2) * (1 - (-1)^n).
G.f.: x/(1 - x^2) + Sum_{m >= 2} x^(m^2)/((1 - x^(2*m)) * (1 - x^(m*(m-1))).
(End)
EXAMPLE
From Petros Hadjicostas, Sep 29 2019: (Start)
a(12) = 3 because we have the following arithmetic progressions of odd numbers, strictly increasing with sum n=12: 1+11, 3+9, and 5+7.
a(13) = 1 because we have only the following arithmetic progressions of odd numbers, strictly increasing with sum n=13: 13.
a(14) = 3 because we have the following arithmetic progressions of odd numbers, strictly increasing with sum n=14: 1+13, 3+11, and 5+9.
a(15) = 3 because we have the following arithmetic progressions of odd numbers, strictly increasing with sum n=15: 15, 3+5+7, and 1+5+9.
(End)
KEYWORD
easy,nonn
AUTHOR
Naohiro Nomoto, Feb 27 2002
STATUS
approved