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Number of odd divisors d of n such that d > 1 and d(1+d/3)/2 <= n <= 3d(d-1)/2.
2

%I #13 Feb 14 2021 20:41:45

%S 0,0,1,0,0,1,0,0,1,1,0,0,0,1,1,0,0,1,0,1,1,0,0,0,1,0,1,1,0,1,0,0,1,0,

%T 1,1,0,0,1,0,0,1,0,1,2,0,0,0,1,0,0,1,0,1,1,1,0,0,0,1,0,0,2,0,1,1,0,1,

%U 0,0,0,1,0,0,1,1,1,1,0,0,1,0,0,1,1,0,0

%N Number of odd divisors d of n such that d > 1 and d(1+d/3)/2 <= n <= 3d(d-1)/2.

%C This sequence is useful for computing A243223, the number of partitions of n into summands in arithmetic progression with common difference 3. The definition follows Nyblom and Evans 2003 (see LINK) with slight modifications and corrections.

%H Jean-Christophe Hervé, <a href="/A243224/b243224.txt">Table of n, a(n) for n = 1..10000</a>

%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.

%e a(6) = 1 because 3, the unique odd divisor > 1 of 6 satisfies 3(1+3/3)/2 <= 6 <= 3.3(3-1)/2.

%o (PARI) a(n) = sumdiv(n, d, (d > 1) && (d % 2) && (d*(1+d/3)/2 <= n) && (n <= 3*d*(d-1)/2)); \\ _Michel Marcus_, Jun 02 2014

%Y Cf. A243223.

%K nonn

%O 1,45

%A _Jean-Christophe Hervé_, Jun 01 2014