login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

Number of pairs of odd divisors of n, (d1,d2), such that d1 <= d2.
2

%I #16 Dec 12 2021 22:53:13

%S 1,1,3,1,3,3,3,1,6,3,3,3,3,3,10,1,3,6,3,3,10,3,3,3,6,3,10,3,3,10,3,1,

%T 10,3,10,6,3,3,10,3,3,10,3,3,21,3,3,3,6,6,10,3,3,10,10,3,10,3,3,10,3,

%U 3,21,1,10,10,3,3,10,10,3,6,3,3,21,3,10,10,3,3,15,3,3,10,10

%N Number of pairs of odd divisors of n, (d1,d2), such that d1 <= d2.

%C Number of distinct rectangles that can be made whose side lengths are odd divisors of n.

%H Antti Karttunen, <a href="/A337333/b337333.txt">Table of n, a(n) for n = 1..20000</a>

%F a(n) = Sum_{d1|n, d2|n, d1 and d2 odd, d1<=d2} 1.

%F From _Bernard Schott_, Aug 24 2020: (Start)

%F a(n) = 1 if and only if n = 2^k, k >= 0 (A000079).

%F a(n) = 3 if n is an odd prime. (End)

%F a(n) = A000217(A001227(n)). - _Antti Karttunen_, Dec 12 2021

%e a(15) = 10; There are 10 pairs of odd divisors of 15, (d1,d2), such that d1<=d2. They are: (1,1), (1,3), (1,5), (1,15), (3,3), (3,5), (3,15), (5,5), (5,15), (15,15). So a(15) = 10.

%e a(16) = 1; (1,1) is the only pair of odd divisors of 16, (d1,d2), such that d1<=d2. So a(16) = 1.

%e a(17) = 3; There are 3 pairs of odd divisors of 17, (d1,d2), such that d1<=d2. They are (1,1), (1,17) and (17,17). So a(17) = 3.

%e a(18) = 6; There are 6 pairs of odd divisors of 18, (d1,d2), such that d1<=d2. They are: (1,1), (1,3), (1,9), (3,3), (3,9) and (9,9). So a(18) = 6.

%t Table[Sum[Sum[Mod[i, 2]*Mod[k, 2] (1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k}], {k, n}], {n, 100}]

%o (PARI) A337333(n) = binomial(numdiv(n>>valuation(n,2))+1,2); \\ _Antti Karttunen_, Dec 12 2021

%Y Cf. A000079, A000217, A001227 (number of odd divisors), A335841.

%K nonn

%O 1,3

%A _Wesley Ivan Hurt_, Aug 23 2020