A337333 Number of pairs of odd divisors of n, (d1,d2), such that d1 <= d2.

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, 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, 3, 21, 1, 10, 10, 3, 3, 10, 10, 3, 6, 3, 3, 21, 3, 10, 10, 3, 3, 15, 3, 3, 10, 10

Offset

1,3

Comments

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

Links

Antti Karttunen, Table of n, a(n) for n = 1..20000

Formula

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

From Bernard Schott, Aug 24 2020: (Start)

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

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

a(n) = A000217(A001227(n)). - Antti Karttunen, Dec 12 2021

Example

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

Mathematica

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}]

Prog

(PARI) A337333(n) = binomial(numdiv(n>>valuation(n, 2))+1, 2); \\ Antti Karttunen, Dec 12 2021

Crossrefs

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

Sequence in context: A059789 A275367 A023136 * A348665 A152774 A275820

Adjacent sequences: A337330 A337331 A337332 * A337334 A337335 A337336

Keyword

nonn

Author

Wesley Ivan Hurt, Aug 23 2020

Status

approved