A335841 Number of distinct rectangles that can be made with one even and one odd side length that are divisors of 2n.

1, 2, 4, 3, 4, 8, 4, 4, 9, 8, 4, 12, 4, 8, 16, 5, 4, 18, 4, 12, 16, 8, 4, 16, 9, 8, 16, 12, 4, 32, 4, 6, 16, 8, 16, 27, 4, 8, 16, 16, 4, 32, 4, 12, 36, 8, 4, 20, 9, 18, 16, 12, 4, 32, 16, 16, 16, 8, 4, 48, 4, 8, 36, 7, 16, 32, 4, 12, 16, 32, 4, 36, 4, 8, 36, 12, 16, 32, 4, 20

Offset: 1

Author

Wesley Ivan Hurt, Sep 13 2020

Keywords

• nonn
• easy

Comments

If p > 2 is prime, a(p) = 4. There are 4 rectangles that can be made with one even and one odd side length that are divisors of 2p: 1 X 2, 1 X 2p, 2 X p, and p X 2p. - Wesley Ivan Hurt, May 21 2021

Examples

			a(6) = 8; The divisors of 2*6 = 12 are {1,2,3,4,6,12}. There are 8 distinct rectangles with one odd and one even side length using these divisors. They are 1 X 2, 1 X 4, 1 X 6, 1 X 12, 2 X 3, 3 X 4, 3 X 6, and 3 X 12.
		

Links

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

Crossrefs

Cf. A337532, A000005.

Programs

Maple

with(numtheory): seq(tau(n)*(tau(2*n)-tau(n)), n=1..100); # Ridouane Oudra, Feb 19 2023

Mathematica

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

PARI

A335841(n) = { my(ds=divisors(2*n)); sum(i=2, #ds, sum(j=1, i-1, (ds[i]+ds[j])%2)); }; \\ Antti Karttunen, Dec 09 2021

Formula

a(n) = Sum_{d1|(2*n), d2|(2*n), d1

a(n) = tau(n)*(tau(2*n) - tau(n)), with tau(n) = A000005(n). - Ridouane Oudra, Feb 19 2023

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