A368684 Number of partitions of n into 2 parts such that the smaller part divides both n and floor(n/2).

0, 1, 1, 2, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 4, 1, 3, 1, 4, 1, 2, 1, 6, 1, 2, 1, 4, 1, 4, 1, 5, 1, 2, 1, 6, 1, 2, 1, 6, 1, 4, 1, 4, 1, 2, 1, 8, 1, 3, 1, 4, 1, 4, 1, 6, 1, 2, 1, 8, 1, 2, 1, 6, 1, 4, 1, 4, 1, 4, 1, 9, 1, 2, 1, 4, 1, 4, 1, 8, 1, 2, 1, 8, 1, 2, 1, 6, 1, 6

Offset: 1

Author

Wesley Ivan Hurt, Jan 03 2024

Keywords

• nonn
• easy

Comments

Essentially, A000005 interspersed with 1's [prepend 0].

Number of divisors of A057979(n+1) for n >= 2.

Links

• Index entries for sequences related to partitions.

Crossrefs

Bisections: A060576, A000005.

Cf. A001620, A057979, A091954.

Programs

Maple

with(numtheory): 0, seq(2*tau(n) - tau(2*n) + (n mod 2), n=2..100); # Ridouane Oudra, Jan 18 2025

Mathematica

Join[{0}, Table[DivisorSigma[0, (n+2+(n-2)*(-1)^n)/4], {n, 2, 100}]]

PARI

a(n) = if(n == 1, 0, numdiv((n+2+(n-2)*(-1)^n)/4)); \\ Amiram Eldar, Jan 28 2025

Formula

a(n) = A000005(A057979(n+1)) for n >= 2.

a(2n-1) = A060576(n), a(2n) = A000005(n).

a(n) = d(floor((n+1)/2))^((n+1) mod 2), for n >= 2.

a(n) = d( (n+2+(n-2)*(-1)^n)/4 ) for n >= 2.

a(n) = Sum_{k=1..floor(n/2)} c(n/k) * c(floor(n/2)/k), where c(m) = 1 - ceiling(m) + floor(m).

a(n) = A000005(n) - A091954(n), for n > 1. - Ridouane Oudra, Jan 18 2025

Sum_{k=1..n} a(k) ~ (log(n/2) + 2*gamma)*n/2, where gamma is Euler's constant (A001620). - Amiram Eldar, Jan 28 2025