A296955 Sum of the smaller parts of the partitions of n into two distinct parts such that the smaller part divides the larger.

0, 0, 1, 1, 1, 3, 1, 3, 4, 3, 1, 10, 1, 3, 9, 7, 1, 12, 1, 12, 11, 3, 1, 24, 6, 3, 13, 14, 1, 27, 1, 15, 15, 3, 13, 37, 1, 3, 17, 30, 1, 33, 1, 18, 33, 3, 1, 52, 8, 18, 21, 20, 1, 39, 17, 36, 23, 3, 1, 78, 1, 3, 41, 31, 19, 45, 1, 24, 27, 39, 1, 87, 1, 3, 49, 26, 19, 51, 1, 66, 40, 3, 1, 98, 23, 3, 33, 48, 1, 99, 21, 30, 35, 3, 25, 108, 1

Offset: 1

Author

Wesley Ivan Hurt, Dec 22 2017

Keywords

• nonn
• easy

Comments

The number of partitions of n into 3 parts whose "middle" part divides n. - Wesley Ivan Hurt, Oct 21 2021

Examples

			a(12) = 10; the partitions of 12 into two distinct parts are (11,1), (10,2), (9,3), (8,4) and (7,5). 1 divides 11, 2 divides 10, 3 divides 9 and 4 divides 8, so the sum of the smaller parts gives 1 + 2 + 3 + 4 = 10.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Joerg Arndt, On computing the generalized Lambert series, arXiv:1202.6525v3 [math.CA], (2012).
• Index entries for sequences related to partitions

Crossrefs

Cf. A297024, A001065, A023645, A000203, A080512.

Programs

Maple

with(numtheory):
a := n -> add( d, d = divisors(n) minus {floor((n+1)/2), n} ):
seq(a(n), n = 1..100); # Peter Bala, Jan 13 2021

Mathematica

Table[Sum[i (Floor[n/i] - Floor[(n - 1)/i]), {i, Floor[(n - 1)/2]}], {n, 100}]
f[n_] := Plus @@ Select[Divisors@n, 2 # < n &]; Array[f, 75] (* Robert G. Wilson v, Dec 23 2017 *)

PARI

A296955(n) = sumdiv(n,d,(d<(n/2))*d); \\ Antti Karttunen, Sep 25 2018

Formula

a(n) = Sum_{i=1..floor((n-1)/2)} i * (floor(n/i) - floor((n-1)/i)).

a(n) = the sum of the divisors < n/2. - Robert G. Wilson v, Dec 23 2017

a(n) = 1 iff n is an odd prime or n=4. - Robert G. Wilson v, Dec 23 2017

G.f.: Sum_{k>=1} k * x^(3*k) / (1 - x^k). - Ilya Gutkovskiy, May 30 2020

G.f.: Sum_{k >= 3} x^k/(1 - x^k)^2. Cf. A023645. - Peter Bala, Jan 13 2021

Faster converging g.f.: Sum_{n >= 1} q^(n*(n+2))*( n*q^(3*n+4) - (n + 1)*q^(2*n+2) - (n - 1)*q^(n+2) + n )/( (1 - q^n )*(1 - q^(n+2))^2 ). (In equation 1 in Arndt, after combining the two n = 0 summands to get t/(1 - t), apply the operator t*d/dt and then set t = q^2 and x = 1. Cf. A001065.) - Peter Bala, Jan 22 2021

a(n) = A000203(n) - A080512(n). - Ridouane Oudra, Aug 15 2024

Extensions

More terms from Antti Karttunen, Sep 25 2018