A352047 Sum of the divisor complements of the odd proper divisors of n.

0, 2, 3, 4, 5, 8, 7, 8, 12, 12, 11, 16, 13, 16, 23, 16, 17, 26, 19, 24, 31, 24, 23, 32, 30, 28, 39, 32, 29, 48, 31, 32, 47, 36, 47, 52, 37, 40, 55, 48, 41, 64, 43, 48, 77, 48, 47, 64, 56, 62, 71, 56, 53, 80, 71, 64, 79, 60, 59, 96, 61, 64, 103, 64, 83, 96, 67, 72, 95, 96, 71, 104

Offset: 1

Author

Wesley Ivan Hurt, Mar 01 2022

Keywords

• nonn
• easy

Examples

			a(10) = 12; a(10) = 10 * Sum_{d|10, d<10, d odd} 1 / d = 10 * (1/1 + 1/5) = 12.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000035, A000593, A002131, A006519, A111003, A300415.

Sum of the k-th powers of the divisor complements of the odd proper divisors of n for k=0..10: A091954 (k=0), this sequence (k=1), A352048 (k=2), A352049 (k=3), A352050 (k=4), A352051 (k=5), A352052 (k=6), A352053 (k=7), A352054 (k=8), A352055 (k=9), A352056 (k=10).

Programs

Maple

A352047 := proc(n)
    local a,d ;
    a := 0 ;
    for d in numtheory[divisors](n) do
        if type(d,'odd') and d < n then
            a := a+n/d ;
        end if;
    end do:
    a ;
end proc:
seq(A352047(n),n=1..30) ; # R. J. Mathar, Mar 09 2022

Mathematica

Table[n DivisorSum[n, 1/# &, # < n && OddQ[#] &], {n, 72}] (* Michael De Vlieger, Mar 02 2022 *)
a[n_] := DivisorSigma[-1, n / 2^IntegerExponent[n, 2]] * n - Mod[n, 2]; Array[a, 100] (* Amiram Eldar, Oct 13 2023 *)

PARI

a(n) = n*sumdiv(n, d, if ((d%2) && (dMichel Marcus, Mar 02 2022

PARI

a(n) = n * sigma(n >> valuation(n, 2), -1) - n % 2; \\ Amiram Eldar, Oct 13 2023

Python

from math import prod
from sympy import factorint
def A352047(n): return prod(p**e if p == 2 else (p**(e+1)-1)//(p-1) for p, e in factorint(n).items()) - n % 2 # Chai Wah Wu, Mar 02 2022

Formula

a(n) = n * Sum_{d|n, d

a(2n+1) = A000593(2n+1) - 1. - Chai Wah Wu, Mar 01 2022

G.f.: Sum_{k>=2} k * x^k / (1 - x^(2*k)). - Ilya Gutkovskiy, May 14 2023

exp( 2*Sum_{n>=1} a(n)*x^n/n ) is the g.f. of A300415. - Paul D. Hanna, May 15 2023

From Amiram Eldar, Oct 13 2023: (Start)

a(n) = A000593(n) * A006519(n) - A000035(n).

Sum_{k=1..n} a(k) = c * n^2 / 2, where c = Pi^2/8 = 1.23370055... (A111003). (End)

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