A366439 The sum of divisors of the exponentially odd numbers (A268335).
1, 3, 4, 6, 12, 8, 15, 18, 12, 14, 24, 24, 18, 20, 32, 36, 24, 60, 42, 40, 30, 72, 32, 63, 48, 54, 48, 38, 60, 56, 90, 42, 96, 44, 72, 48, 72, 54, 120, 72, 120, 80, 90, 60, 62, 96, 84, 144, 68, 96, 144, 72, 74, 114, 96, 168, 80, 126, 84, 108, 132, 120, 180, 90
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Mathematica
f[p_, e_] := (p^(e+1)-1)/(p-1); s[n_] := Module[{fct = FactorInteger[n]}, If[AllTrue[fct[[;;, 2]], OddQ], Times @@ f @@@ fct, Nothing]]; s[1] = 1; Array[s, 100]
-
PARI
lista(max) = for(k = 1, max, my(f = factor(k), isexpodd = 1); for(i = 1, #f~, if(!(f[i, 2] % 2), isexpodd = 0; break)); if(isexpodd, print1(sigma(f), ", ")));
-
Python
from math import prod from itertools import count, islice from sympy import factorint def A366439_gen(): # generator of terms for n in count(1): f = factorint(n) if all(e&1 for e in f.values()): yield prod((p**(e+1)-1)//(p-1) for p,e in f.items()) A366439_list = list(islice(A366439_gen(),30)) # Chai Wah Wu, Oct 11 2023
Formula
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/(2*d^2)) * Product_{p prime} (1 + 1/(p^5-p)) = 1.045911669131479732932..., where d = 0.7044422... (A065463) is the asymptotic density of the exponentially odd numbers.
The asymptotic mean of the abundancy index of the exponentially odd numbers: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k)/A268335(k) = (1/d) * Product_{p prime} (1 + 1/(p^5-p)) = 2 * c * d = 1.4735686365073812503199... .
Comments