OFFSET
1,4
COMMENTS
Are all terms nonnegative? This question is equivalent to the question posed in A285705.
From Antti Karttunen, Aug 05 2020: (Start)
The answer to the above question is yes. Because both A000203 and A003961 are multiplicative sequences, it suffices to prove that for any prime p, and e >= 1, q^e >= sigma(p^e) = ((p^(1+e))-1) / (p-1), where q = A151800(p), i.e., the next larger prime after p. If p is a lesser twin prime, then q = p+2 (and this difference can't be less than 2, apart from case p=2), and it is easy to see that (n+2)^e > ((n^(e+1)) - 1) / (n-1), for all n >= 2, e >= 1.
See comments in A326042.
(End)
This is the inverse Möbius transform of A337549, from which it is even easier to see that all terms are nonnegative. - Antti Karttunen, Sep 22 2020
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..16383
FORMULA
a(n) = Sum_{d|n} A337549(d). - Antti Karttunen, Sep 22 2020
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} ((p^2-p)/(p^2-q(p))) - Pi^2/12 = 1.24152934..., where q(p) = nextprime(p) (A151800). - Amiram Eldar, Dec 21 2023
MATHEMATICA
Array[Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] - DivisorSigma[1, #] &, 96] (* Michael De Vlieger, Oct 05 2020 *)
PROG
(PARI)
A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); }; \\ From A003961
for(n=1, 16384, write("b286385.txt", n, " ", A286385(n)));
(Scheme)
(Python)
from sympy import factorint, nextprime, divisor_sigma as D
from operator import mul
def a048673(n):
f = factorint(n)
return 1 if n==1 else (1 + reduce(mul, [nextprime(i)**f[i] for i in f]))/2
def a(n): return 2*a048673(n) - D(n) - 1 # Indranil Ghosh, May 12 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, May 09 2017
STATUS
approved