OFFSET
1,1
COMMENTS
All terms <= 10^17 are squares. Are there any nonsquare terms? - David A. Corneth, Sep 05 2020
All the terms are squares. Proof: Let f(n) = A068976(n)/n. f(n) is multiplicative with f(p^e) = (p^2 + 1 - 2/p^e)/(p^2-1) if e is even and 2*(p - 1/p^e)/(p^2-1) if e is odd. Both are strictly increasing with e, the limits as e -> oo are f_even(p) = (p^2+1)/(p^2-1) and f_odd(p) = 2*p/(p^2-1), respectively, and f_odd(p) < f_even(p) for all primes p. The upper bound on f(n) is being attained at even exponents: f(n) < lim_{e->oo} Product_{p prime} (p^2 + 1 - 2/p^e)/(p^2-1) = Product_{p prime} f_even(p) = 5/2. If k is not a square, then there is at least one prime q|k with an odd exponent. Replacing the factor f_even(q) with f_odd(q) in the infinite product, we get f(k) < (5/2) * f_odd(q)/f_even(q) = 5*q/(q^2+1) <= 2. Therefore, A068976(k) = f(k) * k < 2*k and k is not a term. - Amiram Eldar, Feb 11 2024
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000 (first 5000 terms from Amiram Eldar)
FORMULA
Numbers k such that A068976(k) > 2*k.
EXAMPLE
3600 is in the sequence as its divisors are 1, 2, 3, 4, 5, 6, 8, 9, 10, ..., 400, 450, 600, 720, 900, 1200, 1800, 3600 and the respective values d/core(d) are 1, 1, 1, 4, 1, 1, 4, 9, ... , 900, 400, 900, 3600. The sum of this latter list is 7722 which is greater than 2*3600. - David A. Corneth, Sep 05 2020
MATHEMATICA
f[p_, e_] := If[OddQ[e], 2*(p^(e + 1) - 1)/(p^2 - 1), (p^(e + 2) + p^e - 2)/(p^2 - 1)]; Select[Range[2, 150000], Times @@ (f @@@ FactorInteger[#]) > 2*# &] (* Amiram Eldar, Sep 05 2020 *)
PROG
(PARI) is(n) = sumdiv(n, d, d/core(d)) > n << 1 \\ David A. Corneth, Sep 05 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Benoit Cloitre, Apr 14 2002
EXTENSIONS
More terms from Amiram Eldar, Sep 05 2020
STATUS
approved