OFFSET
1,2
COMMENTS
Conjecture: if k is in the sequence and k is squarefree then the denominator of the 2k-th Bernoulli's number contains k. E.g., 2310 is squarefree, is in the sequence and A002445(2310)=744535159372016163713900138929458330 is divisible by 2310.
For the first 74 terms, the largest k < n such that a(k) | a(n) is close to n. Is it sufficient to assume a(k) * m = a(n) to find the next terms merely recursively? If so, how large do we choose m? - David A. Corneth, Oct 11 2019
Six other terms are 240998502150, 275082346350, 1660078844550, 2170540451310, 13878528210690, 722754507947850. - David A. Corneth, Oct 21 2019
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..74
MATHEMATICA
aQ[n_] := IntegerQ @ DivisorSum[n, DivisorSigma[1, #]/EulerPhi[#] &]; Select[
Range[10000], aQ] (* Amiram Eldar, Oct 05 2019 *)
PROG
(PARI) isok(n) = denominator(sumdiv(n, d, sigma(d)/eulerphi(d))) == 1; \\ Michel Marcus, Dec 07 2013
(Magma) [k:k in [1..600000]| IsIntegral(&+[ DivisorSigma(1, d)/EulerPhi(d):d in Divisors(k)])]; // Marius A. Burtea, Oct 10 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Apr 06 2002
EXTENSIONS
More terms from Michel Marcus, Dec 07 2013
STATUS
approved