

A068991


Numbers k such that Sum_{d divides k} sigma(d)/phi(d) is an integer.


2



1, 2, 3, 6, 10, 21, 30, 42, 78, 110, 210, 330, 390, 930, 1218, 1830, 2025, 2310, 2530, 4050, 4134, 4290, 6090, 7590, 14175, 14910, 22110, 28350, 51090, 52650, 53130, 66990, 71862, 98670, 118910, 159975, 214650, 319950, 356730, 359310, 442338, 635850, 684450
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OFFSET

1,2


COMMENTS

Conjecture: if k is in the sequence and k is squarefree then the denominator of the 2kth 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

Sequence in context: A120707 A047111 A106741 * A187491 A178852 A215067
Adjacent sequences: A068988 A068989 A068990 * A068992 A068993 A068994


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Apr 06 2002


EXTENSIONS

More terms from Michel Marcus, Dec 07 2013


STATUS

approved



