A068991 Numbers k such that Sum_{d divides k} sigma(d)/phi(d) is an integer.

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

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

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