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A066862
Numbers k such that k divides Sum_{i=1..k} gcd(k,i) = A018804(k).
4
1, 4, 15, 16, 27, 48, 60, 64, 108, 144, 240, 256, 325, 432, 729, 891, 960, 1008, 1024, 1200, 1280, 1296, 1300, 1728, 1875, 2916, 3072, 3125, 3564, 3645, 3840, 3888, 4095, 4096, 5200, 6000, 6237, 6375, 6400, 6912, 7056, 7500, 8775, 9216, 11520, 11664, 12500
OFFSET
1,2
COMMENTS
Also k such that Sum_{d|k} phi(d)/d is an integer. - Benoit Cloitre, Apr 14 2002
If two coprime numbers are terms then their product is as well, because Pillai's function A018804(n) is multiplicative. - Thomas Ordowski, Oct 28 2014
The first six squarefree terms are 1, 15=3*5, 1488251=19*29*37*73, 4464753=3*19*29*37*73, 7441255=5*19*29*37*73 and 22323765=3*5*19*29*37*73. Are there any others? - Michel Marcus and Thomas Ordowski, Nov 01 2014
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..3287
László Tóth, A survey of gcd-sum functions, J. Int. Seq., Vol. 13 (2010), Article 10.8.1.
FORMULA
If n = 4^k with k >= 0, n is in the sequence.
If p is prime and k >= 0 then n = p^(kp) is in the sequence. - Thomas Ordowski, Oct 28 2014
MAPLE
A066862:=n->`if`(add(gcd(n, i), i=1..n) mod n = 0, n, NULL):
seq(A066862(n), n=1..500); # Wesley Ivan Hurt, Oct 28 2014
MATHEMATICA
a066862[n_Integer] := Select[Range[n], Divisible[Sum[GCD[#, i], {i, 1, #}], #] &]; a066862[12500] (* Michael De Vlieger, Nov 23 2014 *)
f[p_, e_] := (e*(p - 1)/p + 1); r[n_] := Times @@ (f @@@ FactorInteger[n]); Select[Range[12500], IntegerQ[r[#]] &] (* Amiram Eldar, Apr 09 2022 *)
PROG
(PARI) isok(n) = sum(i=1, n, gcd(n, i)) % n == 0; \\ Michel Marcus, Nov 20 2013
(PARI) A018804(n)=my(f=factor(n)); prod(i=1, #f~, (f[i, 2]*(f[i, 1]-1)/f[i, 1] + 1)*f[i, 1]^f[i, 2])
is(n)=A018804(n)%n==0 \\ Charles R Greathouse IV, Oct 28 2014
CROSSREFS
Cf. A018804.
Sequence in context: A161769 A135658 A228589 * A228590 A103540 A065159
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Jan 25 2002
EXTENSIONS
More terms from Michel Marcus, Nov 20 2013
STATUS
approved