OFFSET
1,2
COMMENTS
If n > 1, a(n+1) - a(n) == 0 (mod 12), so a(n+1) - a(n) = 12 for n=2,3,4,5,7,8,...; a(n+1) - a(n) = 24 for n=6,9,.... Conjecture: if c > 2 and n > 1, Sum_{d|n} d^c*mu(d) never divides n^c. Hence A063453(n) never divides n^3 for n > 1.
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
FORMULA
Numbers k such that A046970(k) divides k.
MATHEMATICA
f[d_] := d^2*MoebiusMu[d]; ok[n_] := Divisible[n^2, Total[f /@ Divisors[n]]]; Select[Range[3000], ok] (* Jean-François Alcover, Nov 15 2011 *)
PROG
(PARI) for(n=1, 1000, if(n^2%sumdiv(n, d, moebius(d)*d^2)==0, print1(n, ", ")))
(Haskell)
a069056 n = a069056_list !! (n-1)
a069056_list = filter (\x -> x ^ 2 `mod` a046970 x == 0) [1..]
-- Reinhard Zumkeller, Jan 19 2012
CROSSREFS
KEYWORD
easy,nice,nonn
AUTHOR
Benoit Cloitre, Apr 07 2002
STATUS
approved