A069056 Numbers k such that Sum_{d|k} d^2*mu(d) divides k^2.

1, 12, 24, 36, 48, 72, 96, 108, 120, 144, 192, 216, 240, 288, 324, 336, 360, 384, 432, 480, 576, 600, 648, 672, 720, 768, 864, 960, 972, 1008, 1080, 1152, 1200, 1296, 1344, 1440, 1536, 1728, 1800, 1920, 1944, 2016, 2160, 2304, 2352, 2400, 2448, 2592, 2688

Offset: 1

Author

Benoit Cloitre, Apr 07 2002

Keywords

• easy
• nice
• nonn

Comments

Numbers k such that A046970(k) divides k^2. [corrected by Amiram Eldar, Apr 20 2025]

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

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Reinhard Zumkeller)

Crossrefs

Cf. A046970, A063453.

Programs

Haskell

a069056 n = a069056_list !! (n-1)
a069056_list = filter (\x -> x ^ 2 `mod` a046970 x == 0) [1..]
-- Reinhard Zumkeller, Jan 19 2012

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 *)
q[k_] := Divisible[k^2, Times @@ ((First[#]^2-1)& /@ FactorInteger[k])]; q[1] = True; Select[Range[3000], q] (* Amiram Eldar, Apr 20 2025 *)

PARI

for(n=1,1000,if(n^2%sumdiv(n,d,moebius(d)*d^2)==0,print1(n,",")))

PARI

isok(k) = {my(f = factor(k)); !(k^2 % prod(i = 1, #f~, f[i,1]^2-1));} \\ Amiram Eldar, Apr 20 2025