OFFSET
1,1
COMMENTS
If a(n)=p1*p2*..*pk where p1,p2,..pk primes, then a(n)=m(p1^2+p2^2+..+pk^2) with m a positive integer.
For the special case of m=1, a(n) is equal to the sum of the squares of its prime factors.
There are only 5 known numbers to have this property:
16, 27 and three more numbers with 123, 163 and 179 digits found by Giorgos Kalogeropoulos (see Rivera links).
It is not known if any smaller numbers than those three exist for the case of m=1.
From Robert Israel, Oct 16 2020: (Start)
Suppose n is in the sequence with n = k*A067666(n). Then n^m is in the sequence if m divides k^m (in particular for m=k).
For any prime p, p^(p^j) is in the sequence if j >= 1 (except j>=2 if p=2). (End)
LINKS
Robert Israel, Table of n, a(n) for n = 1..2000
Carlos Rivera, Puzzle 625. Sum of squares of prime divisors, The Prime Puzzles and Problems Connection.
Carlos Rivera, Puzzle 1019. Follow-up to Puzzle 625, The Prime Puzzles and Problems Connection.
EXAMPLE
16 = 2*2*2*2 = 1*(2^2 + 2^2 + 2^2 + 2^2).
7920 = 2*2*2*2*3*3*5*11 = 44*(2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 5^2 + 11^2).
MAPLE
filter:= proc(n) local t;
if isprime(n) then return false fi;
n mod add(t[1]^2*t[2], t=ifactors(n)[2]) = 0
end proc:
select(filter, [$4..30000]); # Robert Israel, Oct 16 2020
MATHEMATICA
Select[Range@20000, Mod[#, Total[Flatten[Table@@@FactorInteger@#]^2]]==0&]
PROG
(PARI) isok(m) = if (!isprime(m) && (m>1), my(f=factor(m)); (m % sum(k=1, #f~, f[k, 1]^2*f[k, 2])) == 0); \\ Michel Marcus, Oct 11 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Giorgos Kalogeropoulos, Oct 09 2020
STATUS
approved