OFFSET
1,1
COMMENTS
Numbers m such that for all k, d(k) = prime(d(k-1)), where d(k) is the k-th prime factor of m.
The primitive subsequence b(k), k = 1, 2, ... begins with 6, 15, 30, 55, 110, 165, 330, 341, 533, ... because if d(i) is the i-th prime factor of b(k), so b(k)*d(i)^m is in the sequence, m = 0, 1, 2, ...
Numbers m such that if m = p_1^e_1 * ... * p_k^e_k, p_1 < ... < p_k primes, then for all i > 1, p_i = A000040(p_{i-1}). - Antti Karttunen, Aug 24 2019
LINKS
EXAMPLE
330 is in the sequence because the prime factors are {2, 3, 5, 11} with 3 = prime(2), 5 = prime(3) and 11 = prime(5).
1299210 is in the sequence because the prime factors are {2, 3, 5, 11, 31, 127} with 3 = prime(2), 5 = prime(3), 11 = prime(5), 31 = prime(11) and 127 = prime(31).
MAPLE
with(numtheory):nn:=10^3:
for n from 1 to nn do:
d:=factorset(n):n0:=nops(d):it:=0:
if n0>1
then
for i from 2 to n0 do :
if d[i]=ithprime(d[i-1])
then
it:=it+1:
else fi:
od:
if it=n0-1
then
printf(`%d, `, n):
else fi:fi:
od:
MATHEMATICA
aQ[n_] := (m = Length[(p = FactorInteger[n][[;; , 1]])]) > 1 && NestList[Prime@# &, p[[1]], m - 1] == p; Select[Range[770], aQ] (* Amiram Eldar, Aug 24 2019 *)
PROG
(Magma) sol:=[]; s:=1; for m in [2..1000] do v:=PrimeDivisors(m); if #v ge 2 then nr:=0; for k in [2..#v] do if v[k] eq NthPrime(v[k-1]) then nr:=nr+1; end if; end for; if nr eq #v-1 then sol[s]:=m; s:=s+1; end if; end if; end for; sol; // Marius A. Burtea, Aug 24 2019
(PARI) isok(m) = {my(f=factor(m)[, 1]~); if (#f < 2, return(0)); for (i=2, #f, if (f[i] != prime(f[i-1]), return (0)); ); return (1); } \\ Michel Marcus, Aug 25 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Aug 24 2019
EXTENSIONS
Edited by N. J. A. Sloane, Oct 05 2019, using definition suggested by Antti Karttunen, Aug 24 2019
STATUS
approved