OFFSET
1,2
COMMENTS
If k is not in this sequence, then none of k^(3^t), k^(3^t)+1, ..., (k+1)^(3^t)-1 belong to the sequence. Because (k+1)^(3^k) > 2*k^(3^k), any m > k^(3^k) is not in the sequence, which is a contradiction to {a(n)} is not bounded above. Therefore, this sequence is a permutation of the natural numbers.
LINKS
MATHEMATICA
Nest[Append[#1, If[FreeQ[#1, #2], #2, 2 #1[[-1]] ]] & @@ {#, Floor[#[[-1]]^(1/3)]} &, {1}, 56] (* Michael De Vlieger, Jun 28 2020 *)
PROG
(PARI) lista(nn) = {my(k, v=vector(nn)); v[1]=1; for(n=2, nn, if(vecsearch(vecsort(v), k=sqrtnint(v[n-1], 3)), v[n]=2*v[n-1], v[n]=k)); v; }
CROSSREFS
KEYWORD
nonn
AUTHOR
Jinyuan Wang, Jun 27 2020
STATUS
approved