OFFSET
1,2
COMMENTS
The infinitude of this sequence follows from the fact that lcm(1,2,...,n) = exp(n(1+o(1))).
For each positive integer n, all but finitely many terms are divisible by n.
From Robert Israel, Dec 10 2017: (Start)
Numbers n such that for every m>=2 not dividing n, floor(n^(1/(m-1))) <= n^(1/m).
All terms > 1 are even, all terms > 8 are divisible by 12. (End)
The number of terms < 10^k: 4, 8, 15, 33, 72, 134, 859, 1123, ..., . - Robert G. Wilson v, Dec 11 2017
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
EXAMPLE
The sequence contains 8 because the base-b representation of 8 has 1 digit when b > 8, 2 digits when 2 < b <= 8, and 4 digits when b = 2. In each case, the number of digits is a divisor of 8.
MAPLE
filter:= proc(n) local b, d;
for d from 2 to ilog2(n)+1 do
if n mod d <> 0 then
b:= floor(n^(1/(d-1)));
if b^d > n then return false fi;
fi
od;
true
end proc:
select(filter, [$1..8, seq(i, i=12..10^6, 6)]); # Robert Israel, Dec 10 2017
MATHEMATICA
{1}~Join~Select[Range[2, 10^5], Function[b, AllTrue[Range[2, b], Divisible[b, IntegerLength[b, #]] &]]] (* Michael De Vlieger, Dec 09 2017 *)
fQ[n_] := Block[{b = 2, lmt = Floor[ Sqrt[n +1] +2]}, While[b < lmt, If[ Mod[n, Ceiling[ Log[b, n]]] > 0, b = 0; Break[]]; b++]; b > 0]; Join[{1, 2, 6, 8, 12, 36, 48}, Select[60 Range@425, fQ]] (* Robert G. Wilson v, Dec 11 2017 *)
PROG
(Python) [n for n in range(1, 100000) if all(n%k == 0 or n**(1/k) >= int(n**(1/(k-1))) for k in range(2, len(bin(n))-1))]
(PARI) isok(n) = {for (b=2, n, if (n % #digits(n, b), return (0)); ); return (1); } \\ Michel Marcus, Dec 10 2017
CROSSREFS
KEYWORD
nonn,base
AUTHOR
David Radcliffe, Dec 09 2017
STATUS
approved