OFFSET
1,1
COMMENTS
The number of terms of length k is equal to (9*5^(k-1) - 1)/2. - Bernard Schott, Apr 06 2020
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
MAPLE
seq(seq(j*2^k, j=(5^(k-1)+1)/2 .. 5^k-1), k=1..3); # Robert Israel, Apr 05 2020
MATHEMATICA
Select[Range[360], IntegerQ[#/2^IntegerLength[#]] &] (* Jayanta Basu, May 25 2013 *)
PROG
(PARI) isok(n) = n % (2^#Str(n)) == 0; \\ Michel Marcus, Sep 17 2015
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Patrick De Geest, Jun 15 1999
STATUS
approved