OFFSET
1,2
COMMENTS
Since 9 is not a prime, no element > 1 of the sequence A001019(k)=9^k (having k+1 digits in base 9, but 2k+1 divisors) can be member of this sequence. Also, no power of a prime less than 9 can be in the sequence, since it will always have fewer divisors than digits in base 9. However all powers of 11 up to 11^10 are in this sequence, having the same number of digits (in base 9) than the same power of 9 (since 10 = floor(log(11/9)/log(9))) and also that number of divisors (since 11 is prime).
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1500
EXAMPLE
a(1) = 1 since 1 has 1 divisor and 1 digit (in base 9 as in any other base).
It is followed by the primes (having 2 divisors {1,p}) between 9 and 9^2 - 1 (to have 2 digits in base 9).
Then come the squares of primes (3 divisors) between 9^2 = 100_9 and 9^3 - 1 = 888_9.
These are followed by all semiprimes and cubes of primes (4 divisors) between 9^3 and 9^4 - 1.
MATHEMATICA
Select[Range[500], DivisorSigma[0, #] == IntegerLength[#, 9] &] (* G. C. Greubel, Nov 09 2016 *)
PROG
(PARI) for(d=1, 4, for(n=9^(d-1), 9^d-1, d==numdiv(n)&print1(n", ")))
CROSSREFS
KEYWORD
base,easy,nonn
AUTHOR
M. F. Hasler, Nov 28 2007
STATUS
approved