A135779 Numbers having number of divisors equal to number of digits in base 9.

1, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 121, 169, 289, 361, 529, 731, 734, 737, 745, 746, 749, 753, 755, 758, 763, 766, 767, 771, 778, 779, 781, 785, 789, 791, 793, 794, 799, 802, 803, 807, 813, 815, 817

Offset: 1

Author

M. F. Hasler, Nov 28 2007

Keywords

• base
• easy
• nonn

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).

Examples

			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.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1500

Crossrefs

Cf. A135772-A135778, A095862.

Programs

Mathematica

Select[Range[500], DivisorSigma[0, #] == IntegerLength[#, 9] &] (* G. C. Greubel, Nov 09 2016 *)

PARI

for(d=1,4,for(n=9^(d-1),9^d-1,d==numdiv(n)&print1(n", ")))