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

1,2

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

Cf. A135772-A135778, A095862.

Sequence in context: A347819 A268031 A167847 * A135778 A078875 A244555

Adjacent sequences: A135776 A135777 A135778 * A135780 A135781 A135782

Keyword

base,easy,nonn

Author

M. F. Hasler, Nov 28 2007

Status

approved