

A135779


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


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
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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^(d1), 9^d1, d==numdiv(n)&print1(n", ")))


CROSSREFS

Cf. A135772A135778, A095862.
Sequence in context: A108871 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



