A135775 - OEIS

%I #12 Nov 09 2016 00:48:55

%S 1,5,7,11,13,17,19,23,25,49,121,125,129,133,134,141,142,143,145,146,

%T 155,158,159,161,166,177,178,183,185,187,194,201,202,203,205,206,209,

%U 213,214,215,217,218,219,221,226,235,237,247,249,253,254,259,262,265,267

%N Numbers having number of divisors equal to number of digits in base 5.

%C Since 5 is a prime, any power 5^k has k+1 divisors { 5^i ; i=0..k } and the same number of digits in base 5; thus the sequence A000351(k)=5^k is a subsequence of this one. It also includes the powers of 7 up to 7^4, since (7/5)^4 < 5 < (7/5)^5.

%H G. C. Greubel, <a href="/A135775/b135775.txt">Table of n, a(n) for n = 1..1000</a>

%e a(1) = 1 since 1 has 1 divisor and 1 digit (in base 5).

%e 2,3,4 have 2 resp. 3 divisors but only 1 digit in base 5, so they are not members of the sequence.

%e a(2) = 5 = 10_5 has 2 divisors { 1, 5 } and 2 digits in base 5, so it is (the second term) in this sequence.

%t Select[Range[300],DivisorSigma[0,#]==IntegerLength[#,5]&] (* _Harvey P. Dale_, Mar 14 2013 *)

%o (PARI) for(d=1,4,for(n=5^(d-1),5^d-1,d==numdiv(n)&print1(n", ")))

%Y Cf. A135772-A135779, A095862.

%K base,nonn

%O 1,2

%A _M. F. Hasler_, Nov 28 2007