%I #55 Sep 08 2022 08:46:21

%S 1,10,14,15,21,22,26,27,33,34,35,38,39,46,51,55,57,58,62,65,69,74,77,

%T 82,85,86,87,91,93,94,95,100,196,225,256,441,484,676,1000,1026,1032,

%U 1064,1110,1122,1128,1144,1155,1160,1190,1218,1230,1240,1242,1254,1272,1288,1290,1302,1326,1330,1365,1408

%N Numbers k whose number of divisors is the square of the number of decimal digits of k.

%C The terms with an odd number of digits are squares.

%C The terms with 2 digits are squarefree semiprimes (cf. A006881) Union {27}. The terms with 3 digits belong to A030627 (numbers with 9 divisors) and the ones with 4 digits belong to A030634 (numbers with 16 divisors).

%C The number of terms b(n) with n digits begins with: 1, 30, 7, 753, 3, 11409, 2, ... When there are an odd number of digits, the number of terms decreases from b(3) = 7, b(5) = 3, b(7) = 2. Is there a 2q+1 such that b(2q+1) = 0?

%C The sequence is infinite because 10^k is the term for each k. We have tau(10^k) = tau(2^k)*tau(5^k) = (k + 1)^2 and 10^k has k + 1 digits. - _Marius A. Burtea_, May 09 2019

%C a(n) >= 1, for any n, so b(2q+1)>= 1 for any q. - _Marius A. Burtea_, May 09 2019

%H Sean A. Irvine, <a href="https://github.com/archmageirvine/joeis/blob/master/src/irvine/oeis/a307/A307980.java">Java program</a> (github)

%e 65 is a term with 2 digits and 4 divisors: {1, 5, 13, 65}.

%e 484 is a term with 3 digits and 9 divisors: {1, 2, 4, 11, 22, 44, 121, 242, 484}.

%o (PARI) is(n) = numdiv(n) == #digits(n)^2 \\ _David A. Corneth_, May 08 2019

%o (Magma) [n:n in [1..1500]|NumberOfDivisors(n) eq (#Intseq(n))^2]; // _Marius A. Burtea_, May 09 2019

%Y Cf. A095862 (number of decimal digits = number of divisors).

%Y Cf. A006881 (squarefree semiprimes).

%Y Cf. A030513 (numbers with 4 divisors), A030627 (numbers with 9 divisors), A030634 (numbers with 16 divisors).

%Y Cf. A011557 (subsequence).

%K nonn,base

%O 1,2

%A _Bernard Schott_, May 08 2019