A354180 Numbers k such that d(k) = 3^i*5*j with i,j >= 0, where d(k) is the number of divisors of k (A000005).

1, 4, 9, 16, 25, 36, 49, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 625, 676, 784, 841, 900, 961, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, 3025, 3249

Offset: 1

Author

Amiram Eldar, May 18 2022

Keywords

• nonn

Comments

All the terms are squares since their number of divisors is odd.

Examples

			4 is a term since A000005(4) = 3 = 3^1*5^0;
16 is a term since A000005(16) = 5 = 3^0*5^1;
144 is a term since A000005(144) = 15 = 3^1*5^1;
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Titus Hilberdink, How often is d(n) a power of a given integer?, Journal of Number Theory, Vol. 236 (2022), pp. 261-279.
• Index entries for sequences computed from exponents in factorization of n.

Crossrefs

Cf. A000005, A000290, A003593.

Programs

Mathematica

p35Q[n_] := n == 3^IntegerExponent[n, 3] * 5^IntegerExponent[n, 5]; Select[Range[60]^2, p35Q[DivisorSigma[0, #]] &]

PARI

is(n) = n==3^valuation(n, 3)*5^valuation(n, 5); \\ A003593
isok(m) = is(numdiv(m)); \\ Michel Marcus, May 19 2022

Formula

The number of terms <= x is c*sqrt(x) + O(x^(1/6)), where c = Product_{p prime} (1 - 1/p)*(Sum_{k in A003593} 1/p^((k-1)/2)) = 0.8747347138... (Hilberdink, 2022).