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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).

%I A354180 #15 May 21 2022 14:52:34
%S A354180 1,4,9,16,25,36,49,81,100,121,144,169,196,225,256,289,324,361,400,441,
%T A354180 484,529,625,676,784,841,900,961,1089,1156,1225,1296,1369,1444,1521,
%U A354180 1681,1764,1849,1936,2025,2116,2209,2304,2401,2500,2601,2704,2809,3025,3249
%N 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).
%C A354180 All the terms are squares since their number of divisors is odd.
%H A354180 Amiram Eldar, <a href="/A354180/b354180.txt">Table of n, a(n) for n = 1..10000</a>
%H A354180 Titus Hilberdink, <a href="https://doi.org/10.1016/j.jnt.2021.07.020">How often is d(n) a power of a given integer?</a>, Journal of Number Theory, Vol. 236 (2022), pp. 261-279.
%H A354180 <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>.
%F A354180 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).
%e A354180 4 is a term since A000005(4) = 3 = 3^1*5^0;
%e A354180 16 is a term since A000005(16) = 5 = 3^0*5^1;
%e A354180 144 is a term since A000005(144) = 15 = 3^1*5^1;
%t A354180 p35Q[n_] := n == 3^IntegerExponent[n, 3] * 5^IntegerExponent[n, 5]; Select[Range[60]^2, p35Q[DivisorSigma[0, #]] &]
%o A354180 (PARI) is(n) = n==3^valuation(n, 3)*5^valuation(n, 5); \\ A003593
%o A354180 isok(m) = is(numdiv(m)); \\ _Michel Marcus_, May 19 2022
%Y A354180 Cf. A000005, A000290, A003593.
%K A354180 nonn
%O A354180 1,2
%A A354180 _Amiram Eldar_, May 18 2022