A363059 - OEIS

%I #9 May 16 2023 12:50:59

%S 1,5,57,74,202,292,394,514,652,1354,2114,2125,3145,3208,3395,3723,

%T 3783,4053,4401,5018,5225,5298,5425,5770,6039,6363,6795,6918,7564,

%U 7667,7676,7852,7964,8585,9050,9154,10178,10535,10802,10818,10954,11223,12411,13074,13634

%N Numbers k such that the number of divisors of k^2 equals the number of divisors of phi(k), where phi is the Euler totient function.

%C Numbers k such that A048691(k) = A062821(k).

%C Amroune et al. (2023) characterize solutions to this equation and prove that Dickson's conjecture implies that this sequence is infinite.

%C They show that the only squarefree semiprime terms are 57, 514 and some of the numbers of the form 2*(4*p^2+1), where p and 4*p^2+1 are both primes (a subsequence of A259021).

%H Amiram Eldar, <a href="/A363059/b363059.txt">Table of n, a(n) for n = 1..10000</a>

%H Zahra Amroune, Djamel Bellaouar and Abdelmadjid Boudaoud, <a href="https://doi.org/10.7546/nntdm.2023.29.2.284-309">A class of solutions of the equation d(n^2) = d(phi(n))</a>, Notes on Number Theory and Discrete Mathematics, Vol. 29, No. 2 (2023), pp. 284-309.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Dickson%27s_conjecture">Dickson's conjecture</a>.

%e 5 is a term since both 5^2 = 25 and phi(5) = 4 have 3 divisors.

%t Select[Range[15000], DivisorSigma[0, #^2] == DivisorSigma[0, EulerPhi[#]] &]

%o (PARI) is(n) = numdiv(n^2) == numdiv(eulerphi(n));

%Y Cf. A000005, A000010, A048691, A062821.

%Y Cf. A052291, A259021.

%K nonn

%O 1,2

%A _Amiram Eldar_, May 16 2023