Amiram Eldar, <a href="/A363059/b363059_1.txt">Table of n, a(n) for n = 1..10000</a>

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Amiram Eldar, <a href="/A363059/b363059_1.txt">Table of n, a(n) for n = 1..10000</a>

allocated for Amiram EldarNumbers 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.

1, 5, 57, 74, 202, 292, 394, 514, 652, 1354, 2114, 2125, 3145, 3208, 3395, 3723, 3783, 4053, 4401, 5018, 5225, 5298, 5425, 5770, 6039, 6363, 6795, 6918, 7564, 7667, 7676, 7852, 7964, 8585, 9050, 9154, 10178, 10535, 10802, 10818, 10954, 11223, 12411, 13074, 13634

1,2

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

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

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).

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.

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

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

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

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

Cf. A000005, A000010, A048691, A062821.

Cf. A052291, A259021.

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Amiram Eldar, May 16 2023

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allocated for Jack Braxton

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