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A282391 Numbers j such that d(j) = d(j + 3*d(j)), where d(j) is the number of divisors of j.

%I A282391 #21 Mar 03 2024 10:15:15
%S A282391 5,7,10,11,13,14,15,17,21,22,23,26,27,30,31,32,34,37,39,41,42,45,46,
%T A282391 47,50,53,54,57,60,61,62,65,67,72,73,74,78,82,83,90,94,96,97,98,99,
%U A282391 101,103,104,106,107,111,114,120,122,128,129,130,131,133,134,143
%N A282391 Numbers j such that d(j) = d(j + 3*d(j)), where d(j) is the number of divisors of j.
%C A282391 The sequence contains the smaller member of every pair of sexy primes (A023201).
%C A282391 The sequence contains no perfect squares. Indeed, let a(m) = k^2 for some m. Then, by the definition, d(k^2 + 3*d(k^2)) = d(k^2). Note that d(k^2) is odd. On the other hand, it is known (cf. A046522) that d(k^2) < 2*k. Hence (k+3)^2 - k^2 = 6*k + 1 > 3*d(k^2). Thus k^2 < k^2 + 3*d(k^2) < (k+3)^2. Note that, evidently, k^2 + 3*d(k^2) cannot be (k+2)^2. Let us also show that k^2 + 3*d(k^2) cannot be (k+1)^2, or, equivalently, 3*d(k^2) cannot be equal to 2*k + 1. Indeed, let 3*d(k^2) = 2*k + 1. For some prime p, let p^a || k (that is, p^a | k, but p^(a+1) !| k), a > 0, so 2*k + 1 == 1 (mod p). But now we have 3*p^(a+1) | 3*d(k^2) and thus 3*p^(a+1)|2*k + 1, so 2*k + 1 == 0 (mod p). Contradiction. Therefore, we conclude that k^2 + 3*d(k^2) cannot be a square. Hence, d(k^2 + 3*d(k^2)) is even, which is a contradiction.
%H A282391 Charles R Greathouse IV, <a href="/A282391/b282391.txt">Table of n, a(n) for n = 1..10000</a>
%o A282391 (PARI) isok(n) = numdiv(n) == numdiv(n+3*numdiv(n)); \\ _Michel Marcus_, Feb 14 2017
%o A282391 (PARI) is(n)=my(d=numdiv(n)); d==numdiv(n+3*d) \\ _Charles R Greathouse IV_, Feb 14 2017
%Y A282391 Cf. A000005, A025487, A037916, A175304, A282354.
%K A282391 nonn
%O A282391 1,1
%A A282391 _Vladimir Shevelev_, Feb 14 2017
%E A282391 More terms from _Peter J. C. Moses_, Feb 14 2017