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

5, 7, 10, 11, 13, 14, 15, 17, 21, 22, 23, 26, 27, 30, 31, 32, 34, 37, 39, 41, 42, 45, 46, 47, 50, 53, 54, 57, 60, 61, 62, 65, 67, 72, 73, 74, 78, 82, 83, 90, 94, 96, 97, 98, 99, 101, 103, 104, 106, 107, 111, 114, 120, 122, 128, 129, 130, 131, 133, 134, 143

Offset

1,1

Comments

The sequence contains the smaller member of every pair of sexy primes (A023201).

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.

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Prog

(PARI) isok(n) = numdiv(n) == numdiv(n+3*numdiv(n)); \\ Michel Marcus, Feb 14 2017
(PARI) is(n)=my(d=numdiv(n)); d==numdiv(n+3*d) \\ Charles R Greathouse IV, Feb 14 2017

Crossrefs

Cf. A000005, A025487, A037916, A175304, A282354.

Sequence in context: A364284 A059485 A028786 * A080760 A019470 A161827

Adjacent sequences: A282388 A282389 A282390 * A282392 A282393 A282394

Keyword

nonn

Author

Vladimir Shevelev, Feb 14 2017

Extensions

More terms from Peter J. C. Moses, Feb 14 2017

Status

approved