cp's OEIS Frontend

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.