cp's OEIS Frontend

The sequence contains the smaller member of every pair of twin primes (A001359) and all squarefree semiprimes m such that m+4 is also a squarefree semiprime (A255746). Can one prove that this is an infinite sequence? - Vladimir Shevelev, Jul 11 2015

The sequence does not contain perfect squares. Indeed, let a(m)=k^2. Then d(k^2+d(k^2)) = d(k^2). Note that d(k^2) is odd. On the other hand, it is known (A046522) that d(k^2)<2*k. Hence, (k+1)^2 - k^2 > d(k^2). Thus k^2Vladimir Shevelev, Feb 10 2017

If p is prime and t+1 is odd prime, then p^t is not in the sequence. Indeed, if d(p^t+t+1)=t+1, then p^t+t+1=q^t, where q is prime > p (if p^t+t+1= say q^l*r^m, then (l+1)*(m+1)=t+1 which is impossible by the condition). But q>=p+2 and p^t+t+1>=p^t+2*t*p^(t-1) or t+1>=2*t*p^(t-1) which trivially has only solution t=1; however, by the condition t>=2. - Vladimir Shevelev, Feb 18 2017

If an odd integer k is in this sequence, so is 2k. - Charlie Neder, Jan 14 2019