OFFSET
1,1
COMMENTS
The sequence contains the smaller member of every pair of cousin primes (A023200).
The sequence contains no perfect squares. Indeed, let a(m) = k^2 for some m. Then, by the definition, d(k^2 + 2*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+2)^2 - k^2 = 4*k + 4 > 2*d(k^2). Thus k^2 < k^2 + 2*d(k^2) < (k+2)^2. Since, evidently, k^2 + 2*d(k^2) cannot be (k+1)^2, then k^2 + 2*d(k^2) cannot be a square. Therefore, d(k^2 + 2*d(k^2)) is even, which is a contradiction.
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
MATHEMATICA
Select[Range@ 210, Function[d, DivisorSigma[0, # + 2 d] == d]@ DivisorSigma[0, #] &] (* Michael De Vlieger, Feb 13 2017 *)
PROG
(PARI) is(n)=my(d=numdiv(n)); d==numdiv(n+2*d) \\ Charles R Greathouse IV, Feb 14 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Feb 13 2017
EXTENSIONS
More terms from Peter J. C. Moses, Feb 13 2017
STATUS
approved