%N Numbers k such that A001065(k) = sigma(k) - k is the sum of 2 squares.
%C Troupe (2020) proved that N(x), the number of terms not exceeding x, has an order of magnitude x/sqrt(x), i.e., there are two positive constants c1 and c2 such that c1*x/sqrt(x) < N(x) < c2*x/sqrt(x) for sufficiently large x.
%C All the primes are in this sequence since A001065(p) = 1 = 0^2 + 1^2 for a prime p.
%H Amiram Eldar, <a href="/A342190/b342190.txt">Table of n, a(n) for n = 1..10000</a>
%H Lee Troupe, <a href="https://doi.org/10.1090/proc/15104">Divisor sums representable as the sum of two squares</a>, Proceedings of the American Mathematical Society, Vol. 148, No. 10 (2020), pp. 4189-4202.
%e 1 is a term since A001065(1) = 0 = 0^2 + 0^2.
%e 9 is a term since A001065(9) = 4 = 0^2 + 2^2.
%t s[n_] := DivisorSigma[1, n] - n; Select[Range, SquaresR[2, s[#]] > 0 &]
%Y Cf. A001065, A001481.
%A _Amiram Eldar_, Mar 04 2021