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Odd numbers k such that A053223(k) > 0.

Terms congruent to 5 modulo 6 exist but must be very large: for example A053223(670173643268502741420822977335461337017377351999597045900203591953125) = 1311786588705365455963902347308393766941056366825184647502989937872.

A number m coprime to 2 and 3 such that sigma(m)/m >= 3 (m = A358412(3) = A358413(2) = 5^4*7^3*11^2*13^2*17*...*157 ~ 5.16404*10^66 is the smallest such number; see the link from Mercurial, the Spectre) produces a family of infinitely many terms congruent to 5 modulo 6 in this sequence, by Dirichlet's theorem on arithmetic progressions. Concretely, let k == 5 (mod 6), N(t) = t*k*(k+2) + (k+1)/6 for t >= 0, then:

(i) If sigma(k)/k >= 3. If N(t) is prime and 6*N(t)+1 is composite, then sigma(6*N(t)-1) >= 3*(6*N(t)-1), sigma(6*N(t)) = 12*(N(t)+1) and sigma(6*N(t)+1) >= 1+sqrt(6*N(t)+1)+(6*N(t)+1), so A053223(6*N(t)-1) >= sqrt(6*N(t)+1) - 25 >= sqrt(k+2) - 25 > 0.

(ii) If sigma(k+2)/(k+2) >= 3. If N(t) is prime and 6*N(t)-1 is composite, then sigma(6*N(t)+1) >= 3*(6*N(t)+1), sigma(6*N(t)) = 12*(N(t)+1) and sigma(6*N(t)-1) >= 1+sqrt(6*N(t)-1)+(6*N(t)-1), so A053223(6*N(t)-1) >= sqrt(6*N(t)-1) - 21 >= sqrt(k) - 21 > 0.