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%I #24 Aug 21 2022 11:34:20
%S 3,13,61,127,399403
%N Primes p such that q divides p + 1, r divides q^2 + q + 1, s divides r^2 + r + 1, and p divides s^2 + s + 1 for some primes q, r, and s.
%C There are no other terms below 2^24.
%C If rad(n)^2 = sigma(n), where rad(n) = A007927(n) is the largest squarefree number dividing n and sigma(n) = A000203(n) is the sum of divisors of n, and there exists just one odd prime factor p dividing n exactly once, then p must belong to A354427 or this sequence.
%H Tomohiro Yamada, <a href="https://doi.org/10.2140/moscow.2021.10.249">On a problem of De Koninck</a>, Moscow Journal of Combinatorics and Number Theory, 10:3 (2021), 249-260, <a href="https://doi.org/10.2140/moscow.2021.10.339">correction</a>, 10:4 (2021), 339.
%e 61 is a term since 61 + 1 = 2 * 31, 31^2 + 31 + 1 = 3 * 331, 3^2 + 3 + 1 = 13, and 13^2 + 13 + 1 = 3 * 61.
%o (PARI) is(p)={my(W, V1, V2, V3, V4, q1, q2, q3, q4, i1, i2, i3, i4, l1, l2, l3, l4); W=0; V1=factor(p+1); l1=length(V1[, 1]); for(i1=1, l1, q1=V1[i1, 1]; V2=factor(q1^2+q1+1); l2=length(V2[, 1]); for(i2=1, l2, q2=V2[i2, 1]; V3=factor(q2^2+q2+1); l3=length(V3[, 1]); for(i3=1, l3, q3=V3[i3, 1]; V4=factor(q3^2+q3+1); l4=length(V4[, 1]); for(i4=1, l4, q4=V4[i4, 1];if(q4==p, W=[p, q1, q2, q3]))))); W}
%Y Cf. A101368, A347988, A354427.
%K nonn,more,hard
%O 1,1
%A _Tomohiro Yamada_, Jun 28 2022
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