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A354427
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Primes p such that q divides p + 1, r divides q^2 + q + 1 and p divides r^2 + r + 1 for some primes q and r.
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2
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OFFSET
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1,1
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COMMENTS
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There are no other terms below 2^24.
1794067711 and 10855016833 are also terms, which arise from q = 9829 and q = 5569 in A354426 respectively, since
1794067711 + 1 = 2^8 * 23 * 31 * 9829, 9829^2 + 9829 + 1 = 3 * 439 * 73363, 73363^2 + 73363 + 1 = 3 * 1794067711 and
10855016833 + 1 = 2 * 17 * 5569 * 57329, 5569^2 + 5569 + 1 = 3 * 7 * 1477111, 1477111^2 + 1477111 + 1 = 3 * 67 * 10855016833.
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 this sequence, or 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 (Yamada, 2021). [Corrected by Tomohiro Yamada, Jun 05 2022]
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LINKS
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EXAMPLE
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631 is a term since 631 + 1 = 2^3 * 79, 79^2 + 79 + 1 = 3 * 7^2 * 43, and 43^2 + 43 + 1 = 3 * 631.
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PROG
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(PARI) is(p)={my(W, V1, V2, V3, q1, q2, q3, i1, i2, i3, l1, l2, l3); 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]; if(q3==p, W=[p, q1, q2])))); W}
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CROSSREFS
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Cf. A354426 (q corresponding to primes p in this sequence), A354428 (r corresponding to primes p in this sequence).
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KEYWORD
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nonn,more,hard
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AUTHOR
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STATUS
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approved
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