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A290487
Prime numbers p such that q = p^2+p-1 and r = (p^3+p^2-p+1)/2 are also primes, and thus p*q*r is an extreme 3-Carmichael number.
1
3, 5, 31, 41, 83, 131, 349, 373, 401, 449, 463, 1483, 1801, 1871, 2003, 2011, 2551, 2711, 2861, 3229, 3739, 3779, 4111, 4493, 5413, 5659, 6323, 6421, 7309, 7369, 7523, 8101, 8713, 9091, 9161, 9239, 9719, 9733, 10691, 10709, 11941, 12781, 12893, 15101, 15959
OFFSET
1,1
COMMENTS
Carlos Rivera said that a 3-Carmichael number is "extreme" if its prime factors are the triple (p, p^2+p-1, (p^3+p^2-p+1)/2), since Richard Pinch and Chris Nash showed that given a prime p which is the least in the triple of the prime factors of a 3-Carmichael numbers, p*q*r, with p < q < r, then r <= (p^3+p^2-p+1)/2.
The corresponding extreme 3-Carmichael numbers are 561, 10585, 471905281, 2489462641, 167385219121, 2565387704881, 908658513945001, 1353761995707001, 2089268322242401, 4115059336080001, 4946817525914881, 5326010469307447921, 17081820793305370801, 21472202547687640321, 32321314831390144561,...
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1890 from Vincenzo Librandi)
Carlos Rivera, Conjecture 19, A bound to the largest prime factor of certain Carmichael numbers, The Prime Puzzles and Problems Connection.
EXAMPLE
p = 5, q = p^2+p-1 = 29 and r = (p^3+p^2-p+1)/2 = 73 are all primes, thus 5 is in the sequence. 5*29*73 = 10585 is a 3-Carmichael number.
MATHEMATICA
Select[Prime[Range[2 10^3]], And@@PrimeQ[{#^2 + # - 1, ((#^3 + #^2 - # + 1)/2)}]&] (* Vincenzo Librandi, Oct 13 2017 *)
PROG
(PARI) lista(nn) = forprime(p=3, nn, if (isprime(p^2+p-1) && isprime((p^3+p^2-p+1)/2), print1(p, ", "))); \\ Michel Marcus, Oct 13 2017
(Magma) [p: p in PrimesInInterval(3, 2*10^4) | IsPrime(p^2+p-1) and IsPrime((p^3+p^2-p+1) div 2)]; // Vincenzo Librandi, Oct 13 2017
CROSSREFS
Cf. A087788 (3-Carmichael numbers).
Sequence in context: A189739 A372800 A035410 * A059940 A106893 A175097
KEYWORD
nonn
AUTHOR
Amiram Eldar, Aug 03 2017
STATUS
approved