A300629 a(1) = 561; a(n+1) = smallest Fermat pseudoprime to all natural bases up to lpf(a(n)).

561, 1105, 1729, 29341, 162401, 252601, 1152271, 2508013, 3828001, 6733693, 17098369, 17236801, 29111881, 82929001, 172947529, 216821881, 228842209, 366652201, 413138881, 2301745249, 2438403661, 5255104513, 5781222721, 8251854001, 12173703001, 13946829751, 15906120889, 23224518901, 31876135201, 51436355851, 57274147841, 58094662081

Offset

1,1

Comments

It is sufficient to consider only prime bases: a(n+1) is the least composite number k such that p^(k-1) == 1 (mod k) for every prime p <= lpf(a(n)), with a(1) = 561.

Conjecture: a(n+1) is the smallest Carmichael number k such that lpf(k) > lpf(a(n)), with a(1) = 561. It seems that such Carmichael numbers have exactly three prime factors.

The above conjecture is true if A083876(n) < A285549(n) for all n > 1, but has not been proven; there is no a counterexample up to a(n) < 2^64. - Max Alekseyev and Thomas Ordowski, Mar 13 2018

Carl Pomerance (in a letter to the author) wrote, Mar 13 2018: (Start)

Assuming a strong form of the prime k-tuples conjecture, if there are no small counterexamples, there are likely to be none.

Here's why.

Assuming prime k-tuples, there are infinitely many Carmichael numbers of the form (6k+1)(12k+1)(18k+1), where each factor is prime. And from Bateman-Horn, these are fairly thickly distributed. There are other even better triples such as (60k+41)(90k+61)(150k+101), where "better" means the least prime factor is not so far below the cube root.

So, to get into the sequence, a number needs to be a Fermat pseudoprime for every base up to nearly the cube root.

However, it's a theorem that a sufficiently large number with this property must be a Carmichael number. (End)

Theorem: if lpf(a(n)) < m < a(n), then m is prime if and only if p^(m-1) == 1 (mod m) for every prime p <= lpf(a(n)). - Thomas Ordowski, Mar 13 2018

lpf(a(n)) are listed in A300748. - Max Alekseyev, Mar 13 2018

For m > 1, A135720(m) >= A083876(m-1), with equality iff lpf(a(n)) = prime(m); by this conjecture in the second comment. - Thomas Ordowski, Mar 13 2018

Links

Max Alekseyev, Table of n, a(n) for n = 1..138

Crossrefs

Cf. A002997, A020639, A083876, A271221, A285549, A300748.

Subsequence of A087788 and of A135720.

Sequence in context: A173703 A306338 A367320 * A135720 A263403 A083733

Adjacent sequences: A300626 A300627 A300628 * A300630 A300631 A300632

Keyword

nonn

Author

Thomas Ordowski, Mar 10 2018

Status

approved