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

%I A300629 #57 Apr 29 2025 08:54:10
%S A300629 561,1105,1729,29341,162401,252601,1152271,2508013,3828001,6733693,
%T A300629 17098369,17236801,29111881,82929001,172947529,216821881,228842209,
%U A300629 366652201,413138881,2301745249,2438403661,5255104513,5781222721,8251854001,12173703001,13946829751,15906120889,23224518901,31876135201,51436355851,57274147841,58094662081
%N A300629 a(1) = 561; a(n+1) = smallest Fermat pseudoprime to all natural bases up to lpf(a(n)).
%C A300629 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.
%C A300629 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.
%C A300629 The above conjecture is true if A083876(n) < A285549(n) for all n > 1, but has not been proven; there is no counterexample up to a(n) < 2^64. - _Max Alekseyev_ and _Thomas Ordowski_, Mar 13 2018
%C A300629 Carl Pomerance (in a letter to the author) wrote, Mar 13 2018: (Start)
%C A300629 Assuming a strong form of the prime k-tuples conjecture, if there are no small counterexamples, there are likely to be none.
%C A300629 Here's why.
%C A300629 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.
%C A300629 So, to get into the sequence, a number needs to be a Fermat pseudoprime for every base up to nearly the cube root.
%C A300629 However, it's a theorem that a sufficiently large number with this property must be a Carmichael number. (End)
%C A300629 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
%C A300629 lpf(a(n)) are listed in A300748. - _Max Alekseyev_, Mar 13 2018
%C A300629 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
%H A300629 Max Alekseyev, <a href="/A300629/b300629.txt">Table of n, a(n) for n = 1..138</a>
%Y A300629 Cf. A002997, A020639, A083876, A271221, A285549, A300748.
%Y A300629 Subsequence of A087788 and of A135720.
%K A300629 nonn
%O A300629 1,1
%A A300629 _Thomas Ordowski_, Mar 10 2018