cp's OEIS Frontend

a(n) is the smallest composite k such that p^k == p (mod k) for every prime p <= A000040(n) and A002322(k) does not divide k-1.

If a composite m < a(n) and p^m == p (mod m) for every prime p <= prime(n), then m is a Carmichael number.

a(23) > 2^64. - Max Alekseyev, Apr 22 2017

Conjecture: lpf(a(n)) > prime(n), where lpf = A020639. - Thomas Ordowski, May 13 2017

Except a(19), the listed terms are semiprime. - Thomas Ordowski, Feb 09 2018

a(24) <= 21150412877533909683421, a(362) <= (416*A002110(360) + 1) * (832*A002110(360) + 1). - Daniel Suteu, Nov 13 2022

a(n) is the smallest composite k such that b^(k-1) == 1 (mod (b-1)k) for every b = 2, 3, 4, ..., n. For more comments, see A083876 and A300629. - Max Alekseyev and Thomas Ordowski, Apr 29 2018

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 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

For m > 2, A135720(m) = A083876(m-1) if and only if a(n) = prime(m).

Pseudoprime A001567(n) is a pseudoprime to base 2 through base prime(a(n)), where a(n) is this sequence.

a(n) <= A177415(n).

Each a(n) is an odd prime.

If k = A001567(n) is a Carmichael number, then a(n) = lpf(k).

Conjecture: if k = A001567(n) is semiprime, then a(n) < lpf(k).

The smallest numbers k = A001567(n) such that a(n) = prime(m) for m > 1 are 341, 1105, 1729, 75361, 29341, 162401, 334153, ... See A135720 > 561.

The smallest such semiprimes are 341, 2701, ?, 721801, ... Cf. A285549.

a(n) is coprime to A002110(n).