OFFSET
1,1
COMMENTS
There are very few composite numbers in this sequence: The probability of catching a pseudoprime number (A001567) with this definition is estimated at 1 in 263 billion.
Composite numbers in the sequence include the Carmichael numbers 131314855918751, 23282264781147191, 70122000249565031, 104782993259720471, 583701149409931151, 870012810301712351. - Robert Israel, Nov 28 2017
With the exception of the pseudoprimes, it seems that this is a subsequence of A139035. Primes of this form (A139035) have two special properties. 1. There exists a smallest m of the form m = (A139035 - 1)/2 such that 2^m == 1 (mod A139035). 2. m is odd. The core of this definition is based on these two properties. The term 2^((k-1)/2) == 1 (mod n) is based on the first property, while the term k == 3 (mod 4) is based on the second property. The terms 3^((k-1)/2) == 1 (mod n) and 5^((k-1)/2) == 1 (mod n) I just tried freely to Fermat.
Prime terms are congruent to 71 or 119 modulo 120. - Jianing Song, Nov 22 2018 [This is because 2, 3, and 5 must be quadratic residues modulo every prime number in this sequence. - Jianing Song, Sep 01 2024]
From Jianing Song, Sep 03 2024: (Start)
Compare this sequence to the sequence of absolute Euler pseudoprimes (A033181): odd composite numbers k such that a^((k-1)/2) == +-1 (mod k) for every a coprime to k. Such numbers k satisfy 2*psi(k) | (k-1), where psi = A002322, so we must have k == 1 (mod 4).
All terms in this sequence are congruent to 7 modulo 8. In fact, taking the Jacobi symbol modulo k (which only depends on the remainder modulo k) of both sides of 2^((k-1)/2) == 1 (mod k) yields (2/k)^((k-1)/2) = 1. Since k == 3 (mod 4), we have that (k-1)/2 is odd, so (2/k) = 1, which means that k == 7 (mod 8). (End)
Those numbers given above by Robert Israel are all congruent to 71 modulo 120. There are no known composite terms congruent to 119 modulo 120; cf. A294092. - Bill McEachen and Jianing Song, Sep 05 2024
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
Jonas Kaiser, On the relationship between the Collatz conjecture and Mersenne prime numbers, arXiv:1608.00862 [math.GM], 2016.
MAPLE
filter:= proc(n) [2&^((n-1)/2), 3&^((n-1)/2), 5&^((n-1)/2)] mod n = [1, 1, 1] end proc:
select(filter, [seq(i, i=3..10000, 4)]); # Robert Israel, Nov 28 2017, corrected Feb 26 2018
MATHEMATICA
fQ[n_] := PowerMod[{2, 3, 5}, (n - 1)/2, n] == {1, 1, 1}; Select[3 + 4Range@ 1500, fQ] (* Michael De Vlieger, Nov 28 2017 and slightly modified by Robert G. Wilson v, Feb 26 2018 based on the renaming *)
PROG
(PARI) is(n) = n%4==3 && Mod(2, n)^(n\2)==1 && Mod(3, n)^(n\2)==1 && Mod(5, n)^(n\2)==1 && Mod(2, n)^logint(n+1, 2)!=1 \\ Charles R Greathouse IV, Nov 28 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Jonas Kaiser, Nov 28 2017
EXTENSIONS
Definition corrected by Jonas Kaiser, Feb 05 2018
STATUS
approved