A295835 Numbers k == 3 (mod 4) such that 2^((k-1)/2), 3^((k-1)/2) and 5^((k-1)/2) are congruent to 1 (mod k).

71, 191, 239, 311, 359, 431, 479, 599, 719, 839, 911, 1031, 1151, 1319, 1439, 1511, 1559, 1871, 2039, 2111, 2351, 2399, 2591, 2711, 2879, 2999, 3119, 3191, 3359, 3671, 3719, 3911, 4079, 4271, 4391, 4679, 4751, 4799, 4871, 4919, 5039, 5231, 5279, 5351, 5399, 5471

Offset: 1

Author

Jonas Kaiser, Nov 28 2017

Keywords

• nonn

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.

Crossrefs

Cf. A001567, A293394, A294717, A294993, A294919, A294912.

A294092 is a subsequence.

Programs

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 *)

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

Extensions

Definition corrected by Jonas Kaiser, Feb 05 2018