%I #59 Sep 06 2024 08:07:13
%S 71,191,239,311,359,431,479,599,719,839,911,1031,1151,1319,1439,1511,
%T 1559,1871,2039,2111,2351,2399,2591,2711,2879,2999,3119,3191,3359,
%U 3671,3719,3911,4079,4271,4391,4679,4751,4799,4871,4919,5039,5231,5279,5351,5399,5471
%N 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).
%C 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.
%C Composite numbers in the sequence include the Carmichael numbers 131314855918751, 23282264781147191, 70122000249565031, 104782993259720471, 583701149409931151, 870012810301712351. - _Robert Israel_, Nov 28 2017
%C 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.
%C 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]
%C From _Jianing Song_, Sep 03 2024: (Start)
%C 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).
%C 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)
%C 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
%H Robert Israel, <a href="/A295835/b295835.txt">Table of n, a(n) for n = 1..10000</a>
%H Jonas Kaiser, <a href="https://arxiv.org/abs/1608.00862">On the relationship between the Collatz conjecture and Mersenne prime numbers</a>, arXiv:1608.00862 [math.GM], 2016.
%p filter:= proc(n) [2&^((n-1)/2),3&^((n-1)/2), 5&^((n-1)/2)] mod n = [1,1,1] end proc:
%p select(filter, [seq(i,i=3..10000,4)]); # _Robert Israel_, Nov 28 2017, corrected Feb 26 2018
%t 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 *)
%o (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
%Y Cf. A001567, A293394, A294717, A294993, A294919, A294912.
%Y A294092 is a subsequence.
%K nonn
%O 1,1
%A _Jonas Kaiser_, Nov 28 2017
%E Definition corrected by _Jonas Kaiser_, Feb 05 2018