A294092 Numbers k == 119 (mod 120) such that 2^((k-1)/2), 3^((k-1)/2) and 5^((k-1)/2) are congruent to 1 (mod k).

239, 359, 479, 599, 719, 839, 1319, 1439, 1559, 2039, 2399, 2879, 2999, 3119, 3359, 3719, 4079, 4679, 4799, 4919, 5039, 5279, 5399, 5519, 5639, 5879, 6359, 6599, 6719, 6959, 7079, 7559, 7919, 8039, 8999, 9239, 9479, 9719, 9839, 10079, 10559, 10799, 11159, 11279

Offset: 1

Author

Jonas Kaiser, Feb 09 2018

Keywords

• nonn

Comments

So far no composite numbers have been found in this sequence. There are no pseudoprimes up to 2^64 in this sequence, so a composite term in this sequence has to exceed 18446744066047760377.

Links

• Chai Wah Wu, 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. Subsequence of A295835.

Programs

GAP

Filtered([1..14000],n->n mod 120=119 and 2^((n-1)/2) mod n =1 and 3^((n-1)/2) mod n =1 and 5^((n-1)/2) mod n =1); # Muniru A Asiru, Feb 15 2018

Mathematica

k = 119; lst = {}; While[k < 12000, If[ PowerMod[{2, 3, 5}, (k - 1)/2, k] == {1, 1, 1}, AppendTo[lst, k]];  k += 120]; lst (* Robert G. Wilson v, Feb 11 2018 *)

PARI

is(n) = n%120==119 && Mod(2, n)^((n-1)\2)==1 && Mod(3, n)^((n-1)\2)==1 && Mod(5, n)^((n-1)\2)==1

Python

A294092_list, k, m = [], 119, 59
while len(A294092_list) < 10000:
    if pow(2,m,k) == 1 and pow(3,m,k) == 1 and pow(5,m,k) == 1:
        A294092_list.append(k)
    k += 120
    m += 60 # Chai Wah Wu, Feb 09 2018

Extensions

More terms from Chai Wah Wu, Feb 10 2018