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Numbers k == 77 (mod 120) such that (2*k-1)*2^((k-1)/2), (2*k-1)*3^((k-1)/2) and (2*k-1)*5^((k-1)/2) are congruent to 1 (mod k).
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%I #20 Apr 27 2018 17:06:11

%S 197,317,557,677,797,1277,1637,1877,1997,2237,2357,2477,2837,2957,

%T 3557,3677,3797,3917,4157,4397,4517,4637,4877,5237,5477,5717,6197,

%U 6317,6917,7517,7757,7877,8117,8237,8597,8837,9437,9677,10037

%N Numbers k == 77 (mod 120) such that (2*k-1)*2^((k-1)/2), (2*k-1)*3^((k-1)/2) and (2*k-1)*5^((k-1)/2) are congruent to 1 (mod k).

%C There are no composite numbers up to 2*10^17. The first composite term is 229467972529064957.

%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 a:=k->`if`(k mod 120 = 77 and (2*k-1)*2^((k-1)/2) mod k = 1 and (2*k-1)*3^((k-1)/2) mod k = 1 and (2*k-1)*5^((k-1)/2) mod k = 1,k,NULL): seq(a(k),k=1..50); # _Muniru A Asiru_, Mar 11 2018

%t k = 77; lst = {}; While[k < 12000, If[Mod[(2k -1) PowerMod[{2, 3, 5}, (k -1)/2, k], k] == {1, 1, 1}, AppendTo[lst, k]]; k += 120]; lst (* _Robert G. Wilson v_, Feb 13 2018 *)

%o (PARI) is(n) = n%120==77 &&(2*n-1)* Mod(2, n)^((n-1)\2)==1 &&(2*n-1)* Mod(3, n)^((n-1)\2)==1 &&(2*n-1)* Mod(5, n)^((n-1)\2)==1 \\

%o (GAP) Filtered([1..11000],k->k mod 120 = 77 and (2*k-1)*2^((k-1)/2) mod k = 1 and (2*k-1)*3^((k-1)/2) mod k = 1 and (2*k-1)*5^((k-1)/2) mod k = 1); # _Muniru A Asiru_, Mar 11 2018

%Y Cf. A001567.

%K nonn

%O 1,1

%A _Jonas Kaiser_, Feb 11 2018