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A285300

Numbers k such that 3^(k-1) == 2^(k-1) !== 1 (mod k).

65, 133, 529, 793, 1649, 2059, 2321, 4187, 5185, 6305, 6541, 6697, 6817, 7471, 7613, 8113, 10963, 11521, 13213, 13333, 13427, 14701, 14981, 19171, 19201, 19909, 21349, 21667, 22177, 26065, 26467, 32873, 35443, 36569, 37333, 38897, 42121, 42127, 44023, 47081

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OFFSET

1,1

COMMENTS

All terms are odd composite numbers. There are no pseudoprimes to bases 2 or 3 in this sequence.

Are there infinitely many numbers of this kind?

From Max Alekseyev, Apr 16 2017: (Start)

Also, Fermat pseudoprimes base 2/3 that are not Fermat pseudoprimes base 2.

Also, the set difference of A073631 and either of ({1} U A001567), ({1} U A005935), or ({1} U A052155). (End)

LINKS

Amiram Eldar, Table of n, a(n) for n = 1..1000

EXAMPLE

2^64 = 18446744073709551616 = 65 * 283796062672454640 + 16 and 3^64 = 3433683820292512484657849089281 = 65 * 52825904927577115148582293681 + 16. Therefore 65 is in the sequence.

Note: a(3) = 529 = 23^2 and a(40) = 47081 = 23^2 * 89.

MAPLE

filter:= proc(n) local t;

t:= 3 &^(n-1) mod n;

if t = 1 then return false fi;

t = 2 &^(n-1) mod n;

end proc:

select(filter, [seq(i, i=3..10^5, 2)]); # Robert Israel, Apr 27 2017

MATHEMATICA

Select[Range[2, 10^5], PowerMod[2, # - 1, #] == PowerMod[3, # - 1, #] != 1 &] (* Giovanni Resta, Apr 16 2017 *)

PROG

(PARI) is(n) = Mod(3, n)^(n-1)==2^(n-1) && Mod(2, n)^(n-1)!=1 \\ Felix Fröhlich, Apr 27 2017