A073631 -id:A073631 - cp's OEIS frontend

A130058 Numbers m such that k = m*23^2 divides 3^(k-1) - 2^(k-1).

1, 67, 89, 133, 199, 331, 617, 793, 881, 5281, 8911, 11419, 13333, 22177, 23585, 26467, 35113, 35839, 38897, 40657, 44023, 54913, 65869, 67849, 70819, 92929, 105469, 107185, 114247, 124279, 144673, 153253, 159259, 185329, 196945, 225589

Offset: 1

Alexander Adamchuk, May 04 2007

nonn

All prime p>3 divide 3^(p-1) - 2^(p-1). It appears that 1 and 529 = 23^2 are the only perfect squares in A073631. Most terms of A073631 are squarefree. First 50 nonsquarefree terms of A073631 are the multiples of 23^2.
Conjecture: All nonsquarefree terms of A073631 are the multiples of 23^2.
Prime terms are listed in A130059. Note that the many terms (namely, 1, 133, 793, 8911, 13333, 22177, 26467, 38897, 44023, 54913, 65869, ...) also belong to A073631.

Cf. A073631, A001047, A130059.

Do[ k=n*23^2; f=PowerMod[ 3, k-1, k ] - PowerMod[ 2, k-1, k ]; If[ IntegerQ[ f/k ], Print[ n ] ], {n,1,1000000} ]

A130059 Primes p such that k=p*23^2 divides 3^(k-1) - 2^(k-1); or primes in A130058.

Original entry on oeis.org

67, 89, 199, 331, 617, 881, 5281, 35839, 22270249, 24939553, 395297101, 2414250301, 40834167001, 184879309516177, 207091473814443440700193, 30576308069075829315234744136241, 175651822579831731574054050278935909201, 109606420475170539243380866438311892933511638772789857

Offset: 1

Alexander Adamchuk, May 04 2007

Prime divisors of (3^528 - 2^528) / 23^2 that are congruent to 1 modulo 11.

Cf. A073631, A001047, A130058

Definition clarified by Max Alekseyev, Mar 09 2009

A285300 Numbers k such that 3^(k-1) == 2^(k-1) !== 1 (mod k).

Original entry on oeis.org

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

Offset: 1

Thomas Ordowski, Apr 16 2017

nonn

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)

			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.

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

Cf. A001567, A005935, A052155, A073631.

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

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

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

More terms from Giovanni Resta, Apr 16 2017

A320282 a(n) = (3^(prime(n)-1) - 2^(prime(n)-1))/prime(n).

Original entry on oeis.org

13, 95, 5275, 40565, 2528305, 20376755, 1364211535, 788845655845, 6641614785575, 4056609907500605, 296528399013300025, 2544627551941066235, 188573149984760785495, 121907205372133465501165, 79832689778949397606269355, 694937020886283311634222725, 461241110187445155009340352195

Offset: 3

Jianing Song, Oct 09 2018

nonn

Fermat quotients in base 3/2.
For n > 3, a(n) is divisible by 5.
Primes p such that p^2 divides 3^(p-1) - 2^(p-1) (base-3/2 Wieferich primes) are p = 23, ... What's the next?

			For n = 3, prime(3) = 5 and a(3) = (3^4 - 2^4)/5 = 13.
For n = 4, prime(4) = 7 and a(4) = (3^6 - 2^6)/7 = 95.

G. C. Greubel, Table of n, a(n) for n = 3..317

Cf. A073631 (base-3/2 Fermat pseudoprimes).

[(3^(p-1) - 2^(p-1))  div p: p in PrimesInInterval(4, 100)]; // Vincenzo Librandi, Oct 12 2018

p[n_]:=Prime[n]; a[n_]:=(3^(p[n]-1) - 2^(p[n]-1))/p[n]; Array[a, 50, 3] (* Stefano Spezia, Oct 11 2018 *)

a(n) = my(p=prime(n)); (3^(p-1) - 2^(p-1))/p

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A130058 Numbers m such that k = m*23^2 divides 3^(k-1) - 2^(k-1).

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A130059 Primes p such that k=p*23^2 divides 3^(k-1) - 2^(k-1); or primes in A130058.

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A285300 Numbers k such that 3^(k-1) == 2^(k-1) !== 1 (mod k).

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A320282 a(n) = (3^(prime(n)-1) - 2^(prime(n)-1))/prime(n).

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