[seqfan] An unresolved problem
Tomasz Ordowski
tomaszordowski at gmail.com
Sun Oct 21 07:15:19 CEST 2018
Dear SeqFans!
As is well known:
If p is a prime that does not divide b-1, then (b^p-1)/(b-1) has all
divisors d == 1 (mod p).
Let a(n) be the smallest composite k such that (n^k-1)/(n-1) has all
divisors d == 1 (mod k).
Note that (n^k-1)/(n-1) == 1 (mod k), so n^k == n (mod (n-1)k). Thus a(n)
>= A298908(n).
Cf. https://oeis.org/history/view?seq=A298908&v=25
a(n) is the least composite k such that (n^k-1)/(n-1) has all prime
divisors p == 1 (mod k).
For n > 2, a(n) = 91, 4, 5731, 6, 25, 9, 91, 10, ... Please more.
The basic problem: a(2) = ?
The question: Are there pseudoprimes k = pq such that both 2^p-1 and 2^q-1
are prime?
Theorem: If k is such a pseudoprime, then 2^k-1 has all divisors d == 1
(mod k).
Cf. https://oeis.org/A298076
Can also consider these numbers to negative bases by defining:
Least odd composite k such that (n^k+1)/(n+1) has all divisors d == 1 (mod
k).
Data: 9, 341, 91, ... for n = 1, 2, 3, ... Cf. https://oeis.org/A298310
The most interesting is the base -2, so let's define:
Odd composite numbers k such that (2^k+1)/3 has all divisors d == 1 (mod
k).
I found only two such numbers k = 341 = 11*31 and k = 5461 = 43*127. Please
more.
If both (2^p+1)/3 and (2^q+1)/3 are prime and n = pq is pseudoprime, then n
is such a number k.
Do known the Wagstaf exponents https://oeis.org/A000978 give more such
pseudoprimes?
Does it exhaust all other cases?
Best regards,
Thomas
____________________
The numeric coincidence:
These two numbers 341 and 5461 are also Cipolla pseudoprimes:
https://oeis.org/A210454
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