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Numbers m of the form 2^k + 1 such that tau(m-2) = tau(m-1) - 1.
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%I #13 Sep 08 2022 08:46:26

%S 3,5,17,65,257,65537,4294967297

%N Numbers m of the form 2^k + 1 such that tau(m-2) = tau(m-1) - 1.

%C Corresponding pairs of values [tau(m-2), tau(m-1)]: [1, 2], [2, 3], [4, 5], [6, 7], [8, 9], [16, 17], [32, 33], ...

%C There are no other terms <= 2^1206 + 1 (from A046801 data).

%C The first 5 known Fermat primes from A019434 are in this sequence. Corresponding values of tau(A019434(n - 2)): 1, 2, 4, 8, 16, ...

%C Conjecture 1: Also numbers m of the form 2^k + 1 such that tau(m - 2) = k.

%C Conjecture 2: If 6th Fermat prime F_p6 exists, then tau(F_p6 - 2) is a power of 2 and tau(F_p6 - 1) = tau(F_p6 - 2) + 1.

%C Conjecture 3: Sequence is finite with 7 terms; supersequence of A262534.

%e For number 257 holds: tau(255) = 8, tau(256) = 9.

%o (Magma) [2^k + 1: k in [1..50] | #Divisors(2^k) - #Divisors(2^k-1) eq 1]

%Y Cf. A000005, A019434, A262534, A343144, A347078.

%Y Intersection of (A055927+2) and A000051.

%K nonn,hard,more

%O 1,1

%A _Jaroslav Krizek_, Dec 16 2021