[seqfan] a(n) : 2, 4, 4, 5, 4, 16, 4, 16, 6, 4, 27, ...

[Tomasz Ordowski]

Dear SeqFans!

Let a(n) be the smallest k such that k*2^(prime(n)-1) == 1 (mod
prime(n+1)).

Note that a(n) = 2^(prime(n+1)-prime(n)) mod prime(n+1).

I noticed that a(30) = 2^(127-113) mod 127 = 1.

Are there other such cases that a(n) = 1 ?

Can also search to bases b > 2.

Best regards,

Thomas
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Smallest prime p such that n^(p-1) == 1 (mod q),
where q is the next prime after p. Data: 2, 113, ?
The equivalent congruence: n^(q-p) == 1 (mod q).