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A103795 Minimal base b such that (b^prime(n)+1)/(b+1) is prime. 11

%I #8 Dec 17 2022 08:23:09

%S 2,2,2,2,2,2,2,2,7,2,16,61,2,6,10,6,2,5,46,18,2,49,16,70,2,5,6,12,92,

%T 2,48,89,30,16,147,19,19,2,16,11,289,2,12,52,2,66,9,22,5,489,69,137,

%U 16,36,96,76,117,26,3,159,10,16,209,2,16,23,273,2,460,22,3,36,28,329,43,69,86

%N Minimal base b such that (b^prime(n)+1)/(b+1) is prime.

%C Conjecture: sequence is defined for any n>=2.

%F a(n) = A085398(2*prime(n)) for n >= 2. - _Jinyuan Wang_, Dec 17 2022

%e (2^prime(2)+1)/(2+1) = 3 is prime, so a(2)=2;

%e (2^prime(10)+1)/(2+1) = 178956971 has a factor of 59;

%e (3^prime(10)+1)/(3+1) = 17157594341221 has a factor of 523;

%e ...

%e (7^prime(10)+1)/(7+1) = 402488219476647465854701 is prime, so a(10)=7.

%t Do[p=Prime[k]; n=2; cp=(n^p+1)/(n+1); While[ !PrimeQ[cp], n=n+1; cp=(n^p+1)/(n+1)]; Print[n], {k, 2, 200}]

%Y Cf. A056993, A066180, A085398.

%K nonn

%O 2,1

%A _Lei Zhou_, Feb 23 2005

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