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%I #22 Sep 08 2022 08:46:23
%S 13,95,5275,40565,2528305,20376755,1364211535,788845655845,
%T 6641614785575,4056609907500605,296528399013300025,
%U 2544627551941066235,188573149984760785495,121907205372133465501165,79832689778949397606269355,694937020886283311634222725,461241110187445155009340352195
%N a(n) = (3^(prime(n)-1) - 2^(prime(n)-1))/prime(n).
%C Fermat quotients in base 3/2.
%C For n > 3, a(n) is divisible by 5.
%C 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?
%H G. C. Greubel, <a href="/A320282/b320282.txt">Table of n, a(n) for n = 3..317</a>
%e For n = 3, prime(3) = 5 and a(3) = (3^4 - 2^4)/5 = 13.
%e For n = 4, prime(4) = 7 and a(4) = (3^6 - 2^6)/7 = 95.
%t 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 *)
%o (PARI) a(n) = my(p=prime(n)); (3^(p-1) - 2^(p-1))/p
%o (Magma) [(3^(p-1) - 2^(p-1)) div p: p in PrimesInInterval(4, 100)]; // _Vincenzo Librandi_, Oct 12 2018
%Y Cf. A073631 (base-3/2 Fermat pseudoprimes).
%K nonn
%O 3,1
%A _Jianing Song_, Oct 09 2018
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