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A256236 Smallest b > 1 such that the first n primes p (i.e., A000040(1)-A000040(n)) all satisfy b^(p-1) == 1 (mod p^2), i.e., smallest base b larger than 1 such that any member of the set of first n primes is a base-b Wieferich prime. 18

%I #55 Dec 28 2020 13:53:22

%S 5,17,449,557,19601,132857,4486949,126664001,2363321449,5229752849,

%T 2486195039249,16250570614349,83322586961893,39699586259362801,

%U 8042447016668335049,449320365877347849601,4376479338174582826793

%N Smallest b > 1 such that the first n primes p (i.e., A000040(1)-A000040(n)) all satisfy b^(p-1) == 1 (mod p^2), i.e., smallest base b larger than 1 such that any member of the set of first n primes is a base-b Wieferich prime.

%C There might be bases b where prime(n+1) is also a base-b Wieferich prime. This does not affect the membership of b in the sequence.

%C Are there any terms such that a(n) = a(n+1)?

%C Does b exist for all n?

%C All currently known terms satisfy a(n) >= A255901(n). Are there any terms such that a(n) < A255901(n)?

%C If it exists, a(12) > 6*10^12. - _Robert Price_, Oct 10 2019

%C a(n) <= prime(n)#^2+1 = A189409(n), since any prime p is a Wieferich prime in base k*p^2+1 for all k. - _Jens Kruse Andersen_, Dec 20 2020

%e Values of bases b and the values of first Wieferich primes p to base b:

%e b | p

%e -------------------------------------------------------------------------

%e 5 | 2, 20771, 40487 ...

%e 17 | 2, 3, 46021, 48947 ...

%e 449 | 2, 3, 5, 1789 ...

%e 557 | 2, 3, 5, 7, 23, 39829 ...

%e 19601 | 2, 3, 5, 7, 11, 23, 47 ...

%e 132857 | 2, 3, 5, 7, 11, 13, 73, 257 ...

%e 4486949 | 2, 3, 5, 7, 11, 13, 17, 89, 197 ...

%e 126664001 | 2, 3, 5, 7, 11, 13, 17, 19, 101, 2789 ...

%e 2363321449 | 2, 3, 5, 7, 11, 13, 17, 19, 23 ...

%e 5229752849 | 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 881, 2246969 ...

%e 2486195039249 | 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 ...

%t b = 2; Table[While[fnd = True;

%t For[i = 1, i <= n, i++,

%t p = Prime[i];

%t If[PowerMod[b, (p - 1), p^2] != 1 , fnd = False; Break[]]];

%t b++; ! fnd]; b - 1, {n, 5}] (* _Robert Price_, Oct 10 2019 *)

%o (PARI) a(n) = my(v=primes(n)); for(b=2, oo, for(k=1, #v, if(Mod(b, v[k]^2)^(v[k]-1)!=1, break, if(k==#v, return(b)))))

%Y Cf. A255901.

%K nonn,hard,more

%O 1,1

%A _Felix Fröhlich_, Mar 25 2015

%E a(9)-a(11) from _Robert Price_, Oct 10 2019

%E a(12)-a(17) from _Jens Kruse Andersen_, Dec 28 2020

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