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.

5, 17, 449, 557, 19601, 132857, 4486949, 126664001, 2363321449, 5229752849, 2486195039249, 16250570614349, 83322586961893, 39699586259362801, 8042447016668335049, 449320365877347849601, 4376479338174582826793

Offset: 1

Author

Felix Fröhlich, Mar 25 2015

Keywords

• nonn
• hard
• more

Comments

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.

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

Does b exist for all n?

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

If it exists, a(12) > 6*10^12. - Robert Price, Oct 10 2019

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

Examples

			Values of bases b and the values of first Wieferich primes p to base b:
b             | p
-------------------------------------------------------------------------
5             | 2, 20771, 40487 ...
17            | 2, 3, 46021, 48947 ...
449           | 2, 3, 5, 1789 ...
557           | 2, 3, 5, 7, 23, 39829 ...
19601         | 2, 3, 5, 7, 11, 23, 47 ...
132857        | 2, 3, 5, 7, 11, 13, 73, 257 ...
4486949       | 2, 3, 5, 7, 11, 13, 17, 89, 197 ...
126664001     | 2, 3, 5, 7, 11, 13, 17, 19, 101, 2789 ...
2363321449    | 2, 3, 5, 7, 11, 13, 17, 19, 23 ...
5229752849    | 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 881, 2246969 ...
2486195039249 | 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 ...
		

Crossrefs

Cf. A255901.

Programs

Mathematica

b = 2; Table[While[fnd = True;
  For[i = 1, i <= n, i++,
   p = Prime[i];
   If[PowerMod[b, (p - 1), p^2] != 1 , fnd = False;  Break[]]];
b++; ! fnd]; b - 1, {n, 5}] (* Robert Price, Oct 10 2019 *)

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)))))

Extensions

a(9)-a(11) from Robert Price, Oct 10 2019

a(12)-a(17) from Jens Kruse Andersen, Dec 28 2020