A331810 Smallest integer x such that prime(n) divides 1 + x + x^2 + ... + x^(q-1) where q = A023503(n), or 0 if no such x exists.

2, 4, 2, 3, 3, 16, 7, 2, 7, 2, 10, 10, 4, 2, 10, 3, 9, 9, 20, 8, 8, 3, 2, 35, 36, 8, 3, 45, 16, 2, 39, 16, 6, 5, 8, 14, 58, 2, 6, 3, 42, 5, 84, 36, 18, 58, 2, 3, 16, 2, 6, 87, 20, 256, 2, 5, 10, 16, 59, 4, 16, 9, 7, 27, 10, 74, 8, 3, 31, 22, 2, 7, 12, 86, 2, 5

Offset: 2

Author

Michel Lagneau, Jan 27 2020

Keywords

• nonn

Comments

Conjecture: every integer > 1 is in the sequence.

Theorem: Let p and q be prime numbers for which there exists an integer x such that p divides 1 + x + x^2 + ... + x^(q-1). Then p == 1 (mod q) or p = q.

Examples

			a(8) = 7 because prime(8) = 19 divides 1 + 7^1 + 7^2 = 57 = 3*19, where q = 3 = A023503(8).
a(9) = 2 because prime(9) = 23 divides 1 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + 2^9 + 2^10  = 2047 = 23*89, where q = 11 = A023503(9).
		

Crossrefs

Cf. A023503 (greatest prime divisor of prime(n) - 1).

Programs

Maple

with(numtheory):nn:=100:
for n from 2 to nn do:
p:=ithprime(n):d:=factorset(p-1):n0:=nops(d):q:=d[n0]:ii:=0:
   for x from 1 to 10^5 while(ii=0) do:
    s:=sum('x^(i-1)', 'i'=1..q):
     if irem(s,p)=0 then
      ii:=1:printf(`%d, `,x):
     else fi:
     od:
    od:

Mathematica

Array[Block[{k = 2}, While[Mod[Total[k^#2 ], #1] != 0, k++]; k ] & @@ {Prime@ #, Range[0, FactorInteger[Prime@ # - 1][[-1, 1]] - 1 ]} &, 76, 2] (* Michael De Vlieger, Jan 27 2020 *)

PARI

a(n) = {my(p=prime(n), q=vecmax(factor(p-1)[,1]), x=1); while (sum(k=0, q-1, x^k) % p, x++); x;} \\ Michel Marcus, Jan 30 2020