A359396 a(n) is the least k such that k^j+2 is prime for j = 1 to n but not n+1.

5, 9, 105, 3, 909, 4995825, 28212939

Offset

1,1

Comments

All terms are odd, and all except a(1) = 5 are divisible by 3.

The generalized Bunyakovsky conjecture implies that a(n) exists for all n.

a(8) > 10^10.

a(8) > 10^11. - Lucas A. Brown, Jan 11 2023

Links

Table of n, a(n) for n=1..7.

Example

a(4) = 3 because 3^1 + 2 = 5, 3^2 + 2 = 11, and 3^3 + 2 = 29 and 3^4 + 2 = 83 are prime but 3^5 + 2 = 245 is not.

Maple

f:= proc(n) local j;
 for j from 1 do
     if not isprime(n^j+2) then return j-1 fi
 od
end proc:
V:= Vector(7): V[1]:= 5: count:= 1:
for k from 3 by 6 while count < 7 do
 v:= f(k);
 if v > 0 and V[v] = 0 then V[v]:= k; count:= count+1 fi
od:
convert(V, list);

Prog

(Python)
from sympy import isprime
from itertools import count, islice
def f(k):
    j = 1
    while isprime(k**j + 2): j += 1
    return j-1
def agen():
    adict, n = dict(), 1
    for k in count(2):
        v = f(k)
        if v not in adict: adict[v] = k
        while n in adict: yield adict[n]; n += 1
print(list(islice(agen(), 5))) # Michael S. Branicky, Jan 09 2023

Crossrefs

Cf. A087576.

Sequence in context: A279707 A329002 A280642 * A222583 A222374 A097397

Adjacent sequences: A359393 A359394 A359395 * A359397 A359398 A359399

Keyword

nonn,more

Author

Robert Israel, Dec 29 2022

Status

approved