A174967 Smallest number of the form k^2 + k + 1 with n distinct prime divisors.

1, 3, 21, 273, 10101, 316407, 6914271, 2424626841, 346084535811, 6177672967557, 1741866776384007, 92264158181274807, 103008522046409631057, 22810816825458528984663, 2220066397007943013450011, 545889722100356705628041121, 73293936170018923619553695493

Offset

0,2

Comments

If k == 2 (mod 3), all prime divisors of k^2 + k + 1 are congruent to 1 (mod 3), and if k == 1 (mod 3), the number 3 is divisor, and the other divisors are congruent to 1 (mod 3).

Proof: first case: k == 2 (mod 3): let q divide k^2 + k + 1. Then 4q divides 4*(k^2 + k + 1) = (2k+1)^2 + 3, and (-3/q)=1, where (a/b) is the Legendre symbol. By using the law of quadratic reciprocity, we obtain (-3/q) = (-1/q)(3/q) = (-1/q)(q/3)(-1)^((q-1)/2)(3-1)/2)) = ((-1)^(q-1)/2)((-1)^(q-1)/2)(q/3) = (q/3). Suppose q !== 1 (mod 3). Then k^2 + k + 1 !== 0 (mod 3) => q == 2 (mod 3), and then (q/3) = -1 => (-3/q) = -1, a contradiction. So q == 1 (mod 3).

Second case: k == 1 (mod 3) => 3 is divisor of k^2 + k + 1, and the other divisors q == 1 (mod 3).

a(11) <= 4943071434145592163, a(12) <= 2702887058650660754061, a(13) <= 896265629366361887178273, a(14) <= 72053193593257327979705541. - Michael S. Branicky, Mar 21 2021

Is a(n) squarefree? The first 16 terms are. - David A. Corneth, Mar 21 2021

References

L. Poletti, Le serie dei numeri primi appartenente alle due forme quadratiche (A) n^2+n+1 e (B) n^2+n-1 per l'intervallo compreso entro 121 milioni, e cioè per tutti i valori di n fino a 11000, Atti della Reale Accademia Nazionale dei Lincei, Memorie della Classe di Scienze Fisiche, Matematiche e Naturali, s. 6, v. 3 (1929), pages 193-218.

Links

Table of n, a(n) for n=0..16.

Example

21 = 3*7;
273 = 3*7*13;
10101 = 3*7*13*37;
316407 = 3*7*13*19*61;
6914271 = 3*7*13*19*31*43;
2424626841 = 3*7*13*19*61*79*97;
346084535811 = 3*7*19*37*43*67*79*103;
6177672967557 = 3*7*13*19*31*43*61*97*151;
1741866776384007 = 3*7*13*19*31*37*43*67*151*673.

Maple

A174967 := proc(n)
        for k from 1 do
                a := k^2+k+1 ;
                if A001221(a) = n then
                        return a;
                end if;
        end do:
end proc: # R. J. Mathar, Jul 06 2012

Prog

(Python)
from sympy import primefactors
def a(n):
  k = 1
  while len(primefactors(k**2 + k + 1)) != n: k += 1
  return k**2 + k + 1
print([a(n) for n in range(1, 8)]) # Michael S. Branicky, Mar 21 2021

Crossrefs

Cf. A002384, A002383.

Sequence in context: A336809 A066206 A130032 * A126461 A370741 A000681

Adjacent sequences: A174964 A174965 A174966 * A174968 A174969 A174970

Keyword

nonn

Author

Michel Lagneau, Apr 02 2010

Extensions

a(10) from Michael S. Branicky, Mar 21 2021

a(0), a(11)-a(16) from David A. Corneth, Mar 21 2021

Status

approved