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%I #36 May 25 2023 01:27:58
%S 1,3,21,273,10101,316407,6914271,2424626841,346084535811,
%T 6177672967557,1741866776384007,92264158181274807,
%U 103008522046409631057,22810816825458528984663,2220066397007943013450011,545889722100356705628041121,73293936170018923619553695493
%N Smallest number of the form k^2 + k + 1 with n distinct prime divisors.
%C 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).
%C 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).
%C Second case: k == 1 (mod 3) => 3 is divisor of k^2 + k + 1, and the other divisors q == 1 (mod 3).
%C a(11) <= 4943071434145592163, a(12) <= 2702887058650660754061, a(13) <= 896265629366361887178273, a(14) <= 72053193593257327979705541. - _Michael S. Branicky_, Mar 21 2021
%C Is a(n) squarefree? The first 16 terms are. - _David A. Corneth_, Mar 21 2021
%D 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.
%e 21 = 3*7;
%e 273 = 3*7*13;
%e 10101 = 3*7*13*37;
%e 316407 = 3*7*13*19*61;
%e 6914271 = 3*7*13*19*31*43;
%e 2424626841 = 3*7*13*19*61*79*97;
%e 346084535811 = 3*7*19*37*43*67*79*103;
%e 6177672967557 = 3*7*13*19*31*43*61*97*151;
%e 1741866776384007 = 3*7*13*19*31*37*43*67*151*673.
%p A174967 := proc(n)
%p for k from 1 do
%p a := k^2+k+1 ;
%p if A001221(a) = n then
%p return a;
%p end if;
%p end do:
%p end proc: # _R. J. Mathar_, Jul 06 2012
%o (Python)
%o from sympy import primefactors
%o def a(n):
%o k = 1
%o while len(primefactors(k**2 + k + 1)) != n: k += 1
%o return k**2 + k + 1
%o print([a(n) for n in range(1, 8)]) # _Michael S. Branicky_, Mar 21 2021
%Y Cf. A002384, A002383.
%K nonn
%O 0,2
%A _Michel Lagneau_, Apr 02 2010
%E a(10) from _Michael S. Branicky_, Mar 21 2021
%E a(0), a(11)-a(16) from _David A. Corneth_, Mar 21 2021
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