|
%I #14 Feb 26 2019 03:40:50
%S 2,2,2,2,2,3,2,2,14,4,7,2,38,6,7,3,4,10,2,9,74,6,10,7,4,61,20,4,5,9,6,
%T 16,6,8,2,9,4,10,2,48,44,163,9,2,95,3,27,70,6,26,57,9,6,8,207,2,27,15,
%U 45,7,69,199,55,16,2,5,12,43,137,39,9,57,5,20,4,115,2,103,45,15,20,109
%N Smallest k such that k^n + k^(n-1) - 1 is prime.
%C The polynomial x^n + x^(n-1) - 1 is irreducible over the rationals (see Ljunggren link), so the Bunyakovsky conjecture implies that a(n) always exists. - _Robert Israel_, Nov 16 2016
%H Robert Israel, <a href="/A125973/b125973.txt">Table of n, a(n) for n = 1..800</a>
%H W. Ljunggren, <a href="https://www.jstor.org/stable/24489120">On the irreducibility of certain trinomials and quadrinomials</a>, Math. Scand. 8 (1960) 65-70.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Bunyakovsky_conjecture">Bunyakovsky conjecture</a>.
%e Consider n = 6. k^6 + k^5 - 1 evaluates to 1, 95, 971 for k = 1, 2, 3. Only the last of these numbers is prime, hence a(6) = 3.
%p f:= proc(n) local k;
%p for k from 2 do if isprime(k^n+k^(n-1)-1) then return k fi od
%p end proc:
%p map(f, [$1..100]); # _Robert Israel_, Nov 16 2016
%t a[n_] := For[k = 2, True, k++, If[PrimeQ[k^n + k^(n-1) - 1], Return[k]]];
%t Array[a, 100] (* _Jean-François Alcover_, Feb 26 2019 *)
%o (PARI) {m=82;for(n=1,m,k=1;while(!isprime(k^n+k^(n-1)-1),k++);print1(k,","))} \\ _Klaus Brockhaus_, Dec 17 2006
%Y Cf. A000040, A045546, A125881-A125885, A125965-A125972, A126017.
%Y Cf. A091997 (n such that a(n)=2).
%K nonn
%O 1,1
%A _Artur Jasinski_, Dec 14 2006
%E Edited and extended by _Klaus Brockhaus_, Dec 17 2006
|
