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Least number k such that k^n - k^(n-1) - k^(n-2) - ... - k^2 - k - 1 is prime.
1

%I #12 Nov 09 2024 07:49:10

%S 3,3,4,3,4,4,30,4,4,15,4,27,54,10,4,3,10,11,56,35,4,35,8,34,6,24,32,

%T 17,6,4,18,3,100,191,10,63,54,40,4,129,20,14,474,16,142,330,20,36,116,

%U 4,664,161,32,19,62,7,54,366,132,71,162,5,4,3,204,60,18,30,198,155,28,274,6

%N Least number k such that k^n - k^(n-1) - k^(n-2) - ... - k^2 - k - 1 is prime.

%C a(n) > 2 for all n.

%H Harvey P. Dale, <a href="/A243341/b243341.txt">Table of n, a(n) for n = 1..200</a>

%e 1^1-1 = 0 is not prime. 2^1-1 = 1 is not prime. 3^1-1 = 2 is prime. Thus a(1) = 3.

%t lnk[n_]:=Module[{k=2},While[!PrimeQ[k^n-Total[k^Range[0,n-1]]],k++]; k]; Array[lnk,80] (* _Harvey P. Dale_, Aug 26 2016 *)

%o (PARI) a(n)=for(k=1,oo,if(ispseudoprime(k^n-sum(i=0,n-1,k^i)),return(k)))

%o for(n=1,100,print1(a(n)", "))

%K nonn

%O 1,1

%A _Derek Orr_, Jun 03 2014