|
%I #7 Mar 30 2012 18:35:54
%S 1,3,4,5,7,8,9,11,12,13,15,16,17,19,21,24,25,27,29,31,32,33,35,39,40,
%T 41,43,45,48,49,51,55,56,57,59,60,61,63,64,65,69,71,72,73,75,77,80,81,
%U 84,85,87,88,91,93,95,96,99,101,103,104,105,107,109,111,112,115,120
%N Numbers n such that n*p + 1 is a square for some prime p.
%C The smallest corresponding p are {3, 5, 2, 3, 5, 3, 7, 13, 2, 11, 13, 3, 19, 17, 3, 2, 23, 29, 31, 29, 7, 3, 37,...}
%F Complement of A187884.
%e 21 is in the sequence because 21*3 + 1 = 8^2, with p = 3.
%p with(numtheory): for k from 1 to 120 do : q:=0:for p from 1 to 200 do : x:=sqrt(k*p+1)
%p : if x=trunc(x) and type(p,prime)=true and q=0 then q:=1: printf(`%d, `,k):else
%p fi:od:od:
%Y Cf. A187884, A187886.
%K nonn
%O 1,2
%A _Michel Lagneau_, Mar 15 2011
|
