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%I #8 Oct 01 2013 17:58:24
%S 3,2,3,2,3,7,3,10,3,15,2,19,21,22,2,3,30,5,34,2,37,2,42,5,7,51,2,54,5,
%T 57,2,66,69,70,75,2,79,82,6,87,3,91,6,97,99,7,106,2,3,115,117,3,11,3,
%U 129,132,135,10,139,141,142,147,7,156,157,159,166,13,174,175,177,12,2
%N y in the smallest x such that x^2 + p = y^n over the set of primes p.
%C Conjecture: There will always be an x,y,n such that x^2 + p = y^n for all primes p.
%e 5 is the smallest number that when we add its square to prime 2, we get a perfect power, 3^3. So 3 is the first entry. 498^2 + 997 = 499^2. So 499 is the last entry in the table.
%o (PARI) sqplusp(n) = { local(p,x,y,c=0); forprime(p=2,n, for(x=1,n, y=x^2+p; if(ispower(y), print1(power(y)",");c++;break ) ) ); print(); print(c) } exponent(n) = \ Return the exponent if n is a perfect power { local(x,ln,j,e=0); ln=omega(n); x=factor(n); e=x[1,2]; for(j=2,ln, if(x[j,2] < e,e=x[j,2]) ); return(e) } power(n) = \Return the largest factor for which n is a perfect power { if(ispower(n), return(round(n^(1/exponent(n)))) ) }
%K easy,nonn
%O 1,1
%A _Cino Hilliard_, Feb 26 2006
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