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A115039 Exponent n for the smallest x such that x^2 + p = y^n over the set of primes p. 0

%I #6 Oct 01 2013 17:58:24

%S 3,2,2,3,3,2,4,2,3,2,5,2,2,2,7,6,2,3,2,9,2,7,2,3,4,2,7,2,3,2,7,2,2,2,

%T 2,9,2,2,3,2,5,2,3,2,2,3,2,9,5,2,2,5,4,7,2,2,2,3,2,2,2,2,3,2,2,2,2,4,

%U 2,2,2,3,11,2,2,2,2,2,2,2,2,5,9,3,3,2,4,2,2,9,2,2,9,2,2,9,2,2,2,2,3,3,2,2,2

%N Exponent n for 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.

%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(exponent(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) }

%K easy,nonn

%O 1,1

%A _Cino Hilliard_, Feb 26 2006

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