

A115039


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


0



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, 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, 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
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OFFSET

1,1


COMMENTS

Conjecture: There will always be an x,y,n such that x^2 + p = y^n for all primes p.


LINKS

Table of n, a(n) for n=1..105.


EXAMPLE

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.


PROG

(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) }


CROSSREFS

Sequence in context: A243929 A075392 A069901 * A032536 A115061 A217834
Adjacent sequences: A115036 A115037 A115038 * A115040 A115041 A115042


KEYWORD

easy,nonn


AUTHOR

Cino Hilliard, Feb 26 2006


STATUS

approved



