

A115041


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


0



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, 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, 129, 132, 135, 10, 139, 141, 142, 147, 7, 156, 157, 159, 166, 13, 174, 175, 177, 12, 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



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. 498^2 + 997 = 499^2. So 499 is the last entry in the table.


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(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)))) ) }


CROSSREFS



KEYWORD

easy,nonn


AUTHOR



STATUS

approved



