A115041 - OEIS

A115041

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

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

Table of n, a(n) for n=1..73.

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

Sequence in context: A368276 A039639 A023509 * A039641 A198729 A093489

Adjacent sequences: A115038 A115039 A115040 * A115042 A115043 A115044

KEYWORD

easy,nonn

AUTHOR

Cino Hilliard, Feb 26 2006

STATUS

approved