A115039 a(n) is the greatest k such that A115038(n)^2 + prime(n) is a k-th power.

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

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

Robert Israel, Table of n, a(n) for n = 1..10000

Formula

A115038(n)^2 + prime(n) = A115041(n)^a(n). - Robert Israel, Mar 10 2026

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.

Maple

g:= proc(n) local p, F, x, d;
  p:= ithprime(n);
  for x from 1 do
    F:= ifactors(x^2+p)[2];
    d:= igcd(F[.., 2]);
    if d > 1 then return d fi
  od
end proc:
map(g, [$1..100]); # Robert Israel, Mar 09 2026

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) = { 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) } \\ Return the exponent if n is a perfect power

Crossrefs

Cf. A115038, A115041.

Sequence in context: A075392 A069901 A387144 * A032536 A374167 A115061

Adjacent sequences: A115036 A115037 A115038 * A115040 A115041 A115042

Keyword

easy,nonn

Author

Cino Hilliard, Feb 26 2006

Extensions

Name clarified by Robert Israel, Mar 10 2026

Status

approved