A115041 a(n) = y where A115038(x)^2 + prime(n) = y^A115039(n).

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

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(x)^2 + prime(n) = a(n)^A115039(n). - Robert Israel, Mar 09 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. 498^2 + 997 = 499^2. So 499 is the last entry in the table.

Maple

h:= 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 surd(x^2+p, d) fi
  od
end proc:
map(h, [$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(power(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
power(n) =  { if(ispower(n), return(round(n^(1/exponent(n)))) ) } \\ Return the largest factor for which n is a perfect power

Crossrefs

Cf. A115038, A115039.

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

Extensions

Name clarified by Robert Israel, Mar 10 2026

Status

approved