A115038 a(n) is the smallest x such that x^2 + prime(n) is a perfect power.

5, 1, 2, 1, 4, 6, 8, 9, 2, 14, 1, 18, 20, 21, 9, 26, 29, 8, 33, 21, 36, 7, 41, 6, 48, 50, 5, 53, 4, 56, 1, 65, 68, 69, 74, 19, 78, 81, 7, 86, 8, 90, 5, 96, 98, 12, 105, 17, 4, 114, 116, 2, 120, 44, 128, 131, 134, 27, 138, 140, 141, 146, 6, 155, 156, 158, 165, 168, 173, 174, 176, 37

Offset

1,1

Comments

Conjecture: There will always be an x,y,n such that x^2 + p = y^n for all primes p. In otherwords, there is a one-one mapping of the prime numbers to y^n - x^2 for some x,y,n. Therefore primes of the form y^n - x^2 are infinite in number.

Links

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

Formula

a(n)^2 + prime(n) = A115041(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. 1 is the smallest number that when we add its square to prime 3, we get a perfect power, 2^2. So 5 and 1 are the first two entries in the table.

Maple

ispow:= proc(n) local F;
  F:= ifactors(n)[2];
  igcd(F[.., 2])>1
end proc:
f:= proc(n) local p, x;
  p:= ithprime(n);
  for x from 1 do if ispow(x^2+p) then return x fi od
end proc:
map(f, [$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(x", "); c++; break ) ) ); print(); print(c) }

Crossrefs

Cf. A115039, A115041.

Sequence in context: A265824 A097413 A202352 * A231990 A369915 A010131

Adjacent sequences: A115035 A115036 A115037 * A115039 A115040 A115041

Keyword

easy,nonn

Author

Cino Hilliard, Feb 26 2006

Extensions

Name clarified by Robert Israel, Mar 10 2026

Status

approved