A301941 a(n) is the smallest positive integer k such that n + k divides n^2 + k, or 0 if no such k exists.

1, 1, 0, 3, 2, 5, 4, 7, 6, 3, 5, 11, 10, 13, 12, 6, 4, 17, 16, 19, 18, 7, 11, 23, 22, 5, 24, 12, 8, 29, 28, 31, 30, 11, 17, 35, 6, 37, 36, 18, 12, 41, 40, 43, 42, 10, 23, 47, 46, 7, 20, 24, 16, 53, 52, 11, 14, 19, 29, 59, 58, 61, 60, 30, 8, 15, 12, 67, 66, 23, 35, 71, 70, 73, 72, 36, 19, 56, 13, 79, 78, 9, 41, 83, 82, 17, 84

Offset

0,4

Comments

If even n > 2, such positive k exists and a(n) <= n - 2 since n^2 + n - 2 = (n + 2)*(n - 1) is divisible by n + n - 2 = 2*(n - 1). Additionally, a(2) = 0 because such positive k does not exist. So a(n) <= n - 2 for even positive n.

1 <= a(n) <= n for odd n, in particular, a(p) = p if p is an odd prime and it is also possible that a(t) = t for some composite t. Composite numbers t such that a(t) = t are 35, 95, 119, 143, 203, 215, 247, 275, 299, 335, 395, 403, 437, ...

From Robert Israel, Mar 29 2018: (Start)

If n >= 1 is a square, then a(n) = sqrt(n).

If n is not a square but 8*n+1 is a square, then f(n) = (sqrt(8*n+1)+1)/2.

If n >= 3 and neither n nor 8*n+1 is a square, then a(n) > sqrt(3*n+1).

For n >= 3, n and a(n) are not coprime. (End)

From Bob Andriesse, Jan 02 2024: (Start)

Proof that a(n) = n, if n is an odd prime: Consider any odd n and choose k = n, then (n^2+k)/(n+k) becomes n*(n+1)/(2n), which is an integer. So a(n) = k <= n, for odd n. If n+k = z then k = z-n and n+k | n^2 + k is equivalent to z | n^2 + z-n or z | n^2 - n. So n+k | n*(n-1). If n is an odd prime and k < n, then n+k must divide n-1, which is impossible. Therefore k >= n. We already know that k <= n, so k = n if n is an odd prime p, or a(p) = p.

a(n) is also the smallest k > 0 such that n+k divides n*(k+1), k*(k+1), n*(n-1), k*(n-1) and k^2-n, or 0 if no such k exists. Example: 15+6 divides 15*(6+1), 6*(6+1), 15*(15-1), 6*(15-1) and 6^2-15. (End)

Links

Altug Alkan, Table of n, a(n) for n = 0..10000

Example

a(2) = 0 because there is no positive k such that k + 2 divides k + 4.
a(15) = 6 because 15 + 6 = 3*7 divides 15^2 + 6 = 3*7*11 and 6 is the least positive integer with this property.

Maple

f:= proc(n) local k;
  if issqr(n) then return sqrt(n) fi;
  for k from ceil(sqrt(2*n)) do if (n^2+k) mod (n+k) = 0 then return k fi od
end proc:
f(2):= 0: f(0):= 1:
map(f, [$0..100]); # Robert Israel, Mar 29 2018

Mathematica

a[n_] := If[n == 2, 0, If[PrimeQ[n], n, Module[{k = 1}, While[Mod[n^2+k, n+k] != 0, k++]; k]]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Feb 10 2023, from PARI code *)

Prog

(PARI) a(n) = {if(n==2, 0, if(isprime(n), n, my(k=1); while((n^2+k) % (n+k) != 0, k++); k; ))}

Crossrefs

Cf. A053626.

Sequence in context: A372654 A297208 A372358 * A344547 A135762 A344548

Adjacent sequences: A301938 A301939 A301940 * A301942 A301943 A301944

Keyword

nonn,easy,look

Author

Altug Alkan, Mar 29 2018

Status

approved