A130280 a(n) = smallest integer k>1 such that n(k^2-1)+1 is a perfect square, or 0 if no such number exists.

2, 5, 3, 0, 2, 3, 5, 2, 0, 3, 7, 5, 4, 11, 3, 2, 4, 13, 9, 7, 2, 5, 19, 4, 0, 5, 21, 3, 11, 9, 11, 14, 2, 29, 5, 3, 6, 31, 21, 2, 13, 11, 13, 169, 3, 7, 41, 6, 0, 7, 5, 11, 22, 419, 3, 2, 5, 23, 461, 27, 8, 55, 7, 4, 2, 3, 49, 29

Offset

1,1

Comments

A084702(n) = a(n)^2-1, resp. a(n) = sqrt(A084702(n)+1). See A130283 for values where A130280(n)=0.

Links

M. F. Hasler, Table of n, a(n) for n = 1..1000

Formula

If n=(2k)^2, then A130280(n) <= k, since (2k)^2(k^2-1)+1 = (2k^2-1)^2. See A130281 for the cases where equality does not hold. If n=k^2-1, then A130280(n) <= k-1 since (k^2-1)((k-1)^2-1)+1 = (k^2-k-1)^2. See A130282 for the cases where equality does not hold.

Example

a( (2k)^2 ) <= k since (2k)^2(k^2-1)+1 = (2k^2-1)^2 (but k=1 is excluded since with k^2-1=0 this would be a trivial solution for any n).

Maple

A130280:=proc(n) local x, y, z; if n=1 then return 2 fi; isolve(n*(x^2-1)+1=y^2, z); select(has, `union`(%), x); map(rhs, %); simplify(eval(%, z=1) union eval(%, z=0)) minus {-1, 1}; if %={} then 0 else (min@op@map)(abs, %) fi end;

Mathematica

$MaxExtraPrecision = 100;
r[n_, c_] := Reduce[k > 1 && j > 1 && n*(k^2 - 1) + 1 == j^2, {j, k}, Integers] /. C[1] -> c // Simplify;
a[n_] := If[rn = r[n, 0] || r[n, 1] || r[n, 2]; rn === False, 0, k /. {ToRules[rn]} // Min];
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 1, 800}] (* Jean-François Alcover, May 12 2017 *)

Prog

(PARI) {A130280(n, L=10^15)=if(issquare(n), L=2+sqrtint(n>>2)); for( k=2, L, if( issquare(n*(k^2-1)+1), return(k)))}

Crossrefs

Cf. A084702, A094357, A130281, A130282, A130283, A130284.

See also A306767.

Sequence in context: A055385 A160503 A108429 * A160127 A361226 A011035

Adjacent sequences: A130277 A130278 A130279 * A130281 A130282 A130283

Keyword

easy,nonn

Author

M. F. Hasler, May 20 2007, May 25 2007

Status

approved