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

Author

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

Keywords

• easy
• nonn

Comments

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

Examples

			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).
		

Links

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

Crossrefs

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

See also A306767.

Programs

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 *)

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)))}

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.