A115598 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z-(X+1) values.

1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449, 11309768, 65918161, 384199200, 2239277041, 13051463048, 76069501249, 443365544448, 2584123765441, 15061377048200, 87784138523761, 511643454094368, 2982076586042449, 17380816062160328, 101302819786919521

Offset: 1

Author

N. J. A. Sloane, Mar 14 2006

Keywords

• nonn
• easy

Comments

Old name was A001653(n) - A046090(n).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..200
• L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213.
• Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Crossrefs

Essentially a duplicate of A001108.

Programs

Mathematica

LinearRecurrence[{7,-7,1},{1,8,49},30] (* Harvey P. Dale, Oct 27 2013 *)
CoefficientList[Series[-(x + 1)/((x - 1) (x^2 - 6 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Jul 28 2014 *)

Formula

a(n) = (-2+(3-2*sqrt(2))^n+(3+2*sqrt(2))^n)/4.

a(n) = 7*a(n-1)-7*a(n-2)+a(n-3).

G.f.: -x*(x+1) / ((x-1)*(x^2-6*x+1)).

a(n)^2 + (a(n)+1)^2 = A001542(n)^2 + 1^2. - Hermann Stamm-Wilbrandt, Jul 27 2014

Extensions

Corrected and edited by Colin Barker, Jul 31 2013