A076296 Consider all Pythagorean triples (X,X+7,Z); sequence gives X values.

-3, 0, 5, 8, 21, 48, 65, 140, 297, 396, 833, 1748, 2325, 4872, 10205, 13568, 28413, 59496, 79097, 165620, 346785, 461028, 965321, 2021228, 2687085, 5626320, 11780597, 15661496, 32792613, 68662368, 91281905, 191129372, 400193625, 532029948, 1113983633

Offset: 0

Author

Henry Bottomley, Oct 05 2002

Keywords

• sign
• easy

Comments

First two terms included for consistency with A076293.

From Klaus Brockhaus, Feb 18 2009: (Start)

Lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).

Lim_{n -> infinity} a(n)/a(n-1) = (9+4*sqrt(2))/7 for n mod 3 = {1, 2}.

Lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2)) / ((9+4*sqrt(2))/7)^2 for n mod 3 = 0. (End)

For the generic case x^2 + (x+p)^2 = y^2 with p=2*m^2-1 a prime number in A066436, m >= 2, the x values are given by the sequence defined by: a(n) = 6*a(n-3) - a(n-6) + 2p with a(1)=0, a(2)=2m+1, a(3)=6m^2-10m+4, a(4)=3p, a(5)=6m^2+10m+4, a(6)=40m^2-58m+21. Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(1)=p, b(2)=2m^2+2m+1, b(3)=10m^2-14m+5, b(4)=5p, b(5)=10m^2+14m+5, b(6)=58m^2-82m+29. - Mohamed Bouhamida, Sep 09 2009

For the generic case x^2 + (x + p)^2 = y^2 with p = 2*m^2 - 1 a prime number, m>=2, the first three consecutive solutions are: (0;p), (2*m+1; 2*m^2+2*m+1), (6*m^2-10*m+4; 10*m^2-14*m+5) and the other solutions are defined by: (X(n); Y(n))= (3*X(n-3)+2*Y(n-3)+p; 4*X(n-3)+3*Y(n-3)+2*p). - Mohamed Bouhamida, Aug 20 2019

Examples

			8 is in the sequence as the shorter leg of the (8,15,17) triangle.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Crossrefs

Cf. A001652, A076293, A076294, A076295.

Cf. A156035 (decimal expansion of 3+2*sqrt(2)), A156649 (decimal expansion of (9+4*sqrt(2))/7). - Klaus Brockhaus, Feb 18 2009

Programs

Magma

I:=[-3,0,5,8,21,48,65]; [n le 7 select I[n] else Self(n-1) +6*Self(n-3) -6*Self(n-4) -Self(n-6) +Self(n-7): n in [1..30]]; // G. C. Greubel, May 04 2018

Mathematica

CoefficientList[Series[(3-3x-5x^2-21x^3+5x^4+3x^5+4x^6)/(-1+x+6x^3-6x^4-x^6+x^7),{x,0,50}],x] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2012 *)
LinearRecurrence[{1,0,6,-6,0,-1,1}, {-3,0,5,8,21,48,65}, 50] (* T. D. Noe, Feb 07 2012 *)

PARI

x='x+O('x^30); Vec((-3+3*x+5*x^2+21*x^3-5*x^4-3*x^5-4*x^6)/((1-x)*(1-6*x^3 +x^6))) \\ G. C. Greubel, May 04 2018

Formula

a(n) = 6a(n-3) - a(n-6) + 14 = (A076293(n) - 7)/2.

a(n) = sqrt(A076294(n)^2 - A076295(n)^2) = A076295(n) - 7.

a(3*n+1) = 7*A001652(n).

From Mohamed Bouhamida, Jul 06 2007: (Start)

a(n) = 5*(a(n-3) + a(n-6)) - a(n-9) + 28.

a(n) = 7*(a(n-3) - a(n-6)) + a(n-9). (End)

G.f.: (-3 + 3*x + 5*x^2 + 21*x^3 - 5*x^4 - 3*x^5 - 4*x^6)/((1-x)*(1 - 6*x^3 + x^6)). - Klaus Brockhaus, Feb 18 2009

Extensions

More terms from Klaus Brockhaus, Feb 18 2009