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A076294 Consider all Pythagorean triples (X,X+7,Z); sequence gives Z values. 4
5, 7, 13, 17, 35, 73, 97, 203, 425, 565, 1183, 2477, 3293, 6895, 14437, 19193, 40187, 84145, 111865, 234227, 490433, 651997, 1365175, 2858453, 3800117, 7956823, 16660285, 22148705, 46375763, 97103257, 129092113, 270297755, 565959257, 752403973, 1575410767 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

First two terms included for consistency with A076293.

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)=2*m^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

LINKS

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,-1).

FORMULA

a(n) = 6*a(n-3)-a(n-6) = sqrt((A076293(n)^2+49)/2) = sqrt(A076295(n)^2 + A076296(n)^2).

a(3n+1) = 7*A001653(n).

G.f.: (1 - x)*(5 + 12*x + 25*x^2 + 12*x^3 + 5*x^4) / (1 - 6*x^3 + x^6). - Colin Barker, Apr 25 2017

EXAMPLE

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

MATHEMATICA

LinearRecurrence[{0, 0, 6, 0, 0, -1}, {5, 7, 13, 17, 35, 73}, 40] (* Harvey P. Dale, Mar 19 2019 *)

PROG

(PARI) Vec((1 - x)*(5 + 12*x + 25*x^2 + 12*x^3 + 5*x^4) / (1 - 6*x^3 + x^6) + O(x^50)) \\ Colin Barker, Apr 25 2017

CROSSREFS

Cf. A001653, A076293, A076295, A076296.

Sequence in context: A285886 A106069 A339691 * A073574 A092110 A171518

Adjacent sequences:  A076291 A076292 A076293 * A076295 A076296 A076297

KEYWORD

nonn,easy

AUTHOR

Henry Bottomley, Oct 05 2002

STATUS

approved

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Last modified July 29 09:31 EDT 2021. Contains 346344 sequences. (Running on oeis4.)