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%I #11 Feb 15 2020 10:52:26
%S 5,7,13,17,35,73,97,203,425,565,1183,2477,3293,6895,14437,19193,40187,
%T 84145,111865,234227,490433,651997,1365175,2858453,3800117,7956823,
%U 16660285,22148705,46375763,97103257,129092113,270297755,565959257,752403973,1575410767
%N Consider all Pythagorean triples (X,X+7,Z); sequence gives Z values.
%C First two terms included for consistency with A076293.
%C 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
%H Colin Barker, <a href="/A076294/b076294.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,6,0,0,-1).
%F a(n) = 6*a(n-3)-a(n-6) = sqrt((A076293(n)^2+49)/2) = sqrt(A076295(n)^2 + A076296(n)^2).
%F a(3n+1) = 7*A001653(n).
%F 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
%e 17 is in the sequence as the hypotenuse of the (8,15,17) triangle.
%t LinearRecurrence[{0,0,6,0,0,-1},{5,7,13,17,35,73},40] (* _Harvey P. Dale_, Mar 19 2019 *)
%o (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
%Y Cf. A001653, A076293, A076295, A076296.
%K nonn,easy
%O 0,1
%A _Henry Bottomley_, Oct 05 2002
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