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%I #14 Dec 02 2017 11:10:00
%S 13,17,73,97,425,565,2477,3293,14437,19193,84145,111865,490433,651997,
%T 2858453,3800117,16660285,22148705,97103257,129092113,565959257,
%U 752403973,3298652285,4385331725,19225954453,25559586377,112057074433,148972186537,653116492145
%N Consider Pythagorean triples which satisfy X^2+(X+7)^2=Z^2; sequence gives increasing values of Z.
%C The sequence gives the values of Z in X^2 + Y^2 = Z^2 where Y = X + 7 and gcd(X,Y,Z)=1. The values of X are given by the formula: X(1)=5, X(2)=8, X(3)=48, X(4)=65, X(n) = 6*X(n-2) - X(n-4) + 14 for n >= 5 - see A117474. Also, Y - X = 7, which is the second term in A058529. We have Z(1)=13, Z(2)=17, Z(3)=73, Z(4)=97, Z(n)=6*Z(n-2) - Z(n-4) for n >= 5. - Andras Erszegi (erszegi.andras(AT)chello.hu), Mar 19 2006
%H Robert Israel, <a href="/A060569/b060569.txt">Table of n, a(n) for n = 1..2608</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,6,0,-1).
%F G.f.: (13 + 17 x - 5 x^2 - 5 x^3)/(1 - 6 x^2 + x^4). - _Robert Israel_, Jul 17 2017
%p f:=proc(n) option remember; if n=1 then RETURN(13) fi; if n=2 then RETURN(17) fi; if n=3 then RETURN(73) fi; if n=4 then RETURN(97) fi; 6*f(n-2)-f(n-4); end;
%t LinearRecurrence[{0,6,0,-1},{13,17,73,97},30] (* _Harvey P. Dale_, Dec 02 2017 *)
%K nonn
%O 1,1
%A _Sebastiao Antonio da Silva_, Apr 12 2001
%E Edited by _N. J. A. Sloane_, Oct 06 2007
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