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A157092 Consider all consecutive integer Pythagorean 9-tuples (X, X+1, X+2, X+3, X+4, Z-3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values.

%I A157092 #43 Nov 09 2024 14:23:36
%S A157092 0,36,680,12236,219600,3940596,70711160,1268860316,22768774560,
%T A157092 408569081796,7331474697800,131557975478636,2360712083917680,
%U A157092 42361259535039636,760141959546795800,13640194012307284796,244763350261984330560,4392100110703410665316,78813038642399407645160
%N A157092 Consider all consecutive integer Pythagorean 9-tuples (X, X+1, X+2, X+3, X+4, Z-3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values.
%C A157092 In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0, first(1) = k*(2k+1) and, for n > 1, first(n) = (4k+2)*first(n-1) - first(n-2) + 2*k^2; e.g., if k=5, then first(2) = 1260 = 22*55 - 0 + 50.
%C A157092 In general, the first and last terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0 and last(0)=k; for n > 0, let first(n) = (2k+1)*first(n-1) + 2k*last(n-1) + k and last(n) = (2k+2)*first(n-1) + (2k+1)*last(n-1) + 2k; e.g., if k=5 and n=2, then first(2) = 1260 = 11*55 + 10*65 + 5 and last(2) = 1385 = 12*55 + 11*65 + 10.
%C A157092 In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(n) = (k^(n+1)((1+sqrt((k+1)/k))^(2n+1) + (1-sqrt((k+1)/k))^(2n+1)) - 2*k)/4; e.g., if k=5 and n=2, then first(2) = 1260 = (5^3((1+sqrt((6/5))^5 + (1-sqrt(6/5))^5) - 2*5)/4.
%C A157092 In general, if u(n) is the numerator and e(n) is the denominator of the n-th continued fraction convergent to sqrt((k+1)/k), then the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(2n+1) = k*u(2n)*u(2n+1) and, for n > 0, first(2n) = (k+1)*e(2n-1)*e(2n); e.g., a(1) = 36 = 4*1*9 and a(2) = 680 = 5*8*17.
%C A157092 In general, if first(n) is the first term of the n-th consecutive integer Pythagorean 2k+1-tuple, then lim_{n->inf} first(n+1)/first(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).
%D A157092 A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.
%D A157092 L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.
%D A157092 W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.
%H A157092 G. C. Greubel, <a href="/A157092/b157092.txt">Table of n, a(n) for n = 0..790</a>
%H A157092 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H A157092 Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%H A157092 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (19,-19,1).
%F A157092 For n > 1, a(n) = 18*a(n-1) - a(n-2) + 32.
%F A157092 For n > 0, a(n) = 9*a(n-1) + 8*A157093(n-1) + 4.
%F A157092 a(n) = (4^(n+1)((1+sqrt(5/4))^(2n+1) + (1-sqrt(5/4))^(2n+1)) - 2*4)/4.
%F A157092 Lim_{n->inf} a(n+1)/a(n) = 4*(1+sqrt(5/4))^2 = 9 + 2*sqrt(20).
%F A157092 From _R. J. Mathar_, Mar 19 2009: (Start)
%F A157092 G.f.: 4*x*(-9+x)/((x-1)*(x^2-18*x+1)).
%F A157092 a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3).
%F A157092 a(n) = 4*A119032(n+1). (End)
%F A157092 For n > 0, 1/a(n) = Sum_{k>=1} F(3*k)/phi^(6*k*n + 3*k), where F(n) = A000045(n) and phi = A001622 = (sqrt(5)+1)/2. - _Diego Rattaggi_, Dec 28 2019
%F A157092 E.g.f.: (1/2)*((2 + sqrt(5))*exp((9+4*sqrt(5))*x) + (2 - sqrt(5))*exp((9-4*sqrt(5))*x) - 4*exp(x)). - _Stefano Spezia_, Dec 29 2019
%e A157092 a(2)=680 since 680^2 + 681^2 + 682^2 + 683^2 + 684^2 = 761^2 + 762^2 + 763^2 + 764^2.
%t A157092 RecurrenceTable[{a[0]==0,a[1]==36,a[n]==18a[n-1]-a[n-2]+32},a,{n,20}] (* or *) LinearRecurrence[{19,-19,1},{0,36,680},20] (* _Harvey P. Dale_, Oct 09 2012 *)
%o A157092 (PARI) x='x+O('x^50); concat([0], Vec(4*x*(-9+x)/((x-1)*(x^2-18*x+1)))) \\ _G. C. Greubel_, Nov 04 2017
%o A157092 (Magma) [Round((4^(n+1)*((1+Sqrt(5/4))^(2*n+1) + (1-Sqrt(5/4))^(2*n+1)) - 2*4)/4): n in [0..50]]; // _G. C. Greubel_, Nov 04 2017
%Y A157092 Cf. A001652, A157084, A157088, A157096.
%K A157092 nonn
%O A157092 0,2
%A A157092 _Charlie Marion_, Mar 12 2009
%E A157092 Terms a(15) onward added by _G. C. Greubel_, Nov 06 2017