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A157089 Consider all consecutive integer Pythagorean septuples (X, X+1, X+2, X+3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives Z values.

%I A157089 #45 Aug 24 2025 23:08:55
%S A157089 3,27,363,5043,70227,978123,13623483,189750627,2642885283,36810643323,
%T A157089 512706121227,7141075053843,99462344632563,1385331749802027,
%U A157089 19295182152595803,268747218386539203,3743165875258953027
%N A157089 Consider all consecutive integer Pythagorean septuples (X, X+1, X+2, X+3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives Z values.
%C A157089 The last terms of consecutive integer Pythagorean (2k+1)-tuples may be found as follows: let last(0)=0, last(1)=k*(2k+3) and, for n > 1, last(n) = (4k+2)*last(n-1) - last(n-2) - 2*k*(k-1); e.g., if k=4, then last(2) = 764 = 18*44 - 4 - 24.
%C A157089 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=4 and n=2, then first(2) = 680 = 9*36 + 8*44 + 4 and last(2) = 764 = 10*36 + 9*44 + 8.
%C A157089 The last terms of consecutive integer Pythagorean (2k+1)-tuples may be found as follows: if q = (k+1)/k, then last(n) = k^n*(k+1)*((1+sqrt(q))^(2*n+1) - (1-sqrt(q))^(2*n+1))/(4*sqrt(q)) + (k-1)/2; e.g., if k=4 and n=2, then last(2) = 764 = 4^2*5*((1+sqrt(5/4))^5 - (1-sqrt(5/4))^5)/(4*sqrt(5/4)) + 3/2.
%C A157089 If u(n) is the numerator and e(n) is the denominator of the n-th continued fraction to sqrt((k+1)/k), then the last terms of consecutive integer Pythagorean (2k+1)-tuples may be found as follows:
%C A157089 last(2n+1) = (e(2n+1)^2 + k^2*e(2n)^2 + k*(k-1)*e(2n+1)*e(n))/k and, for n > 0, last(2n) = (k*(u(2n)^2 + u(2n-1)^2 + (k-1)*u(2n)*u(2n-1)))/(k+1); e.g., a(3) = 5043 = (84^2 + 3^2*13^2 + 3*2*84*13)/3 and a(4) = 70227 = (3*(209^2 + 97^2 + 2*209*97))/4.
%C A157089 If b(0)=1, b(1)=4k+2 and, for n > 1, b(n) = (4k+2)*b(n-1) - b(n-2), and last(n) is the last term of the n-th consecutive integer Pythagorean (2k+1)-tuple as defined above, then Sum_{i=0..n} (k*last(i) - k(k-1)/2) = k(k+1)/2*b(n); e.g., if n=3, then 1+2+3+25+26+27+361+362+363 = 1170 = 6*195.
%C A157089 Lim_{n->infinity} a(n+1)/a(n) = 3*(1+sqrt(4/3))^2 = 7 + 2*sqrt(12).
%C A157089 If first(n) is the first term of the n-th consecutive integer Pythagorean (2k+1)-tuple, then lim_{n->infinity} first(n+1)/first(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).
%C A157089 The generalization using the binomial formula is : a(n,k)= ((k-1) + (k+1)*k^n*(Sum_{j=0..n} binomial(2*n+1, 2*j+1)*(1+1/k)^j)))/2. - _Gerry Martens_, Aug 19 2025
%D A157089 A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.
%D A157089 L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.
%D A157089 W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.
%H A157089 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H A157089 Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%H A157089 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-15,1).
%F A157089 For n > 1, a(n) = 14*a(n-1) - a(n-2) - 12.
%F A157089 a(n) = 8*a(n-1) + 7*A157088(n-1) + 6.
%F A157089 G.f.: (3-18*x+3*x^2)/(1-15*x+15*x^2-x^3). - _Colin Barker_, Jan 01 2012
%F A157089 From _Gerry Martens_, Aug 19 2025: (Start)
%F A157089 a(n) = ((1 + sqrt(3))^(2*n+1) - (1 - sqrt(3))^(2*n+1))^2/4^(n+1).
%F A157089 a(n) = 1/(3*4^n)*(Sum_{k=0..n} binomial(2*n+1, 2*k+1)*3^(k+1))^2. (End)
%e A157089 a(3) = 363 since 312^2 + 313^2 + 314^2 + 315^2 = 361^2 + 362^2 + 363^2.
%t A157089 LinearRecurrence[{15,-15,1},{3,27,363},20] (* _Harvey P. Dale_, May 14 2022 *)
%Y A157089 Cf. A001653, A157085, A157093, A157097.
%K A157089 nonn,easy,changed
%O A157089 0,1
%A A157089 _Charlie Marion_, Mar 12 2009