%I #22 Feb 07 2022 11:11:53

%S 2,14,134,1322,13082,129494,1281854,12689042,125608562,1243396574,

%T 12308357174,121840175162,1206093394442,11939093769254,

%U 118184844298094,1169909349211682,11580908647818722,114639177128975534

%N Consider all Consecutive Integer Pythagorean quintuples (X, X+1, X+2, Z-1, Z) ordered by increasing Z; sequence gives Z values.

%C "Consecutive Integer Pythagorean quintuple" means that X^2 + (X+1)^2 + (X+2)^2 = (Z-1)^2 + Z^2. - _M. F. Hasler_, Oct 04 2014

%C 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=3, then last(2) = 363 = 14*27 - 3 - 12.

%C For n > 0, a(n) = 6*a(n-1) + 5*A157084(n-1) + 4; e.g., 1322 = 6*108 + 5*134 + 4.

%C 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=3 and n=2, then first(2) = 312 = 7*21 + 6*27+3 and last(2) = 363 = 8*21 + 7*27 + 6.

%C a(n) = 2^n*3((1+sqrt(3/2))^(2n+1) - (1-sqrt(3/2))^(2n+1))/(4*sqrt(3/2)) + 1/2; e.g., 134 = 2^2*3((1+sqrt(3/2))^5 - (1-sqrt(3/2))^5)/(4*(sqrt(3/2)) + 1/2.

%C 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=3 and n=2, then last(2) = 363 = 3^2*4((1+sqrt((4/3))^5 - (1-sqrt(4/3))^5)/(4*sqrt(4/3)) + 2/2.

%C 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 last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows:

%C 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) = 1322 = (40^2 + 2^2*9^2 + 2*1*40*9)/2 and a(4) = 13082 = (2*(109^2 + 49^2 + 1*109*49))/3.

%C 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 + 13 + 14 + 133 + 134 = 297 = 3*99.

%C 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).

%D A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.

%D L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.

%D W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (11, -11, 1).

%F For n > 1, a(n) = 10*a(n-1) - a(n-2) - 4; e.g., 1322 = 10*134 - 14 - 4.

%F Lim_{n->infinity} a(n+1)/a(n) = 2*(1+sqrt(3/2))^2 = 5 + 2*sqrt(6).

%F g.f.: 2*(1-4*x+x^2)/((1-x)*(x^2-10*x+1)) - Colin Barker, Jan 01 2012

%F a(n) = 2*A253175(n+1). - _R. J. Mathar_, Feb 07 2022

%e a(3)=134 since 108^2 + 109^2 + 110^2 = 133^2 + 134^2.

%o (PARI) for(y=0,9e9,yy=(y+2)^2+(y+1)^2;for(x=sqrtint(yy\3),ceil(sqrt(yy/3)),x^2+(x-1)^2+(x-2)^2==yy&&print1(y+2,","))) \\ For illustrative purpose. - _M. F. Hasler_, Oct 04 2014

%Y Cf. A001653, A157089, A157093, A157097.

%K nonn

%O 0,1

%A _Charlie Marion_, Mar 12 2009

%E Edited by _M. F. Hasler_, Oct 04 2014