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%I #27 May 24 2021 07:38:09
%S 5,65,1385,30365,666605,14634905,321301265,7053992885,154866542165,
%T 3400009934705,74645352021305,1638797734533965,35978904807725885,
%U 789897108035435465,17341757471971854305,380728767275345359205
%N Consider all Consecutive Integer Pythagorean 11-tuples (X, X+1, X+2, X+3, X+4, X+5, Z-4, Z-3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives Z values.
%C For n > 1, a(n) = 22*a(n-1) - a(n-2) - 40; e.g., 30365 = 22*1385 - 65 - 40.
%C In general, 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=6, then last(2) = 2274 = 26*90 - 6 - 60.
%C For n > 0, a(n) = 12*A157096(n-1) + 11*a(n-1) + 10; e.g., 30365 = 12*1260 + 11*1385 + 10.
%C 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=6 and n=2, then first(2) = 2100 = 13*78 + 12*90 + 6 and last(2) = 2274 = 14*78 + 13*90 + 12.
%C a(n) = 5^n*6((1+sqrt(6/5))^(2n+1) - (1-sqrt(6/5))^(2n+1))/(4*sqrt(6/5)) + 4/2; e.g., 1385 = 5^2*6((1+sqrt((6/5))^5 - (1-sqrt(6/5))^5)/(4*sqrt(6/5)) + 4/2.
%C In general, 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=6 and n=2, then last(2) = 2274 = 6^2*7((1+sqrt((7/6))^5 - (1-sqrt(7/6))^5)/(4*sqrt(7/6)) + 5/2.
%C 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 last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: 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) = 30365 = (220^2 + 5^2*21^2 + 5*4*220*21)/5 and a(4) = 666605 = (5*(505^2 + 241^2 + 4*505*241))/6.
%C In general, 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+4+5+61+62+63+64+65+1381+1382+1383+1384+1385 = 7245 = 15*483.
%C Lim_{n->infinity} a(n+1)/a(n) = 5*(1+sqrt(6/5))^2 = 11 + 2*sqrt(30).
%C In general, if last(n) is the last term of the n-th Consecutive Integer Pythagorean 2k+1-tuple, then lim_{n->infinity} last(n+1)/last(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 (23, -23, 1).
%F G.f.: 5*(1-10*x+x^2)/((1-x)*(1-22*x+x^2)). - _Colin Barker_, Mar 27 2012
%F a(n) = 23*a(n-1) - 23*a(n-2) + a(n-3). - _Wesley Ivan Hurt_, Oct 26 2020
%e a(2)=65 since 55^2 + 56^2 + 57^2 + 58^2 + 59^2 + 60^2 = 61^2 + 62^2 + 63^2 + 64^2 + 65^2.
%Y Cf. A001653, A157085, A157089, A157093.
%K nonn,uned,easy
%O 0,1
%A _Charlie Marion_, Mar 12 2009
%E a(13), a(15) corrected by _Georg Fischer_, Oct 26 2020
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