A157093 - OEIS

%I #13 May 24 2021 07:38:35

%S 4,44,764,13684,245524,4405724,79057484,1418628964,25456263844,

%T 456794120204,8196837899804,147086288076244,2639356347472564,

%U 47361327966429884,849864547048265324,15250200518902345924

%N 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 Z values.

%C For n > 1, a(n) = 18*a(n-1) - a(n-2) - 24; e.g., 13684 = 18*764 - 44 - 24.

%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=5, then last(2) = 1385 = 22*65 - 5 - 40.

%C For n > 0, a(n) = 10*A157092(n-1) + 9*a(n-1) + 8; e.g., 13684 = 10*680 + 9*764 + 8. 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 + 10*65 + 10.

%C a(n) = 4^n*5((1+sqrt(5/4))^(2n+1) - (1-sqrt(5/4))^(2n+1))/(4*sqrt(5/4)) + 3/2; e.g., 764 = 4^2*5((1+sqrt((5/4))^5 - (1-sqrt(5/4))^5)/(4*sqrt(5/4)) + 3/2.

%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=5 and n=2, then last(2) = 1385 = 5^2*6((1+sqrt((6/5))^5 - (1-sqrt(6/5))^5)/(4*sqrt(6/5)) + 4/2.

%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) = 13684 = (144^2 + 4^2*17^2 + 4*3*144*17)/4 and a(4) = 245524 = (4*(341^2 + 161^2 + 3*341*161))/5.

%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+41+42+43+44+761+762+763+764 = 3230 = 10*323.

%C Lim_{n->infinity} a(n+1)/a(n) = 4*(1+sqrt(5/4))^2 = 9 + 2*sqrt(20). 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 (19, -19, 1).

%F Empirical g.f.: 4*(1-8*x+x^2)/((1-x)*(1-18*x+x^2)). - _Colin Barker_, Mar 27 2012

%e a(2)=764 since 680^2 + 681^2 + 682^2 + 683^2 + 684^2 = 761^2 + 762^2 + 763^2 + 764^2.

%Y Cf. A001653, A157085, A157089, A157097.

%K nonn

%O 0,1

%A _Charlie Marion_, Mar 12 2009