|
%I #14 Jun 03 2018 11:35:26
%S 29,18799189,7663249255406115433,
%T 3605093400349900568684962740253251161,
%U 4991502287564231140564742546889815977689154940104978501999141,43835167264777185998985243579910029928546583864541049798936281152692136028486139208681333389
%N a(n) = ((2n+1)^(2n^2+2n)+(2n^2+2n)^(2n+1))/(2n^2+2n+1).
%C If (a, b, c) is a primitive Pythagorean triple such that a^2+b^2 = c^2, where b is the even leg and b = c-1 then a^b + b^a is divisible by c. See Joardar link. To generate the n-th such Pythagorean triple we take a = 2n+1, b = 2n^2+2n and c = 2n^2+2n+1. This sequence is the quotient (a^b+b^a)/c, i.e., a(n) = ((2n+1)^(2n^2+2n)+(2n^2+2n)^(2n+1))/(2n^2+2n+1).
%H B. Joardar, <a href="http://azimpremjiuniversity.edu.in/SitePages/resources-ara-august-2017-a-property-of-primitive-pythagorean-triples.aspx">A Property of Primitive Pythagorean Triples</a>, At Right Angles, August 2017, Azim Premji University, India.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Pythagorean_triple#Almost-isosceles_Pythagorean_triples">Almost-isosceles Pythagorean triples</a>.
%e a(2) = 18799189 because n = 2 generates the primitive Pythagorean triple (5, 12, 13) and (5^12+12^5)/13 = 18799189.
%t a[n_] := ((2n+1)^(2n^2+2n)+(2n^2+2n)^(2n+1))/(2n^2+2n+1); Array[a, 6]
%Y Cf. A001844.
%K nonn
%O 1,1
%A _Frank M Jackson_, Apr 29 2018
| 