OFFSET
1,1
COMMENTS
The sequence is infinite, even if it is restricted to terms that end with a nonzero digit, as any term generates an infinite number of other terms by the following scheme: If m is a term of the sequence and d(m) denotes the number of digits of m, then set m' = m*10^d'+m with d' >= d(m). For d' >= d(m) we have reverse(m') = reverse(m)*10^d' + reverse(m) and thus (m')^2 + reverse(m')^2 = (m*10^d'+m)^2 + (reverse(m)*10^d'+reverse(m))^2 = (m^2+reverse(m)^2)*(10^d'+1)^2. As m^2+reverse(m)^2 is a perfect square by assumption, the product on the right-hand side of the equation is also a perfect square and m' is part of the sequence. The calculation works also with m' = m*(10^(k*d')+...+10^d'+1). As an example take a(1)=88209. All numbers 8820988209, 882098820988209, 88209882098820988209, ... and 88209088209, 882090088209, 8820900088209, ... are also terms of the sequence. - Matthias Baur, May 01 2020
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..60
Sheng Jiang and Rui-Chen Chen, Digits reversed Pythagorean triples, International Journal of Mathematical Education in Science and Technology, volume 29, number 5, 1998, pages 689-696.
EXAMPLE
88209^2 + 90288^2 = 126225^2, so 88209 belongs to the sequence.
MATHEMATICA
Do[If[IntegerQ[Sqrt[n^2 + FromDigits[Reverse[IntegerDigits[n]]]^2]], Print[n]], {n, 1, 10^6}]
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Joseph L. Pe, Mar 22 2002
EXTENSIONS
a(7)-a(15) from Donovan Johnson, Apr 09 2010
a(16)-a(25) from Donovan Johnson, Jul 15 2011
STATUS
approved