A113798 - OEIS

A113798

Numbers k such that k^2 plus the reverse of k is a square.

117, 119817, 13101687, 119819817, 13101801687, 119819819817, 11983101689817, 13101801801687, 119819819819817

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

a(10) > 1.36*10^14. The sequence contains two infinite sets, 117*1000^k + 2817*(1000^k - 1)/999 and 13101687*1000^k + 114687*(1000^k - 1)/999, for k >= 0. Terms of both sets satisfy the equation rev(x) = 6*x + 9, and thus x^2 + rev(x) = (x+3)^2. - Giovanni Resta, Aug 26 2019

LINKS

Table of n, a(n) for n=1..9.

EXAMPLE

117^2 + 711 = 120^2, thus 117 is a term.

CROSSREFS

Cf. A004086, A068536, A202386.

Sequence in context: A220600 A194613 A079196 * A151593 A263425 A066923

Adjacent sequences: A113795 A113796 A113797 * A113799 A113800 A113801

KEYWORD

nonn,base,more

AUTHOR

Giovanni Resta, Jan 21 2006

EXTENSIONS

a(5)-a(6) from Giovanni Resta, May 10 2017

a(7)-a(9) from Giovanni Resta, Aug 26 2019

STATUS

approved