A113798 Numbers k such that k^2 plus the reverse of k is a square.

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

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