A113798 - OEIS

%I #11 Aug 30 2019 06:10:55

%S 117,119817,13101687,119819817,13101801687,119819819817,

%T 11983101689817,13101801801687,119819819819817

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

%C 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

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

%Y Cf. A004086, A068536, A202386.

%K nonn,base,more

%O 1,1

%A _Giovanni Resta_, Jan 21 2006

%E a(5)-a(6) from _Giovanni Resta_, May 10 2017

%E a(7)-a(9) from _Giovanni Resta_, Aug 26 2019