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Numbers k such that k^3 divides 9^(k^2) + 1.
20

%I #13 Jun 08 2021 02:22:43

%S 1,5,505,5905,596405,1971005,199071505,515151005,581457505,618181105,

%T 2327756905,11855474405,19555113485,31361954905,101901244565,

%U 235103447405,305354088005,395331063745,600905819065,608393336905,686701313405,730071885005

%N Numbers k such that k^3 divides 9^(k^2) + 1.

%C 5 divides a(n) for n > 1.

%t Select[Range[5*10^6], Mod[ 9^(#^2)+1, #^3]==0 &] (* _G. C. Greubel_, Jan 18 2018 *)

%Y Cf. A127263, A128677, A128678, A128679, A128680, A128681, A128683, A128684, A128685.

%K nonn

%O 1,2

%A _Alexander Adamchuk_, Mar 31 2007

%E a(5)-a(6) from J. Mulder (jasper.mulder(AT)planet.nl), Jan 26 2010

%E Terms a(7) onward from _Max Alekseyev_, May 14 2010