%I #9 Jul 22 2021 07:51:46

%S 1,21,63,270,291,1065,2991,6102,10087,10279,10654,16528,144069,

%T 1440069,10253179,10657654,14400069,16163608,46676013,69460293,

%U 2346534651,2369436474,6313047393,10270341004,10657657654,14406534069,23400000651

%N Numbers k such that the reverse of the representation of phi(k) is a substring of k, in base 10.

%C If p = 480...023 is prime, then k = 3*p = 1440...069 and phi(k) = 960...044, thus k belongs to the sequence.

%e phi(1_440069) = 960044.

%t lst = {}; Do[If[{}!= StringPosition[ToString@n, StringReverse@ToString@EulerPhi@n], AppendTo[lst, n], {n, 2*10^7}]; lst

%Y Cf. A000010, A069215.

%K base,nonn

%O 1,2

%A _Giovanni Resta_, Jan 26 2006

%E a(19)-a(27) from _Donovan Johnson_, May 11 2010