%I #8 Oct 28 2012 04:45:52
%S 42,402,492,4000002,57906504,400000002,4000000002,6279090751,
%T 62698513951,400000000002
%N Numbers n such that sigma(n)=4*reversal(n).
%C If p=(2*10^n+1)/3 is prime then m=6*p is in the sequence because sigma(m)=sigma(6*p)=12*(2*10^n+4)/3=4*(2*10^n+4)=4* reversal(4*10^n+2)=4*reversal(6*(2*10^n+1)/3)=4*reversal(6*p) =4*reversal(m). Next term is greater than 5*10^8.
%C a(11) > 10^12. - _Giovanni Resta_, Oct 28 2012
%e 492 is in the sequence because sigma(492)=sigma(4*3*41)=7*4*42
%e =4*294=4*reversal(492).
%t Do[If[DivisorSigma[1, n]==4*FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 500000000}]
%Y Cf. A069216, A105324, A114927.
%K base,more,nonn
%O 1,1
%A _Farideh Firoozbakht_, Jan 28 2006
%E a(7)-a(9) from _Donovan Johnson_, Dec 21 2008
%E a(10) from _Giovanni Resta_, Oct 28 2012