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%I #9 Oct 28 2012 04:46:29
%S 861,951,2070,8241,900051,8864151,9000051,82000041,8200000041,
%T 82000000041
%N Numbers n such that sigma(n)=8*reversal(n).
%C If p=3*10^n+17 is prime then 3*p is in the sequence because sigma(3*p)=4*(3*10^n+18)=12*10^n+72=8*(15*10^(n-1)+9)=8* reversal(9*10^n+51)=8*reversal(3*p). Also if p=(2*10^n+1)/3 is prime then 123*p is in the sequence (the proof is easy). Next term is greater than 13*10^7.
%C a(11) > 10^12. - _Giovanni Resta_, Oct 28 2012
%e 82000041 is in the sequence because sigma(82000041)
%e =112000224=8*14000028=8*reversal(82000041).
%t Do[If[DivisorSigma[1,n]==8*FromDigits[Reverse[IntegerDigits[n]]],Print[n]],{n,130000000}]
%Y Cf. A069216, A105324, A114928, A115747, A115748.
%K base,more,nonn
%O 1,1
%A _Farideh Firoozbakht_, Feb 12 2006
%E a(9)-a(10) from _Donovan Johnson_, Dec 21 2008
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