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%I #5 Apr 25 2017 13:02:01
%S 1,10,831,8310
%N Numbers n such that d(n)*reversal(n)=phi(n), where d(n) is number of positive divisors of n.
%C If n is a term of this sequence and gcd(10,n)=1 then 10*n is also in the sequence because reversal(10*n)=reversal(n); d(10)=phi(10) and both functions d & phi are multiplicative. No further terms up to 350000000.
%C a(5) > 10^12. - _Giovanni Resta_, Apr 25 2017
%e 8310 is in the sequence because d(8310)=16; reversal(8310)=138;
%e phi(8310)=2208 & 16*138=2108.
%t reversal[n_]:= FromDigits[Reverse[IntegerDigits[n]]]; Do[If[DivisorSigma[0, n]*reversal[n] == EulerPhi[n], Print[n]], {n, 350000000}]
%Y Cf. A063903, A104904, A104905, A104907.
%K more,nonn
%O 1,2
%A _Farideh Firoozbakht_, Apr 14 2005
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