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%I #20 Mar 29 2017 08:58:41
%S 1,37,12,15,199,189,124,1004,18,168,126,12048,10426,1358,11638,1078,
%T 1011868,112518,108018288,1076768,1012998
%N a(n) is the least number such that phi(a(n)) = phi(R(a(n)))/n, where R(n) is the digit reverse of n and phi(n) is the Euler totient function of n.
%C a(22) > 10^9. - _Giovanni Resta_, Mar 29 2017
%F Solutions of the equation A000010(n) = A000010(A004086(n))/n.
%e phi(1) = 1. Its digit reverse is again 1 and phi(1) = 1 = 1 * 1;
%e phi(37) = 36 and phi(73) = 72 = 2 * 36;
%e phi(12) = 4 and phi(21) = 12 = 3 * 4;
%e phi(15) = 8 and phi(51) = 32 = 4 * 8; etc.
%p with(numtheory): R:=proc(w) local x, y, z; x:=w; y:=0;for z from 1 to ilog10(x)+1 do y:=10*y+(x mod 10); x:=trunc(x/10); od; y; end:
%p P:=proc(q) local k,n; for k from 1 to q do for n from 1 to q do
%p if k*phi(n)=phi(R(n)) then print(n); break; fi; od; od; end: P(10^9);
%t rev[n_] := FromDigits@ Reverse@ IntegerDigits@ n; a[n_] := Block[{k}, For[k = 1, EulerPhi@ rev@ k != n EulerPhi@ k, k++]; k]; Array[a, 18] (* _Giovanni Resta_, Mar 29 2017 *)
%Y Cf. A000010, A004086, A284495, A284496, A284498.
%K nonn,more,base
%O 1,2
%A _Paolo P. Lava_, Mar 28 2017
%E a(19)-a(21) from _Giovanni Resta_, Mar 29 2017
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