%I
%S 1,3,4,8,12,15,24,32,60,96,120,128,255,384,480,1020,1920,2040,2418,
%T 8160,8192,24576,32640,65535,122880,131072,262140,370986,393216,
%U 524280,524288,1572864,1966080,2088960,2097120,7864320,8388480,33423360,110771178,133693440
%N Numbers n such that sigma(uphi(n)) = n where uphi is the unitary totient (or unitary phi) function (see A047994).
%C Theorem: If 2^p1 is prime (a Mersenne prime) and 0<=k<=5 then m=2^p*(2^2^k1) is in the sequence. Proof: If k=0 then m=2^p and sigma(uphi(m))=sigma(uphi(2^p))=sigma(2^p1)=2^p=m. Now if k>0 then m=2^p*(2^2^0+1)*(2^2^1+1)*...*(2^2^(k1)+1) since 2^2^i+1 for i=0,1,...,4 are primes(Fermat primes) we conclude that uphi(m)=(2^p1)*(2^2^0)*(2^2^1)*...*(2^2^(k1))=(2^p1)*2^ (2^0+2^1+...+2^(k1))=(2^p1)*2^(2^k1) hence sigma(uphi(m))= sigma(2^p1)*sigma(2^(2^k1))=2^p*(2^2^k1)=m and the proof is complete. 2418, 370986 & 110771178 are the only terms up to 15*10^8 which aren't of the form 2^p*(2^2^k1).
%H Donovan Johnson, <a href="/A120116/b120116.txt">Table of n, a(n) for n = 1..50</a> (terms < 2^35)
%e 110771178 = 2*3*7*19*127*1093 is in the sequence because sigma(uphi(2*3*7*19*127*1093)) = sigma((21)*(31)*(71)*(191)*(1271)*(10931) = sigma(2*6*18*126*1092) = sigma(2^6*3^6*7^2*13) = (2^71)*((3^71)/2)*((7^31)/6)*14 = 110771178.
%t uphi[n_] := (A = FactorInteger[n]; l = Length[A]; Product[A[[k]][[1]] ^A[[k]][[2]]  1, {k, l}]); Do[If[DivisorSigma[1, uphi[n]] == n, Print[n]], {n, 150000000}]
%Y Cf. A047994, A019434.
%K nonn
%O 1,2
%A _Farideh Firoozbakht_, Jul 11 2006
%E Name corrected by _Donovan Johnson_, May 01 2013
