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%I #5 May 15 2013 01:33:44
%S 2,6,8,24,128,384,32768,98304,2147483648,6442450944
%N Numbers n such that phi(sigma(n))-sigma(phi(n))=1.
%C If k is a natural number less than 6 then 2^(2^k-1) is in the sequence because phi(sigma(2^(2^k-1)) = phi(2^(2^k)-1) = phi((2^(2^0)+1)*(2^(2^1)+1)*...*(2^(2^(k-1))+1)) = 2^(2^0)*2^(2^1)*...* 2^(2^(k-1)) = 2^(2^0+2^1+...+2^(k-1)) = 2^(2^k-1) and sigma(phi(2^(2^k-1))) = sigma(2^(2^k-2)) = 2^(2^k-1)-1 so phi(sigma(2^(2^k-1))) -sigma(phi(2^(2^k-1))) = 1(note that for i = 0,1,2,3 & 4 the Fermat number 2^2^i+1 is prime). Next term is greater than 7*10^8.
%C Also if n is a natural number less than 6 then 3*2^(2^n-1) is in the sequence, the proof is similar to the case m=2^(2^n-1). Note that all known terms of the sequence are the ten numbers m*2^(2^n-1) m=1 & 3 and n=1,2,3,4 & 5. Conjecture: There is no other term. - _Farideh Firoozbakht_, Mar 24 2006
%C There are no more terms up to 11*10^9. [From _Farideh Firoozbakht_, Apr 13 2010]
%F For n<11, a(n)=((-1)^n+2)*2^(2^Floor[(n+1)/2]-1). - _Farideh Firoozbakht_, Mar 24 2006
%e phi(sigma(2147483648)) = 2147483648 and sigma(phi(2147483648)) = 2147483647 so phi(sigma(2147483648))-sigma(phi(2147483648)) = 1. Hence 2147483648 is in the sequence.
%t Do[If[EulerPhi[DivisorSigma[1,n]]-DivisorSigma[1,EulerPhi[n]]==1, Print[n]],{n,700000000}]
%t Table[((-1)^n + 2)*2^(2^Floor[(n + 1)/2] - 1), {n, 10}] - _Farideh Firoozbakht_, Mar 24 2006
%o (PARI) is(n)=eulerphi(sigma(n))-sigma(eulerphi(n))==1 \\ _Charles R Greathouse IV_, May 15 2013
%Y Cf. A001229.
%K nonn
%O 1,1
%A _Farideh Firoozbakht_, Mar 12 2006
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