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 A116083 Numbers n such that phi(sigma(n))-sigma(phi(n))=1. 0
 2, 6, 8, 24, 128, 384, 32768, 98304, 2147483648, 6442450944 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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. 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 There are no more terms up to 11*10^9. [From Farideh Firoozbakht, Apr 13 2010] LINKS FORMULA For n<11, a(n)=((-1)^n+2)*2^(2^Floor[(n+1)/2]-1). - Farideh Firoozbakht, Mar 24 2006 EXAMPLE phi(sigma(2147483648)) = 2147483648 and sigma(phi(2147483648)) = 2147483647 so phi(sigma(2147483648))-sigma(phi(2147483648)) = 1. Hence 2147483648 is in the sequence. MATHEMATICA Do[If[EulerPhi[DivisorSigma[1, n]]-DivisorSigma[1, EulerPhi[n]]==1, Print[n]], {n, 700000000}] Table[((-1)^n + 2)*2^(2^Floor[(n + 1)/2] - 1), {n, 10}] - Farideh Firoozbakht, Mar 24 2006 PROG (PARI) is(n)=eulerphi(sigma(n))-sigma(eulerphi(n))==1 \\ Charles R Greathouse IV, May 15 2013 CROSSREFS Cf. A001229. Sequence in context: A193946 A189850 A189358 * A115506 A243552 A057852 Adjacent sequences:  A116080 A116081 A116082 * A116084 A116085 A116086 KEYWORD nonn AUTHOR Farideh Firoozbakht, Mar 12 2006 STATUS approved

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Last modified April 7 01:11 EDT 2020. Contains 333291 sequences. (Running on oeis4.)