OFFSET
1,2
COMMENTS
Theorem: If 2^p-1 is prime (a Mersenne prime) and 0<=k<=5 then m=2^p*(2^2^k-1) is in the sequence. Proof: If k=0 then m=2^p and sigma(uphi(m))=sigma(uphi(2^p))=sigma(2^p-1)=2^p=m. Now if k>0 then m=2^p*(2^2^0+1)*(2^2^1+1)*...*(2^2^(k-1)+1) since 2^2^i+1 for i=0,1,...,4 are primes(Fermat primes) we conclude that uphi(m)=(2^p-1)*(2^2^0)*(2^2^1)*...*(2^2^(k-1))=(2^p-1)*2^ (2^0+2^1+...+2^(k-1))=(2^p-1)*2^(2^k-1) hence sigma(uphi(m))= sigma(2^p-1)*sigma(2^(2^k-1))=2^p*(2^2^k-1)=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^k-1).
LINKS
Donovan Johnson, Table of n, a(n) for n = 1..50 (terms < 2^35)
EXAMPLE
110771178 = 2*3*7*19*127*1093 is in the sequence because sigma(uphi(2*3*7*19*127*1093)) = sigma((2-1)*(3-1)*(7-1)*(19-1)*(127-1)*(1093-1) = sigma(2*6*18*126*1092) = sigma(2^6*3^6*7^2*13) = (2^7-1)*((3^7-1)/2)*((7^3-1)/6)*14 = 110771178.
MATHEMATICA
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Farideh Firoozbakht, Jul 11 2006
EXTENSIONS
Name corrected by Donovan Johnson, May 01 2013
STATUS
approved