A070814 Solutions to phi(gpf(x)) - gpf(phi(x)) = 14 = c are special multiples of 17, x = 17k, where greatest prime factors of factor k were observed from {2, 3, 5}, i.e., it is smaller than 17. See solutions to other even cases of c (=A070813): A007283 for 0, A070004 for 2, A070815 for 254, A070816 for 65534. Gpf = greatest prime factor.

17, 34, 51, 68, 85, 102, 136, 170, 204, 255, 272, 340, 408, 510, 544, 680, 816, 1020, 1088, 1360, 1632, 2040, 2176, 2720, 3264, 4080, 4352, 5440, 6528, 8160, 8704, 10880, 13056, 16320, 17408, 21760, 26112, 32640, 34816, 43520, 52224, 65280

Offset: 1

Author

Labos Elemer, May 09 2002

Keywords

• nonn

Comments

For n > 10, a(n) = 2a(n-4). First, it is easy to show that with i >= 0 and k,m in {0,1}, a(n) are of the form 2^i*3^k*5^m. Factoring this sequence reveals the regular pattern 2^i, 2^(i-2)*5, 2^(i-1)*3, 2^(i-3)*3*5, 2^(i+1), ... which obviously has the property a(n) = 2a(n-4) for n > 10. - Lambert Herrgesell (lambert.herrgesell(AT)googlemail.com), Jan 09 2007

Examples

			For n = 32640 = 128*3*5*17, gpf(n) = 17, phi(n) = 16384, commutator[32640] = phi(17) - gpf(16384) = 16 - 2 = 14.
		

Crossrefs

Cf. A000010, A000215, A006530, A007283, A070002, A070003, A070004, A070777, A070812, A070813.

Programs

Mathematica

pf[x_] := Part[Reverse[Flatten[FactorInteger[x]]], 2] Do[s=EulerPhi[pf[n]]-pf[EulerPhi[n]]; If[Equal[s, 14], Print[{n, n/17, pf[n/17]}]], {n, 3, 1000000}] (* Terms of sequence are n *)

Formula

For n > 10, a(n) = 2a(n-4) (conjectured). - Ralf Stephan, May 09 2004