A081377 Numbers n such that the set of prime divisors of phi(n) is equal to the set of prime divisors of sigma(n).

1, 3, 14, 35, 42, 70, 105, 119, 209, 210, 238, 248, 297, 357, 418, 477, 594, 595, 616, 627, 714, 744, 954, 1045, 1178, 1190, 1240, 1254, 1463, 1485, 1672, 1674, 1736, 1785, 1848, 1863, 2079, 2090, 2376, 2385, 2540, 2728, 2926, 2945, 2970, 3080, 3135, 3302

Offset: 1

Author

Labos Elemer, Mar 26 2003

Keywords

• nonn

Comments

The multiplicities of the divisors are to be ignored.

Is it true that 1 is the only term in both this sequence and A055744? - Farideh Firoozbakht, Jul 01 2008. Answer from Luke Pebody, Jul 10 2008: No! In fact the numbers 103654150315463023813006470 and 6534150553412193640795377701190 are in both sequences.

Examples

			n=418=2*11*19: sigma(418)=720, phi[418]=180, common prime factor set ={2,3,5}
k = 477 = 3*3*53: sigma(477) = 702 = 2*3*3*3*13; phi(477) = 312 = 2*2*2*3*13; common factor set: {2,3,13}.
phi(89999)=66528=2^5*3^3*7*11 and sigma(89999)=118272=2^9*3*7*11 so 89999 is in the sequence.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000
• Prime Puzzles, Puzzle 451

Crossrefs

Cf. A000010, A000203, A076533, A065642, A081378, A110751, A110819, A027598, A141718.

Programs

Mathematica

ffi[x_] := Flatten[FactorInteger[x]] lf[x_] := Length[FactorInteger[x]] ba[x_] := Table[Part[ffi[x], 2*w-1], {w, 1, lf[x]}] Do[s=ba[DivisorSigma[1, n]]; s1=ba[EulerPhi[n]]; If[Equal[s, s1], k=k+1; Print[n]], {n, 1, 10000}]

PARI

is(n)=factor(eulerphi(n=factor(n)))[,1]==factor(sigma(n))[,1] \\ Charles R Greathouse IV, Nov 27 2013

Extensions

Edited by N. J. A. Sloane, Jul 11 2008 at the suggestion of Farideh Firoozbakht