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

%I A081377 #20 Sep 09 2017 23:18:51
%S A081377 1,3,14,35,42,70,105,119,209,210,238,248,297,357,418,477,594,595,616,
%T A081377 627,714,744,954,1045,1178,1190,1240,1254,1463,1485,1672,1674,1736,
%U A081377 1785,1848,1863,2079,2090,2376,2385,2540,2728,2926,2945,2970,3080,3135,3302
%N A081377 Numbers n such that the set of prime divisors of phi(n) is equal to the set of prime divisors of sigma(n).
%C A081377 The multiplicities of the divisors are to be ignored.
%C A081377 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.
%H A081377 T. D. Noe, <a href="/A081377/b081377.txt">Table of n, a(n) for n = 1..1000</a>
%H A081377 Prime Puzzles, <a href="http://www.primepuzzles.net/puzzles/puzz_451.htm">Puzzle 451</a>
%e A081377 n=418=2*11*19: sigma(418)=720, phi[418]=180, common prime factor set ={2,3,5}
%e A081377 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}.
%e A081377 phi(89999)=66528=2^5*3^3*7*11 and sigma(89999)=118272=2^9*3*7*11 so 89999 is in the sequence.
%t A081377 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}]
%o A081377 (PARI) is(n)=factor(eulerphi(n=factor(n)))[,1]==factor(sigma(n))[,1] \\ _Charles R Greathouse IV_, Nov 27 2013
%Y A081377 Cf. A000010, A000203, A076533, A065642, A081378, A110751, A110819, A027598, A141718.
%K A081377 nonn
%O A081377 1,2
%A A081377 _Labos Elemer_, Mar 26 2003
%E A081377 Edited by _N. J. A. Sloane_, Jul 11 2008 at the suggestion of Farideh Firoozbakht