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Numbers n such that n and tau(n) = A000005(n) have the same prime factors (ignoring multiplicity).
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%I #18 Oct 19 2017 10:56:31

%S 1,2,8,9,12,18,72,80,96,108,128,288,448,486,625,720,768,864,972,1152,

%T 1200,1250,1620,1944,2000,2025,2560,4032,4050,5000,5625,6144,6561,

%U 6912,7500,7776,8748,9408,10800,11250,11264,12960,13122,16200,18000,18432,19440

%N Numbers n such that n and tau(n) = A000005(n) have the same prime factors (ignoring multiplicity).

%H Donovan Johnson and Giovanni Resta, <a href="/A081381/b081381.txt">Table of n, a(n) for n = 1..1096</a> (terms < 10^11, first 500 terms from Donovan Johnson)

%e n = 5000 = 2*2*2*5*5*5*5, tau(5000) = 20 = 2*2*5, common prime factors: {2,5}

%t 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[0, n]]; If[Equal[s, ba[n]], Print[n]], {n, 1, 10000}]

%o (PARI) is(n)=my(f=factor(n)); factor(numdiv(f))[,1]==f[,1] \\ _Charles R Greathouse IV_, Oct 19 2017

%Y Cf. A000005, A027598, A055744, A065642, A081377-A081380.

%K nonn

%O 1,2

%A _Labos Elemer_, Mar 26 2003