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Numbers k such that 1/(1/phi(k) + 1/phi(k+1) + 1/phi(k+2) + 1/phi(k+3)) is an integer.
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%I #12 Feb 18 2021 14:34:15

%S 13,75,111,144,192,558,559,1683,2016,3624,7129,10369,11658,18362,

%T 19442,19800,19801,32772,47627,60482,82082,133988,143642,229321,

%U 291721,312483,352846,390603,395136,436801,465482,600601,711936,806736,819729

%N Numbers k such that 1/(1/phi(k) + 1/phi(k+1) + 1/phi(k+2) + 1/phi(k+3)) is an integer.

%e 1/phi(75)+1/phi(76)+1/phi(77)+1/phi(78) = 1/40+1/36+1/60+1/24 = 1/9 so 75 is in the sequence.

%t Select[Range[900000], IntegerQ[1/Sum[1/EulerPhi[ #+i], {i, 0, 3}]]&]

%o (PARI) isok(k) = numerator(1/eulerphi(k) + 1/eulerphi(k+1) + 1/eulerphi(k+2) + 1/eulerphi(k+3)) == 1; \\ _Michel Marcus_, Feb 18 2021

%Y Cf. A000010, A073542, A073543.

%K nonn

%O 1,1

%A _Benoit Cloitre_, Aug 27 2002

%E Edited by _Dean Hickerson_, Sep 03 2002