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A346277 Primitive terms of A108569.

%I A346277 #41 Dec 23 2024 22:48:56
%S A346277 4,110,506,550,1830,2162,2750,3422,4114,4746,5490,5566,6806,7782,9150,
%T A346277 11342,13750,14238,16470,16762,23346,27450,27722,31862,33222,42714,
%U A346277 43378,45254,45750,49410,49726,51302,61226,68750,68906,70038,82350,99238,99666,112110,115650
%N A346277 Primitive terms of A108569.
%C A346277 If k is an even term of A108569 then 2k is another term.
%C A346277 This sequence lists the initial term k_0 of each infinite subsequence of A108569 that is solution of the equation phi(k) = phi(k + phi(k)).
%C A346277 As 2 is no solution, A108569(1) = 1 is not primitive.
%C A346277 Each k_0 > 4 is of the form k_0 = 2*m with m odd.
%C A346277 If p > 3 is a Sophie Germain prime, then every m = 2*p^q*(2p+1), q >=1 is a term because in this case, phi(m) = phi(m+phi(m)) = 2*(p-1)*p^q; the first terms that are not of this form are 4, 1830, 4114, ...
%e A346277 a(1) = 4 because every k = 2^m, m >= 2 satisfies phi(k) = phi(k+phi(k)) = 2^(m-1), and k_0 = 4 is the smallest term of this subsequence of A108569.
%e A346277 a(2) = 110 because every k = 5*11*2^m, m >= 1 satisfies phi(k) = phi(k+phi(k)) = 5*2^(m+2) and k_0 = 110 is the smallest term of this subsequence of A108569 (note that 5 is a Sophie Germain prime).
%e A346277 a(5) = 1830 because every k = 3*5*61*2^m, m >= 1 satisfies phi(k) = phi(k+phi(k)) = 3*5*2^(m+4) and k_0 = 1830 is the smallest term of this subsequence of A108957.
%p A346277 with(numtheory):
%p A346277 for m from 2 to 116000 by 2 do
%p A346277 u:=phi(m+phi(m)) - phi(m);
%p A346277 if u=0 and phi(m/2 + phi(m/2)) <> phi(m/2) then print(m); else fi; od:
%o A346277 (PARI) f(m) = eulerphi(m+eulerphi(m)) - eulerphi(m);
%o A346277 isok(m) = !f(m) && !(m % 2) && f(m/2); \\ _Michel Marcus_, Aug 31 2021
%Y A346277 Cf. A000010, A005384.
%Y A346277 Subsequence of A108569.
%Y A346277 Similar to A346694 (with phi(k) = phi(k-phi(k))).
%K A346277 nonn
%O A346277 1,1
%A A346277 _Bernard Schott_, Aug 22 2021