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Integers k such that phi(Catalan(n+1)) = 4*phi(Catalan(n)) where phi is A000010 and Catalan is A000108.
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%I #27 Mar 04 2019 23:12:16

%S 2,8,19,20,36,42,44,55,56,76,91,109,116,120,140,143,152,156,176,184,

%T 200,204,213,216,224,235,242,260,289,296,300,380,384,400,401,415,436,

%U 464,469,476,524,547,553,564,595,602,616,624,630,631,660,685,704,716,744,776,800

%N Integers k such that phi(Catalan(n+1)) = 4*phi(Catalan(n)) where phi is A000010 and Catalan is A000108.

%C Integers k such that A062624(n+1) = 4*A062624(n).

%C Consists of integers k (see p. 1405 of Luca link):

%C k = 2p-2, where p >= 5 is a prime such that q = 4p-3 is also prime (see A157978);

%C k = 3p-2, where p > 5 is a prime such that q = 2p-1 is also prime (see A005382).

%H Florian Luca, Pantelimon Stanica, <a href="https://doi.org/10.1016/j.jnt.2012.02.002">On the Euler function of the Catalan numbers</a>, Journal of Number Theory 132(7):1404-1424.

%e phi(C(2)) = phi(2) = 1 and phi(C(3)) = phi(5) = 4 so 2 is a term.

%t Select[Range[1000], EulerPhi[CatalanNumber[#+1]]== 4*EulerPhi[CatalanNumber[#]] &] (* _G. C. Greubel_, Mar 02 2019 *)

%o (PARI) C(n) = binomial(2*n,n)/(n+1);

%o isok(n) = eulerphi(C(n+1)) == 4*eulerphi(C(n));

%o (Sage) [n for n in (1..1000) if euler_phi(catalan_number(n+1)) == 4*euler_phi(catalan_number(n))] # _G. C. Greubel_, Mar 02 2019

%Y Cf. A000010, A000108, A062624.

%Y Cf. A005382, A157978.

%K nonn

%O 1,1

%A _Michel Marcus_, Mar 01 2019