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A306616
Integers k such that phi(Catalan(n+1)) = 4*phi(Catalan(n)) where phi is A000010 and Catalan is A000108.
0
2, 8, 19, 20, 36, 42, 44, 55, 56, 76, 91, 109, 116, 120, 140, 143, 152, 156, 176, 184, 200, 204, 213, 216, 224, 235, 242, 260, 289, 296, 300, 380, 384, 400, 401, 415, 436, 464, 469, 476, 524, 547, 553, 564, 595, 602, 616, 624, 630, 631, 660, 685, 704, 716, 744, 776, 800
OFFSET
1,1
COMMENTS
Integers k such that A062624(n+1) = 4*A062624(n).
Consists of integers k (see p. 1405 of Luca link):
k = 2p-2, where p >= 5 is a prime such that q = 4p-3 is also prime (see A157978);
k = 3p-2, where p > 5 is a prime such that q = 2p-1 is also prime (see A005382).
LINKS
Florian Luca, Pantelimon Stanica, On the Euler function of the Catalan numbers, Journal of Number Theory 132(7):1404-1424.
EXAMPLE
phi(C(2)) = phi(2) = 1 and phi(C(3)) = phi(5) = 4 so 2 is a term.
MATHEMATICA
Select[Range[1000], EulerPhi[CatalanNumber[#+1]]== 4*EulerPhi[CatalanNumber[#]] &] (* G. C. Greubel, Mar 02 2019 *)
PROG
(PARI) C(n) = binomial(2*n, n)/(n+1);
isok(n) = eulerphi(C(n+1)) == 4*eulerphi(C(n));
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Marcus, Mar 01 2019
STATUS
approved