A306616 Integers k such that phi(Catalan(n+1)) = 4*phi(Catalan(n)) where phi is A000010 and Catalan is A000108.

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

Table of n, a(n) for n=1..57.

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

Cf. A000010, A000108, A062624.

Cf. A005382, A157978.

Sequence in context: A222361 A134789 A058217 * A183183 A072675 A033711

Adjacent sequences: A306613 A306614 A306615 * A306617 A306618 A306619

Keyword

nonn

Author

Michel Marcus, Mar 01 2019

Status

approved