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a(n) = (1/3)*(n+2)^2*(3*n+3)!/(n+2)!^3.
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%I #12 Sep 08 2022 08:46:23

%S 1,10,140,2310,42042,816816,16628040,350574510,7595781050,

%T 168212023980,3792416540640,86787993910800,2011383287449200,

%U 47123837020238400,1114478745528638160,26575401262863040830,638330716607984804250,15431925043610580004500,375239440534109892741000

%N a(n) = (1/3)*(n+2)^2*(3*n+3)!/(n+2)!^3.

%C Number of Schröder paths of length 2n+1 having n peaks.

%F a(n) = (n+2)*(3*n+2)!/((n+2)!^2*n!).

%F a(n) = A060693(2n+1,n).

%p a := n -> (n+2)*(3*n+2)!/((n+2)!^2*n!): seq(a(n), n = 0..18);

%t Table[(n+2) (3*n+2)! / ((n+2)!^2 n!), {n, 0, 30}] (* _Vincenzo Librandi_, Oct 01 2018 *)

%o (PARI) a(n) = (1/3)*(n+2)^2*(3*n+3)!/(n+2)!^3; \\ _Michel Marcus_, Oct 01 2018

%o (Magma) [(1/3)*(n+2)^2*Factorial(3*n+3)/Factorial(n+2)^3: n in [0..20]]; // _Vincenzo Librandi_, Oct 01 2018

%Y Cf. A007004, A060693, A215287.

%K nonn

%O 0,2

%A _Peter Luschny_, Sep 30 2018