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Third column of triangle in A234950.
2

%I #10 Apr 06 2018 02:39:11

%S 2,20,135,770,4004,19656,92820,426360,1918620,8498776,37182145,

%T 161056350,691945800,2952675600,12527780760,52895074320,222399744300,

%U 931689977400,3890668331550,16201562020644,67298796085752,278927990831600,1153747598439800,4763749454427600,19637233862140440

%N Third column of triangle in A234950.

%C Remmel (2014) asks for a formula.

%H Vincenzo Librandi, <a href="/A244887/b244887.txt">Table of n, a(n) for n = 2..500</a>

%H Jeffrey B. Remmel, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v21i3p2/0">Consecutive Up-down Patterns in Up-down Permutations</a>, Electron. J. Combin., 21 (2014), #P3.2.

%F a(n) = A234950(n, 2).

%t Table[Sum[Binomial[s, 2] Binomial[n+s, n] (n - s + 1) / (n + 1), {s, 2, n}], {n, 2, 15}] (* _Vincenzo Librandi_, Apr 06 2018 *)

%o (PARI) a(n) = sum(s=2, n, binomial(s, 2)*binomial(n+s, n)*(n-s+1)/(n+1)); \\ _Michel Marcus_, Apr 06 2018

%Y Cf. A234950.

%K nonn

%O 2,1

%A _N. J. A. Sloane_, Jul 12 2014

%E More terms from _Michel Marcus_, Apr 06 2018