|
%I #12 Nov 07 2016 12:02:47
%S 1,1,5,132,16796,9694845,24466267020,263747951750360,
%T 11959798385860453492,2257117854077248073253720,
%U 1759414616608818870992479875972,5632681584560312734993915705849145100
%N Catalan numbers at triangular positions: a(n) = A000108(n(n+1)/2).
%H G. C. Greubel, <a href="/A135758/b135758.txt">Table of n, a(n) for n = 0..50</a>
%F a(n) = C(n(n+1), n(n+1)/2) / [n(n+1)/2 + 1].
%t With[{nn=100},Join[{1},Pick[CatalanNumber[Range[nn]],Array[ IntegerQ[ (Sqrt[1+8#]-1)/2]&,nn]]]] (* _Harvey P. Dale_, Sep 24 2011 *)
%t Table[Binomial[n*(n + 1), n*(n + 1)/2]/(1 + Binomial[n + 1, 2]), {n,0,10}] (* _G. C. Greubel_, Nov 07 2016 *)
%o (PARI) a(n)=binomial(n*(n+1),n*(n+1)/2)/(n*(n+1)/2+1)
%Y Cf. A000108, A135757.
%K nonn,easy
%O 0,3
%A _Paul D. Hanna_, Dec 02 2007
|
