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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A256799 Catalan number analogs for A099927, the generalized binomial coefficients for Pell numbers (A000129).

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%I A256799 #18 Jan 23 2016 21:51:19
%S A256799 1,1,6,203,40222,46410442,312163223724,12237378320283699,
%T A256799 2796071362211148193590,3723566980632561787914135870,
%U A256799 28901575272390972687956930234335380,1307480498356321410289575304307661963042110,344746842780849469098742541704318199701366091840620
%N A256799 Catalan number analogs for A099927, the generalized binomial coefficients for Pell numbers (A000129).
%C A256799 One definition of the Catalan numbers is binomial(2*n,n) / (n+1); the current sequence models this definition using the generalized binomial coefficients arising from Pell numbers (A000129).
%H A256799 Alois P. Heinz, <a href="/A256799/b256799.txt">Table of n, a(n) for n = 0..50</a>
%H A256799 Tom Edgar and Michael Z. Spivey, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Edgar/edgar3.html">Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers</a>, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.
%F A256799 a(n) = Pell(2n)Pell(2n-1)...Pell(n+2)/Pell(n)Pell(n-1)...Pell(1) = A099927(2*n,n)/Pell(n+1) = A099929(n)/Pell(n+1), where Pell(k) = A000129(k).
%F A256799 a(n) ~ 2^(3/2) * (1+sqrt(2))^(n^2-n-1) / c, where c = A256831 = 1.141982569667791206028... . - _Vaclav Kotesovec_, Apr 10 2015
%e A256799 a(5) = Pell(10)..Pell(7)/Pell(5)..Pell(1) = (2378*985*408*169)/(29*12*5*2*1) = 46410442.
%e A256799 a(3) = A099927(6,3)/Pell(3) = 2436/12 = 203.
%p A256799 p:= n-> (<<2|1>, <1|0>>^n)[1, 2]:
%p A256799 a:= n-> mul(p(i), i=n+2..2*n)/mul(p(i), i=1..n):
%p A256799 seq(a(n), n=0..12);  # _Alois P. Heinz_, Apr 10 2015
%t A256799 Pell[m_]:=Expand[((1+Sqrt[2])^m-(1-Sqrt[2])^m)/(2*Sqrt[2])]; Table[Product[Pell[k],{k,1,2*n}]/(Product[Pell[k],{k,1,n}])^2 / Pell[n+1],{n,0,15}] (* _Vaclav Kotesovec_, Apr 10 2015 *)
%o A256799 (Sage)
%o A256799 P=[lucas_number1(n, 2, -1) for n in [0..30]]
%o A256799 [1/P[n+1]*prod(P[1:2*n+1])/(prod(P[1:n+1]))^2 for n in [0..14]]
%Y A256799 Cf. A000129, A099927, A003150, A099929, A256831, A256832.
%K A256799 nonn
%O A256799 0,3
%A A256799 _Tom Edgar_, Apr 10 2015