A067337 -id:A067337 - cp's OEIS frontend

A067336 a(0)=1, a(1)=2, a(n) = a(n-1)*9/2 - Catalan(n-1) where Catalan(n) = binomial(2n,n)/(n+1) = A000108(n).

1, 2, 8, 34, 148, 652, 2892, 12882, 57540, 257500, 1153888, 5175700, 23231864, 104335376, 468766292, 2106773874, 9470787588, 42583186476, 191494694352, 861248485884, 3873850923288, 17425765034376, 78391476387672, 352670161180884, 1586672665700328, 7138737091504152

Offset: 0

Henry Bottomley, Jan 15 2002

nonn

Note that while a(n) is even (for n > 0), it is a multiple of 4 except when n = 2^m-1, i.e., when Catalan(n) is odd.
Result of applying the Riordan matrix ((1+sqrt(1-4*x))/2, (1-sqrt(1-4*x))/2) (inverse of (1/(1-x), x*(1-x))) to 3^n. - Paul Barry, Mar 12 2005
Hankel transform is A001787(n+1). - Paul Barry, Mar 15 2010

			a(2) =   2*9/2 -  1 =   8;
a(3) =   8*9/2 -  2 =  34;
a(4) =  34*9/2 -  5 = 148;
a(5) = 148*9/2 - 14 = 652.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000108, A001045, A039599, A067337, A088218.

CoefficientList[Series[(1+Sqrt[1-4*x])/(3*Sqrt[1-4*x]-1), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *)

a(n) = A067337(2n, n).
G.f.: (1+sqrt(1-4*x))/(3*sqrt(1-4*x)-1). - Paul Barry, Mar 12 2005
a(n) = Sum_{k=0..n} A039599(n,k)*A001045(k+1). - Philippe Deléham, Jun 10 2007
G.f.: (1-x*c(x))/(1-3*x*c(x)), where c(x) is the g.f. of A000108. - Paul Barry, Mar 15 2010
Conjecture: 2*n*a(n) + (-17*n+12)*a(n-1) + 18*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 30 2012
The above conjecture is true. - Nguyen Tuan Anh, Mar 15 2025
G.f.: 1 + 2*x/(Q(0)-3*x), where Q(k) = 2*x + (k+1)/(2*k+1) - 2*x*(k+1)/(2*k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 03 2013
a(n) ~ 3^(2*n-1) / 2^n. - Vaclav Kotesovec, Feb 13 2014

A135299 Pascal's triangle, but the last element of the row is the sum of all the previous terms.

Original entry on oeis.org

1, 1, 2, 1, 3, 8, 1, 4, 11, 32, 1, 5, 15, 43, 128, 1, 6, 20, 58, 171, 512, 1, 7, 26, 78, 229, 683, 2048, 1, 8, 33, 104, 307, 912, 2731, 8192, 1, 9, 41, 137, 411, 1219, 3643, 10923, 32768, 1, 10, 50, 178, 548, 1630, 4862, 14566, 43691, 131072

Offset: 0

Jose Ramon Real, Dec 04 2007

			T(2,1) = T(1,0) + T(1,1) = 1 + 2 = 3;
T(2,2) = T(0,0) + T(1,0) + T(1,1) + T(2,0) + T(2,1) = 1 + 1 + 2 + 1 + 3 = 8.
From _G. C. Greubel_, Oct 09 2016: (Start)
The triangle is:
  1;
  1, 2;
  1, 3,  8;
  1, 4, 11, 32;
  1, 5, 15, 43, 128;
  1, 6, 20, 58, 171, 512;
  ... (End)

G. C. Greubel, Table of n, a(n) for the first 25 rows

Cf. A007318, A067337.

T[0, 0] := 1; T[n_, 0] := 1; T[n_, k_] := T[n - 1, k] + T[n - 1, k - 1]; T[n_, n_] := 2^(2*n - 1); Table[T[n, k], {n, 0, 5}, {k, 0, n}] (* G. C. Greubel, Oct 09 2016 *)

T(0,0) = 1;
T(n,0) = 1;
T(n,k) = T(n-1, k-1) + T(n-1, k) if k < n;
T(n,n) = (Sum_{j=0..n-1} Sum_{i=0..j} T(j,i)) + Sum_{i=0..n-1} T(n,i) [i.e., sum of all earlier terms of the triangle].
T(n,n) = (4^n)/2 for n > 0;
T(n,n) = 2*Sum_{i=0..n-1} T(n,i).

cp's OEIS Frontend

A067336 a(0)=1, a(1)=2, a(n) = a(n-1)*9/2 - Catalan(n-1) where Catalan(n) = binomial(2n,n)/(n+1) = A000108(n).

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