A059348 -id:A059348 - cp's OEIS frontend

A104722 Self-convolution of repeated Catalan numbers.

1, 2, 3, 4, 7, 10, 19, 28, 56, 84, 174, 264, 561, 858, 1859, 2860, 6292, 9724, 21658, 33592, 75582, 117572, 266798, 416024, 950912, 1485800, 3417340, 5348880, 12369285, 19389690, 45052515, 70715340, 165002460, 259289580, 607283490

Offset: 0

Paul Barry, Mar 20 2005

This is the same as A059348 after the first term. [Proof by James Sellers, seqfan 19 May 2008: The generating functions are the same, ignoring the constant terms which cause the difference between the two sequences. If the g.f. in the formula here is expanded, the constant term ignored, we obtain ( 1 + 2x - x^2 - 4x^3 - (x+1)^2*sqrt(1-4*x^2) )/(2*x^4) ].
From the Bernhart Reference in A059348 we see that A059348 originates from A000108 padded with zeros, 1 0 1 0 2 0 5 0 14 0 42 0 132 ... with g.f. C(x^2). Taking the sum of each pair of consecutive values we get the auxiliary sequence 1 1 1 2 2 5 5 14 14 42 42 132 132 ... with g.f. ((1+x)C(x^2) - 1)/x . Sum pairs of consecutive values once more to obtain 2 2 3 4 7 10 19 28 56 ... which is A059348.
So this generating function is (1+x)*(((1+x)*C(x^2) - 1)/x -1) / x again ignoring the constant term. Straightforward algebraic manipulations show that this quantity equals (1+2x-...)/(2x^4) above, again ignoring the constant term.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000108, A059348, A104721.

CoefficientList[Series[( (1 + x)*(1 - Sqrt[1 - 4*x^2])/(2*x^2))^2, {x, 0, 100}], x] (* G. C. Greubel, Jan 07 2017 *)

Vec( ((1 + x)*(1 - sqrt(1 - 4*x^2))/(2*x^2))^2 + O(x^20)) \\ G. C. Greubel, Jan 07 2017

G.f.: (1+x)^2*c(x^2)^2, c(x) the g.f. of the Catalan numbers A000108;
Let b(n) = (binomial(n-1, (n-1)/2)/((n-1)/2+1))*(1-(-1)^n)/2 + (binomial(n, n/2)/(n/2+1))*(1+(-1)^n)/2, then a(n) = Sum_{k=0..n} b(k)*b(n-k).
Conjecture: (n+4)*a(n) + (n+1)*a(n-1) - 4*(n+1)*a(n-2) + 4*(2-n)*a(n-3) = 0. - R. J. Mathar, Nov 09 2012
a(n) ~ 2^(n + 1/2) * (9 + (-1)^n) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Mar 10 2018

A059347 Difference array of Motzkin numbers A001006 read by antidiagonals.

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 0, 1, 2, 4, 2, 2, 3, 5, 9, 0, 2, 4, 7, 12, 21, 5, 5, 7, 11, 18, 30, 51, 0, 5, 10, 17, 28, 46, 76, 127, 14, 14, 19, 29, 46, 74, 120, 196, 323, 0, 14, 28, 47, 76, 122, 196, 316, 512, 835, 42, 42, 56, 84, 131, 207, 329, 525, 841, 1353, 2188, 0, 42, 84, 140, 224

Offset: 0

N. J. A. Sloane, Jan 27 2001

Row sums of odd rows (e.g., 4 = 1+1+2 for 3rd row) equal the Motzkin number of next row. Row sums of even rows equal the Motzkin number of the next row - n!/((n/2)!((n/2)+1)!) (i.e., A001006(n) - A000108(n/2) where A000108 are the Catalan numbers). - Gerald McGarvey, Dec 05 2004

			Triangle begins:
1;
0,1;
1,1,2;
0,1,2,4;
2,2,3,5,9;
0,2,4,7,12,21;
5,5,7,11,18,30,51;
...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
F. R. Bernhart, Catalan, Motzkin and Riordan numbers, Discr. Math., 204 (1999), 73-112.

Top row is A001006, leading diagonals give A000108 (interspersed with 0's), A000108 doubled up, A059348.

max = 12; A001006 = CoefficientList[ Series[ (1-x-(1-2x-3x^2)^(1/2))/(2x^2), {x, 0, max}], x] ; row[0] = A001006; row[n_] := Differences[A001006, n]; Flatten[ Table[ row[n-k][[k]], {n, 1, max}, {k, 1, n}]] (* Jean-François Alcover, Nov 12 2012, from formula *)

More terms from Larry Reeves (larryr(AT)acm.org), Feb 16 2001

cp's OEIS Frontend

A104722 Self-convolution of repeated Catalan numbers.

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