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A104722
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Self-convolution of repeated Catalan numbers.
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3
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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
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OFFSET
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0,2
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COMMENTS
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This is the same as A059348 after the first term. [Proof by James A. 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.]
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LINKS
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FORMULA
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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
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MATHEMATICA
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CoefficientList[Series[( (1 + x)*(1 - Sqrt[1 - 4*x^2])/(2*x^2))^2, {x, 0, 100}], x] (* G. C. Greubel, Jan 07 2017 *)
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PROG
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(PARI) Vec( ((1 + x)*(1 - sqrt(1 - 4*x^2))/(2*x^2))^2 + O(x^20)) \\ G. C. Greubel, Jan 07 2017
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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