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a(n) = 2^n*binomial(4*n, n)/(3*n+1).
3

%I #51 Sep 08 2022 08:46:04

%S 1,2,16,176,2240,31008,453376,6888960,107707392,1721477120,

%T 28000141312,461964898304,7712495058944,130050777006080,

%U 2211737871974400,37892693797109760,653389823437701120,11330548232319664128,197475886172892823552

%N a(n) = 2^n*binomial(4*n, n)/(3*n+1).

%C Old name was: Series reversion of x - 2*x^4.

%C Regular blocks of 2 intermediate zeros have been removed from the sequence: If y = x - 2*x^4, then x = y + 2*y^4 + 16*y^7 + 176*y^10 + 2240*y^13 + 31008*y^16 + ...

%C a(n) is the number of lattice paths (Schroeder paths) from (0,0) to (n,4n) with unit steps N=(0,1), E=(1,0) and D=(1,1) staying weakly above the line y = 4x with the total number of occurrences of NE and D equal to n. - _Michael D. Weiner_, Jul 25 2019

%H Jinyuan Wang, <a href="/A217360/b217360.txt">Table of n, a(n) for n = 0..100</a>

%H D. Birmajer, J. B. Gil, J. D. Gil and M. D. Weiner, <a href="http://arxiv.org/abs/1908.08103">Schröder Coloring and Applications</a>, arXiv:1908.08103 [math.CO], 2019.

%F D-finite with recurrence 3*n*(3*n-1)*(3*n+1)*a(n)- 8*(4*n-1)*(4*n-3)*(4*n-2)*a(n-1) = 0, so a(n) = 8^n*A060706(n)/A100089(n) = 2^n*A002293(n).

%F a(n) = [x^(3*n)](f(x)/x) where f(x) is the reversion of x - 2*x^4.

%F G.f.: F([1/4, 1/2, 3/4], [2/3, 4/3], 512*x/27), where F is the generalized hypergeometric function. - _Stefano Spezia_, Aug 18 2019

%F G.f. A(x) satisfies: A(x) = 1 / (1 - 2 * x * A(x)^3). - _Ilya Gutkovskiy_, Nov 12 2021

%p A100089 := proc(n)

%p (3*n+1)! ;

%p end proc:

%p A060706 := proc(n)

%p (4*n)!/n!/4^n ;

%p end proc:

%p A217360 := proc(n)

%p 8^(n)*A060706(n)/A100089(n) ;

%p end proc:

%p seq(A217360(n), n=0..20);

%t Table[2^n Binomial[4 n, n] / (3 n + 1), {n, 0, 20}] (* _Vincenzo Librandi_, Jul 26 2019 *)

%o (Magma) [2^n*Binomial(4*n, n)/(3*n+1): n in [0..25]]; // _Vincenzo Librandi_, Jul 26 2019

%Y Cf. A153231 (x+2*x^3).

%K nonn,easy

%O 0,2

%A _R. J. Mathar_, Oct 01 2012

%E Offset decreased by 1 and name changed by _Michael D. Weiner_, Jul 25 2019