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%I #40 Oct 11 2017 05:22:55
%S 1,6,35,210,1287,8008,50388,319770,2042975,13123110,84672315,
%T 548354040,3562467300,23206929840,151532656696,991493848554,
%U 6499270398159,42671977361650,280576272201225
%N a(n) = binomial(3*n+1, n+1).
%C a(n) = A258993(2*n+1, n). - _Reinhard Zumkeller_, Jun 22 2015
%H Reinhard Zumkeller, <a href="/A117671/b117671.txt">Table of n, a(n) for n = 0..1000</a>
%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Two Enumerative Functions</a>
%F G.f.: (2*(-1+Hypergeometric2F1[-(1/3),1/3,-(1/2),(27*x)/4]))/(3*x). - _Harvey P. Dale_, Jul 19 2011
%F G.f.: A(x) = B'(x)/B(x)-B'(x)-1/x, where B(x) = 4/3*sin(1/3*asin(sqrt((27*x)/4)))^2. - _Vladimir Kruchinin_, Nov 26 2014
%F From _Peter Bala_, Nov 04 2015: (Start)
%F With an extra initial term equal to 1, the o.g.f. equals f(x)/g(x)^2, where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764.
%F More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1). (End)
%F a(n) = [x^(2*n)] 1/(1 - x)^(n+2). - _Ilya Gutkovskiy_, Oct 10 2017
%F a(n+1) = 3*(3*n+2)*(3*n+4)*a(n)/(2*(n+2)*(2*n+1)). - _Robert Israel_, Oct 10 2017
%e if n=0 then C(3*0+1,0+1) = C(1,1) = 1.
%e if n=10 then C(3*10+1,10+1) = C(31,11) = 84672315.
%p seq(binomial(3*n+1,n+1),n=0..30); # _Robert Israel_, Oct 10 2017
%t Table[Binomial[3n+1,n+1],{n,0,20}] (* _Harvey P. Dale_, Jul 19 2011 *)
%o (Haskell)
%o a117671 n = a258993 (2 * n + 1) n -- _Reinhard Zumkeller_, Jun 22 2015
%o (PARI) vector(30, n, n--; binomial(3*n+1, n+1)) \\ _Altug Alkan_, Nov 04 2015
%Y Cf. A025174: binomial(3n-1,n-1), A006013.
%Y Cf. A258993, A001764, A004319, A005809, A013698, A045721, A165817, A236194.
%K nonn,easy
%O 0,2
%A _Zerinvary Lajos_, Apr 12 2006
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