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%I #119 Apr 09 2023 03:08:31
%S 0,1,5,28,165,1001,6188,38760,245157,1562275,10015005,64512240,
%T 417225900,2707475148,17620076360,114955808528,751616304549,
%U 4923689695575,32308782859535,212327989773900,1397281501935165
%N a(n) = binomial(3n-1, n-1).
%C Number of standard tableaux of shape (2n-1,n). Example: a(2)=5 because in the top row we can have 123, 124, 125, 134, or 135. - _Emeric Deutsch_, May 23 2004
%C Number of peaks in all generalized {(1,2),(1,-1)}-Dyck paths of length 3n.
%C Positive terms in this sequence are the numbers k such that k and 2k are consecutive terms in a row of Pascal's triangle. 1001 is the only k such that k, 2k, and 3k are consecutive terms in a row of Pascal's triangle. - _J. Lowell_, Mar 11 2023
%D B. C. Berndt, Ramanujan's Notebooks Part I, Springer-Verlag, see Entry 14, Corollary 1, p. 71.
%H Robert Israel, <a href="/A025174/b025174.txt">Table of n, a(n) for n = 0..1190</a>
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL23/Barry/barry444.html">On the Central Antecedents of Integer (and Other) Sequences</a>, J. Int. Seq., Vol. 23 (2020), Article 20.8.3.
%H D. Kruchinin and V. Kruchinin, <a href="http://cs.uwaterloo.ca/journals/JIS/VOL18/Kruchinin/kruch9.html">A Generating Function for the Diagonal T2n,n in Triangles</a>, Journal of Integer Sequence, Vol. 18 (2015), article 15.4.6.
%H W. Mlotkowski and K. A. Penson, <a href="http://arxiv.org/abs/1309.0595">Probability distributions with binomial moments</a>, arXiv preprint arXiv:1309.0595 [math.PR], 2013.
%H Emanuele Munarini, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL20/Munarini/muna4.pdf">Shifting Property for Riordan, Sheffer and Connection Constants Matrices</a>, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.2.
%F G.f.: z*g^2/(1-3*z*g^2), where g=g(z) is given by g=1+z*g^3, g(0)=1, that is, (in Maple command) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - _Emeric Deutsch_, May 22 2003
%F a(n) = Sum_{k=0..n} ((3k+1)/(2n+k+1))C(3n, 2n+k)*A001045(k). - _Paul Barry_, Oct 07 2005
%F Hankel transform of a(n+1) is A005156(n+1). - _Paul Barry_, Apr 14 2008
%F G.f.: x*B'(x)/B(x) where B(x) is the g.f. of A001764. - _Vladimir Kruchinin_ Feb 03 2013
%F D-finite with recurrence: 2*n*(2*n-1)*a(n) -3*(3*n-1)*(3*n-2)*a(n-1)=0. - _R. J. Mathar_, Feb 05 2013
%F Logarithmic derivative of A001764; g.f. of A001764 satisfies G(x) = 1 + x*G(x)^3. - _Paul D. Hanna_, Jul 14 2013
%F G.f.: (2*cos((1/3)*arcsin((3/2)*sqrt(3*x)))-sqrt(4-27*x))/(3*sqrt(4-27*x)). - _Emanuele Munarini_, Oct 14 2014
%F a(n) = Sum_{k=1..n} binomial(n-1,n-k)*binomial(2*n,n-k). - _Vladimir Kruchinin_, Nov 12 2014
%F a(n) = [x^n] C(x)^n for n >= 1, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the generating function for A000108 (Ramanujan). - _Peter Bala_, Jun 24 2015
%F From _Peter Bala_, Nov 04 2015: (Start)
%F Without the initial term 0, 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. g(x)^2 is the o.g..f for A006013. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1), A117671 (k = -2). (End)
%F G.f.: ( 2F1(1/3,2/3;1/2;27*x/4)-1)/3. - _R. J. Mathar_, Jan 27 2020
%F O.g.f. without the initial term 0, in the form g(x)=(2*cos(arcsin((3*sqrt(3)*sqrt(x))/2)/3)/sqrt(4-27*x)-1)/(3*x), satisfies the following algebraic equation: 1+(9*x-1)*g(x)+x*(27*x-4)*g(x)^2+x^2*(27*x-4)*g(x)^3=0. - _Karol A. Penson_, Oct 11 2021
%F O.g.f. equals f(x)/(1 - 2*f(x)), where f(x) = series reversion (x/(1 + x)^3) = x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... is the o.g.f. of A001764 with the initial term omitted. Cf. A224274. - _Peter Bala_, Feb 03 2022
%F Right-hand side of the identities (1/2)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+2)*n+k-1,k) = C(3*n-1,n-1) and (1/3)*Sum_{k = 0..n} (-1)^k* C(x*n,n-k)*C((x-3)*n+k-1,k) = C(3*n-1,n-1), both valid for n >= 1 and x arbitrary. - _Peter Bala_, Feb 28 2022
%e L.g.f.: L(x) = x + 5*x^2/2 + 28*x^3/3 + 165*x^4/4 + 1001*x^5/5 + 6188*x^6/6 + ...
%e where G(x) = exp(L(x)) satisfies G(x) = 1 + x*G(x)^3, and begins:
%e exp(L(x)) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... + A001764(n)*x^n + ...
%p with(combinat):seq(numbcomp(3*i,i), i=0..20); # _Zerinvary Lajos_, Jun 16 2007
%t Table[ GegenbauerC[ n, n, 1 ]/2, {n, 0, 24} ]
%t Join[{0},Table[Binomial[3n-1,n-1],{n,20}]] (* _Harvey P. Dale_, Oct 19 2022 *)
%o (Magma) [Binomial(3*n-1,n-1): n in [0..30]]; // _Vincenzo Librandi_, Nov 12 2014
%o (PARI) vector(30, n, n--; binomial(3*n-1, n-1)) \\ _Altug Alkan_, Nov 04 2015
%Y Cf. A001764 (binomial(3n,n)/(2n+1)), A117671 (C(3n+1,n+1)), A004319, A005809, A006013, A013698, A045721, A117671, A165817, A224274, A236194.
%K nonn,easy
%O 0,3
%A _Wouter Meeussen_, _Emeric Deutsch_
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