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A047098 a(n) = 2*binomial(3*n, n) - Sum_{k=0..n} binomial(3*n, k).

%I A047098 #42 Jan 17 2024 19:07:21
%S A047098 1,2,8,38,196,1062,5948,34120,199316,1181126,7080928,42860534,
%T A047098 261542752,1607076200,9934255472,61732449648,385393229460,
%U A047098 2415935640198,15200964233864,95962904716402,607640599286276,3858198001960438,24559243585545644,156692889782067712
%N A047098 a(n) = 2*binomial(3*n, n) - Sum_{k=0..n} binomial(3*n, k).
%C A047098 T(2n,n), array T as in A047089. [Corrected Dec 08 2006]
%C A047098 Let B_3^+ denote the semigroup with presentation <a,b | aba=bab>. Let D=aba be the 'fundamental word'. Then this sequence is also equal to the number of words in B_3^+ equal in B_3^+ to D^n, n >= 0. - _Stephen P. Humphries_, Jan 20 2004
%C A047098 In the language of Riordan arrays, row sums of (1/(1+x), x/(1+x)^3)^-1, where (1/(1+x), x/(1+x)^3) has general term (-1)^(n-k)*binomial(n+2k, 3k). - _Paul Barry_, May 09 2005
%C A047098 Hankel transform is 2^n*A051255(n) where A051255 is the Hankel transform of C(3n,n)/(2n+1). - _Paul Barry_, Jan 21 2007
%H A047098 Michael De Vlieger, <a href="/A047098/b047098.txt">Table of n, a(n) for n = 0..1211</a>
%H A047098 Paul Barry, <a href="https://arxiv.org/abs/2104.01644">Centered polygon numbers, heptagons and nonagons, and the Robbins numbers</a>, arXiv:2104.01644 [math.CO], 2021.
%H A047098 Christopher R. Cornwell and Stephen P. Humphries, <a href="https://doi.org/10.1142/S0218216508006051">Counting fundamental paths in certain Garside semigroups</a>, Journal of Knot Theory and Its Ramifications, Vol. 17 (2008), No. 02, pp. 191-211.
%F A047098 G.f. A(x)=y satisfies (8x-1)y^3-y^2+y+1=0. - _Michael Somos_, Jan 28 2004
%F A047098 Coefficient of x^n in ((1+10x-2x^2+(1-4x)^(3/2))/2)^n. - _Michael Somos_, Sep 25 2003
%F A047098 a(n) = Sum_{k = 0..n} A109971(k)*2^k; a(0) = 1, a(n) = Sum_{k = 0..n} 2^k*C(3n-k,n-k)*2*k/(3*n-k), n > 0. - _Paul Barry_, Jan 21 2007
%F A047098 Conjecture: 2*n*(2*n-1)*a(n) +(-71*n^2+112*n-48)*a(n-1) +3*(131*n^2-391*n+296)*a(n-2) -72*(3*n-7)*(3*n-8)*a(n-3)=0. - _R. J. Mathar_, Nov 30 2012
%F A047098 a(n) = A321957(n) + 2*binomial(3*n, n) - 8^n. - _Peter Luschny_, Nov 22 2018
%F A047098 The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. - _Peter Bala_, Mar 05 2022
%p A047098 A047098 := n -> 2*binomial(3*n, n)-add(binomial(3*n, k), k=0..n);
%t A047098 Table[2Binomial[3n,n]-Sum[Binomial[3n,k],{k,0,n}],{n,0,35}] (* _Harvey P. Dale_, Jul 27 2011 *)
%o A047098 (PARI) a(n)=if(n<0,0,polcoeff((((1+10*x-2*x^2)+(1-4*x)*sqrt(1-4*x+x*O(x^n)))/2)^n,n))
%o A047098 (PARI) a(n)=if(n<0,0, 2*binomial(3*n,n)-sum(k=0,n,binomial(3*n,k)))
%Y A047098 Column k=2 of A213028.
%Y A047098 Cf. A047089, A047099, A107026, A321957.
%K A047098 nonn
%O A047098 0,2
%A A047098 _Clark Kimberling_, Aug 15 1998
%E A047098 _Clark Kimberling_, Dec 08 2006, changed "T(3n,2n)" to "T(2n,n)" in the comment line, but observes that some of the other comments seem to apply to the sequence T(3n,2n) rather than to the sequence T(2n,n).
%E A047098 Edited by _N. J. A. Sloane_, Dec 21 2006, replacing the old definition in terms of A047089 by an explicit formula supplied by _Benoit Cloitre_, Oct 25 2003.