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

1, 2, 8, 38, 196, 1062, 5948, 34120, 199316, 1181126, 7080928, 42860534, 261542752, 1607076200, 9934255472, 61732449648, 385393229460, 2415935640198, 15200964233864, 95962904716402, 607640599286276, 3858198001960438, 24559243585545644, 156692889782067712

Offset: 0

Author

Clark Kimberling, Aug 15 1998

Keywords

• nonn

Comments

T(2n,n), array T as in A047089. [Corrected Dec 08 2006]

Let B_3^+ denote the semigroup with presentation . 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

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

Hankel transform is 2^n*A051255(n) where A051255 is the Hankel transform of C(3n,n)/(2n+1). - Paul Barry, Jan 21 2007

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..1211
• Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
• Christopher R. Cornwell and Stephen P. Humphries, Counting fundamental paths in certain Garside semigroups, Journal of Knot Theory and Its Ramifications, Vol. 17 (2008), No. 02, pp. 191-211.

Crossrefs

Column k=2 of A213028.

Cf. A047089, A047099, A107026, A321957.

Programs

Maple

A047098 := n -> 2*binomial(3*n, n)-add(binomial(3*n, k), k=0..n);

Mathematica

Table[2Binomial[3n,n]-Sum[Binomial[3n,k],{k,0,n}],{n,0,35}] (* Harvey P. Dale, Jul 27 2011 *)

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))

PARI

a(n)=if(n<0,0, 2*binomial(3*n,n)-sum(k=0,n,binomial(3*n,k)))

Formula

G.f. A(x)=y satisfies (8x-1)y^3-y^2+y+1=0. - Michael Somos, Jan 28 2004

Coefficient of x^n in ((1+10x-2x^2+(1-4x)^(3/2))/2)^n. - Michael Somos, Sep 25 2003

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

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

a(n) = A321957(n) + 2*binomial(3*n, n) - 8^n. - Peter Luschny, Nov 22 2018

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

Extensions

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).

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.