A054335 - OEIS

A054335

A convolution triangle of numbers based on A000984 (central binomial coefficients of even order).

1, 2, 1, 6, 4, 1, 20, 16, 6, 1, 70, 64, 30, 8, 1, 252, 256, 140, 48, 10, 1, 924, 1024, 630, 256, 70, 12, 1, 3432, 4096, 2772, 1280, 420, 96, 14, 1, 12870, 16384, 12012, 6144, 2310, 640, 126, 16, 1, 48620, 65536, 51480, 28672, 12012, 3840, 924, 160, 18, 1

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. The g.f. for the row polynomials p(n,x) (increasing powers of x) is 1/(sqrt(1-4*z)-x*z).

The column sequences are for m=0..20: A000984, A000302 (powers of 4), A002457, A002697, A002802, A038845, A020918, A038846, A020920, A040075, A020922, A045543, A020924, A054337, A020926, A054338, A020928, A054339, A020930, A054340, A020932.

Riordan array (1/sqrt(1-4x),x/sqrt(1-4x)). - Paul Barry, May 06 2009

The matrix inverse is apparently given by deleting the leftmost column from A206022. - R. J. Mathar, Mar 12 2013

LINKS

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.

Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.

FORMULA

a(n, 2*k+1) = binomial(n-k-1, k)*4^(n-2*k-1), a(n, 2*k) = binomial(2*(n-k), n-k)*binomial(n-k, k)/binomial(2*k, k), k >= 0, n >= m >= 0; a(n, m) := 0 if n<m.

Column recursion: a(n, m)=2*(2*n-m-1)*a(n-1, m)/(n-m), n>m >= 0, a(m, m) := 1.

G.f. for column m: cbie(x)*((x*cbie(x))^m, with cbie(x) := 1/sqrt(1-4*x).

G.f.: 1/(1-xy-2x/(1-x/(1-x/(1-x/(1-x/(1-... (continued fraction). - Paul Barry, May 06 2009

Sum_{k>=0} T(n,2k)*(-1)^k*A000108(k) = A000108(n+1). - Philippe Deléham, Jan 30 2012

Sum_{k=0..floor(n/2)} T(n-k,n-2k) = A098615(n). - Philippe Deléham, Feb 01 2012

T(n,k) = 4*T(n-1,k) + T(n-2,k-2) for k>=1. - Philippe Deléham, Feb 02 2012

Vertical recurrence: T(n,k) = 1*T(n-1,k-1) + 2*T(n-2,k-1) + 6*T(n-3,k-1) + 20*T(n-4,k-1) + ... for k >= 1 (the coefficients 1, 2, 6, 20, ... are the central binomial coefficients A000984). - Peter Bala, Oct 17 2015

EXAMPLE

Triangle begins:

2, 1;

6, 4, 1;

20, 16, 6, 1;

70, 64, 30, 8, 1;

252, 256, 140, 48, 10, 1;

924, 1024, 630, 256, 70, 12, 1; ...

Fourth row polynomial (n=3): p(3,x) = 20 + 16*x + 6*x^2 + x^3.

From Paul Barry, May 06 2009: (Start)

Production matrix begins

2, 1;

2, 2, 1;

0, 2, 2, 1;

-2, 0, 2, 2, 1;

0, -2, 0, 2, 2, 1;

4, 0, -2, 0, 2, 2, 1;

0, 4, 0, -2, 0, 2, 2, 1;

-10, 0, 4, 0, -2, 0, 2, 2, 1;

0, -10, 0, 4, 0, -2, 0, 2, 2, 1; (End)

MAPLE

A054335 := proc(n, k)

if k <0 or k > n then

0 ;

elif type(k, odd) then

kprime := floor(k/2) ;

binomial(n-kprime-1, kprime)*4^(n-k) ;

else

kprime := k/2 ;

binomial(2*n-k, n-kprime)*binomial(n-kprime, kprime)/binomial(k, kprime) ;

end if;

end proc: # R. J. Mathar, Mar 12 2013

# Uses function PMatrix from A357368. Adds column 1, 0, 0, 0, ... to the left.

PMatrix(10, n -> binomial(2*(n-1), n-1)); # Peter Luschny, Oct 19 2022

MATHEMATICA

Flatten[ CoefficientList[#1, x] & /@ CoefficientList[ Series[1/(Sqrt[1 - 4*z] - x*z), {z, 0, 9}], z]] (* or *)

a[n_, k_?OddQ] := 4^(n-k)*Binomial[(2*n-k-1)/2, (k-1)/2]; a[n_, k_?EvenQ] := (Binomial[n-k/2, k/2]*Binomial[2*n-k, n-k/2])/Binomial[k, k/2]; Table[a[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 08 2011, updated Jan 16 2014 *)

PROG

(PARI) T(n, k) = if(k%2==0, binomial(2*n-k, n-k/2)*binomial(n-k/2, k/2)/binomial(k, k/2), 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2));

for(n=0, 10, for(k=0, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Jul 20 2019

(Magma)

T:= func< n, k | (k mod 2) eq 0 select Binomial(2*n-k, n-Floor(k/2))* Binomial(n-Floor(k/2), Floor(k/2))/Binomial(k, Floor(k/2)) else 4^(n-k)*Binomial(n-Floor((k-1)/2)-1, Floor((k-1)/2)) >;

[[T(n, k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jul 20 2019

(Sage)

def T(n, k):

if (mod(k, 2)==0): return binomial(2*n-k, n-k/2)*binomial(n-k/2, k/2)/binomial(k, k/2)

else: return 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2)

[[T(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Jul 20 2019

(GAP)

T:= function(n, k)

if k mod 2=0 then return Binomial(2*n-k, n-Int(k/2))*Binomial(n-Int(k/2), Int(k/2))/Binomial(k, Int(k/2));

else return 4^(n-k)*Binomial(n-Int((k-1)/2)-1, Int((k-1)/2));

fi;

end;