A078812 Triangle read by rows: T(n, k) = binomial(n+k-1, 2*k-1).

1, 2, 1, 3, 4, 1, 4, 10, 6, 1, 5, 20, 21, 8, 1, 6, 35, 56, 36, 10, 1, 7, 56, 126, 120, 55, 12, 1, 8, 84, 252, 330, 220, 78, 14, 1, 9, 120, 462, 792, 715, 364, 105, 16, 1, 10, 165, 792, 1716, 2002, 1365, 560, 136, 18, 1, 11, 220, 1287, 3432, 5005, 4368, 2380, 816, 171, 20, 1

Offset: 0

Author

Michael Somos, Dec 05 2002

Keywords

• easy
• nice
• nonn
• tabl

Comments

Warning: formulas and programs sometimes refer to offset 0 and sometimes to offset 1.

Apart from signs, identical to A053122.

Coefficient array for Morgan-Voyce polynomial B(n,x); see A085478 for references. - Philippe Deléham, Feb 16 2004

T(n,k) is the number of compositions of n having k parts when there are q kinds of part q (q=1,2,...). Example: T(4,2) = 10 because we have (1,3),(1,3'),(1,3"), (3,1),(3',1),(3",1),(2,2),(2,2'),(2',2) and (2',2'). - Emeric Deutsch, Apr 09 2005

T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k (height(alpha) = |Im(alpha)|). - Abdullahi Umar, Oct 02 2008

This sequence is jointly generated with A085478 as a triangular array of coefficients of polynomials v(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = u(n-1,x) + x*v(n-1)x and v(n,x) = u(n-1,x) + (x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012

Concerning Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016

Subtriangle of the triangle T(n,k), 0 <= k <= n, read by rows, given by (0, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 27 2012

From Wolfdieter Lang, Aug 30 2012: (Start)

With offset [0,0] the triangle with entries R(n,k) = T(n+1,k+1):= binomial(n+k+1, 2*k+1), n >= k >= 0, and zero otherwise, becomes the Riordan lower triangular convolution matrix R = (G(x)/x, G(x)) with G(x):=x/(1-x)^2 (o.g.f. of A000027). This means that the o.g.f. of column number k of R is (G(x)^(k+1))/x. This matrix R is the inverse of the signed Riordan lower triangular matrix A039598, called in a comment there S.

The Riordan matrix with entries R(n,k), just defined, provides the transition matrix between the sequence entry F(4*m*(n+1))/L(2*l), with m >= 0, for n=0,1,... and the sequence entries 5^k*F(2*m)^(2*k+1) for k = 0,1,...,n, with F=A000045 (Fibonacci) and L=A000032 (Lucas). Proof: from the inverse of the signed triangle Riordan matrix S used in a comment on A039598.

For the transition matrix R (T with offset [0,0]) defined above, row n=2: F(12*m) /L(2*m) = 3*5^0*F(2*m)^1 + 4*5^1*F(2*m)^3 + 1*5^2*F(2*m)^5, m >= 0. (End)

From R. Bagula's comment in A053122 (cf. Damianou link p. 10), this array gives the coefficients (mod sign) of the characteristic polynomials for the Cartan matrix of the root system A_n. - Tom Copeland, Oct 11 2014

For 1 <= k <= n, T(n,k) equals the number of (n-1)-length ternary words containing k-1 letters equal 2 and avoiding 01. - Milan Janjic, Dec 20 2016

The infinite sum (Sum_{i >= 0} (T(s+i,1+i) / 2^(s+2*i)) * zeta(s+1+2*i)) = 1 allows any zeta(s+1) to be expressed as a sum of rational multiples of zeta(s+1+2*i) having higher arguments. For example, zeta(3) can be expressed as a sum involving zeta(5), zeta(7), etc. The summation for each s >= 1 uses the s-th diagonal of the triangle. - Robert B Fowler, Feb 23 2022

The convolution triangle of the nonnegative integers. - Peter Luschny, Oct 07 2022

Examples

			Triangle begins, 1 <= k <= n:
                          1
                        2   1
                      3   4   1
                    4  10   6   1
                  5  20  21   8   1
                6  35  56  36  10   1
              7  56 126 120  55  12   1
            8  84 252 330 220  78  14   1
From _Peter Bala_, Feb 11 2025: (Start)
The array factorizes as an infinite product of lower triangular arrays:
  / 1               \    / 1              \ / 1              \ / 1             \
  | 2    1           |   | 2   1          | | 0  1           | | 0  1          |
  | 3    4   1       | = | 3   2   1      | | 0  2   1       | | 0  0  1       | ...
  | 4   10   6   1   |   | 4   3   2  1   | | 0  3   2  1    | | 0  0  2  1    |
  | 5   20  21   8  1|   | 5   4   3  2  1| | 0  4   3  2  1 | | 0  0  3  2  1 |
  |...               |   |...             | |...             | |...            |
Cf. A092276. (End)
		

Links

• T. D. Noe, Rows n = 0..50 of triangle, flattened
• J.-P. Allouche and M. Mendès France, Stern-Brocot polynomials and power series, arXiv preprint arXiv:1202.0211 [math.NT], 2012. - From _N. J. A. Sloane_, May 10 2012
• Peter Bala, Factorisations of some Riordan arrays as infinite products
• Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
• Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.
• Paul Barry, Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers, arXiv:2011.10827 [math.CO], 2020.
• P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv preprint arXiv:1110.6620 [math.RT], 2014.
• Nour-Eddine Fahssi, On the combinatorics of exclusion in Haldane fractional statistics, arXiv:1808.00045 [cond-mat.stat-mech], 2018.
• Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.
• A. Laradji, and A. Umar, Combinatorial results for semigroups of order-preserving full transformations, Semigroup Forum 72 (2006), 51-62.
• Yidong Sun, Numerical triangles and several classical sequences, Fib. Quart. 43, no. 4, (2005) 359-370.

Crossrefs

This triangle is formed from odd-numbered rows of triangle A011973 read in reverse order.

Row sums give A001906. With signs: A053122.

The column sequences are A000027, A000292, A000389, A000580, A000582, A001288 for k=1..6, resp. For k=7..24 they are A010966..(+2)..A011000 and for k=25..50 they are A017713..(+2)..A017763.

Cf. A128908, A053123, A049310, A119900, A085478, A029653, A106195.

Programs

GAP

Flat(List([0..12], n-> List([0..n], k-> Binomial(n+k+1, 2*k+1) ))); # G. C. Greubel, Aug 01 2019

Haskell

a078812 n k = a078812_tabl !! n !! k
a078812_row n = a078812_tabl !! n
a078812_tabl = [1] : [2, 1] : f [1] [2, 1] where
   f us vs = ws : f vs ws where
     ws = zipWith (-) (zipWith (+) ([0] ++ vs) (map (* 2) vs ++ [0]))
                      (us ++ [0, 0])
-- Reinhard Zumkeller, Dec 16 2013

Magma

/* As triangle */ [[Binomial(n+k-1, 2*k-1): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Jun 01 2018

Maple

for n from 1 to 11 do seq(binomial(n+k-1,2*k-1),k=1..n) od; # yields sequence in triangular form; Emeric Deutsch, Apr 09 2005
# Uses function PMatrix from A357368. Adds a row and column above and to the left.
PMatrix(10, n -> n); # Peter Luschny, Oct 07 2022

Mathematica

(* First program *)
u[1, x_]:= 1; v[1, x_]:= 1; z = 13;
u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%] (* A085478 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%] (* A078812 *) (* Clark Kimberling, Feb 25 2012 *)
(* Second program *)
Table[Binomial[n+k+1, 2*k+1], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 01 2019 *)

Maxima

T(n,m):=sum(binomial(2*k,n-m)*binomial(m+k,k)*(-1)^(n-m+k)*binomial(n+1,m+k+1),k,0,n-m); /* Vladimir Kruchinin, Apr 13 2016 */

PARI

{T(n, k) = if( n<0, 0, binomial(n+k-1, 2*k-1))};

PARI

{T(n, k) = polcoeff( polcoeff( x*y / (1 - (2 + y) * x + x^2) + x * O(x^n), n), k)};

Sage

@cached_function
def T(k,n):
    if k==n: return 1
    if k==0: return 0
    return sum(i*T(k-1,n-i) for i in (1..n-k+1))
A078812 = lambda n,k: T(k,n)
[[A078812(n,k) for k in (1..n)] for n in (1..8)] # Peter Luschny, Mar 12 2016

Sage

[[binomial(n+k+1, 2*k+1) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019

Formula

G.f.: x*y / (1 - (2 + y)*x + x^2). To get row n, expand this in powers of x then expand the coefficient of x^n in increasing powers of y.

From Philippe Deléham, Feb 16 2004: (Start)

If indexing begins at 0 we have

T(n,k) = (n+k+1)!/((n-k)!*(2k+1))!.

T(n,k) = Sum_{j>=0} T(n-1-j, k-1)*(j+1) with T(n, 0) = n+1, T(n, k) = 0 if n < k.

T(n,k) = T(n-1, k-1) + T(n-1, k) + Sum_{j>=0} (-1)^j*T(n-1, k+j)*A000108(j) with T(n,k) = 0 if k < 0, T(0, 0)=1 and T(0, k) = 0 for k > 0.

G.f. for the column k: Sum_{n>=0} T(n, k)*x^n = (x^k)/(1-x)^(2k+2).

Row sums: Sum_{k>=0} T(n, k) = A001906(n+1). (End)

Antidiagonal sums are A000079(n) = Sum_{k=0..floor(n/2)} binomial(n+k+1, n-k). - Paul Barry, Jun 21 2004

Riordan array (1/(1-x)^2, x/(1-x)^2). - Paul Barry, Oct 22 2006

T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n, T(n,k) = T(n-1,k-1) + 2*T(n-1,k) - T(n-2,k). - Philippe Deléham, Jan 26 2010

For another version see A128908. - Philippe Deléham, Mar 27 2012

T(n,m) = Sum_{k=0..n-m} (binomial(2*k,n-m)*binomial(m+k,k)*(-1)^(n-m+k)* binomial(n+1,m+k+1)). - Vladimir Kruchinin, Apr 13 2016

T(n, k) = T(n-1, k) + (T(n-1, k-1) + T(n-2, k-1) + T(n-3, k-1) + ...) for k >= 2 with T(n, 1) = n. - Peter Bala, Feb 11 2025

From Peter Bala, May 04 2025: (Start)

With the column offset starting at 0, the n-th row polynomial B(n, x) = 1/sqrt(x + 4) * Chebyshev_U(2*n+1, (1/2)*sqrt(x + 4)) = (-1)^n * Chebyshev_U(n, -(1/2)*(x + 2)).

B(n, x) / Product_{k = 1..2*n} (1 + 1/B(k, x)) = b(n, x), the n-th row polynomial of A085478. (End)

Extensions

Edited by N. J. A. Sloane, Apr 28 2008