A010966 -id:A010966 - cp's OEIS frontend

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

Michael Somos, Dec 05 2002

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

			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)

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.

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.

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

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

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

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

(* 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 *)

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 */

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

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

@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

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

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)

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

A053122 Triangle of coefficients of Chebyshev's S(n,x-2) = U(n,x/2-1) polynomials (exponents of x in increasing order).

Original entry on oeis.org

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

Offset: 0

Wolfdieter Lang

Apart from signs, identical to A078812.
Another version with row-leading 0's and differing signs is given by A285072.
G.f. for row polynomials S(n,x-2) (signed triangle): 1/(1+(2-x)*z+z^2). Unsigned triangle |a(n,m)| has g.f. 1/(1-(2+x)*z+z^2) for row polynomials.
Row sums (signed triangle) A049347(n) (periodic(1,-1,0)). Row sums (unsigned triangle) A001906(n+1)=F(2*(n+1)) (even-indexed Fibonacci).
In the language of Shapiro et al. (see A053121 for the reference) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group.
The (unsigned) column sequences are A000027, A000292, A000389, A000580, A000582, A001288 for m=0..5, resp. For m=6..23 they are A010966..(+2)..A011000 and for m=24..49 they are A017713..(+2)..A017763.
Riordan array (1/(1+x)^2,x/(1+x)^2). Inverse array is A039598. Diagonal sums have g.f. 1/(1+x^2). - Paul Barry, Mar 17 2005. Corrected by Wolfdieter Lang, Nov 13 2012.
Unsigned version is in A078812. - Philippe Deléham, Nov 05 2006
Also row n gives (except for an overall sign) coefficients of characteristic polynomial of the Cartan matrix for the root system A_n. - Roger L. Bagula, May 23 2007
From Wolfdieter Lang, Nov 13 2012: (Start)
The A-sequence for this Riordan triangle is A115141, and the Z-sequence is A115141(n+1), n>=0. For A- and Z-sequences for Riordan matrices see the W. Lang link under A006232 with details and references.
S(n,x^2-2) = sum(r(j,x^2),j=0..n) with Chebyshev's S-polynomials and r(j,x^2) := R(2*j+1,x)/x, where R(n,x) are the monic integer Chebyshv T-polynomials with coefficients given in A127672. Proof from comparing the o.g.f. of the partial sum of the r(j,x^2) polynomials (see a comment on the signed Riordan triangle A111125) with the present Riordan type o.g.f. for the row polynomials with x -> x^2.  (End)
S(n,x^2-2) = S(2*n+1,x)/x, n >= 0, from the odd part of the bisection of the o.g.f. - Wolfdieter Lang, Dec 17 2012
For a relation to a generator for the Narayana numbers A001263, see A119900, whose columns are unsigned shifted rows (or antidiagonals) of this array, referring to the tables in the example sections. - Tom Copeland, Oct 29 2014
The unsigned rows of this array are alternating rows of a mirrored A011973 and alternating shifted rows of A030528 for the Fibonacci polynomials. - Tom Copeland, Nov 04 2014
Boas-Buck type recurrence for column k >= 0 (see Aug 10 2017 comment in A046521 with references): a(n, m) =  (2*(m + 1)/(n - m))*Sum_{k = m..n-1} (-1)^(n-k)*a(k, m), with input a(n, n) = 1, and a(n,k) = 0 for n < k. - Wolfdieter Lang, Jun 03 2020
Row n gives the characteristic polynomial of the (n X n)-matrix M where M[i,j] = 2 if i = j, -1 if |i-j| = 1 and 0 otherwise. The matrix M is positive definite and has 2-condition number (cot(Pi/(2*n+2)))^2. - Jianing Song, Jun 21 2022
Also the convolution triangle of (-1)^(n+1)*n. - Peter Luschny, Oct 07 2022

			The triangle a(n,m) begins:
n\m   0    1    2     3     4     5     6    7    8  9
0:    1
1:   -2    1
2:    3   -4    1
3:   -4   10   -6     1
4:    5  -20   21    -8     1
5:   -6   35  -56    36   -10     1
6:    7  -56  126  -120    55   -12     1
7:   -8   84 -252   330  -220    78   -14    1
8:    9 -120  462  -792   715  -364   105  -16    1
9:  -10  165 -792  1716 -2002  1365  -560  136  -18  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 13 2012
E.g., fourth row (n=3) {-4,10,-6,1} corresponds to the polynomial S(3,x-2) = -4+10*x-6*x^2+x^3.
From _Wolfdieter Lang_, Nov 13 2012: (Start)
Recurrence: a(5,1) = 35 = 1*5 + (-2)*(-20) -1*(10).
Recurrence from Z-sequence [-2,-1,-2,-5,...]: a(5,0) = -6 = (-2)*5 + (-1)*(-20) + (-2)*21 + (-5)*(-8) + (-14)*1.
Recurrence from A-sequence [1,-2,-1,-2,-5,...]: a(5,1) = 35 = 1*5  + (-2)*(-20) + (-1)*21 + (-2)*(-8) + (-5)*1.
(End)
E.g., the fourth row (n=3) {-4,10,-6,1} corresponds also to the polynomial S(7,x)/x = -4 + 10*x^2 - 6*x^4 + x^6. - _Wolfdieter Lang_, Dec 17 2012
Boas-Buck type recurrence: -56 = a(5, 2) = 2*(-1*1 + 1*(-6) - 1*21) = -2*28 = -56. - _Wolfdieter Lang_, Jun 03 2020

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.
R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 62.
Sigurdur Helgasson, Differential Geometry, Lie Groups and Symmetric Spaces, Graduate Studies in Mathematics, volume 34. A. M. S.: ISBN 0-8218-2848-7, 1978, p. 463.

T. D. Noe, Rows n=0..50 of triangle, flattened
Wolfdieter Lang, First rows of the triangle.
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
P. Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Naiomi T. Cameron and Asamoah Nkwanta, On some (pseudo) involutions in the Riordan group, J. of Integer Sequences, 8(2005), 1-16.
P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv preprint arXiv:1110.6620 [math.RT], 2014 (p. 10). - _Tom Copeland_, Oct 11 2014
Pentti Haukkanen, Jorma Merikoski, Seppo Mustonen, Some polynomials associated with regular polygons, Acta Univ. Sapientiae, Mathematica, 6, 2 (2014) 178-193.
Eric Weisstein's World of Mathematics, Cartan Matrix
Eric Weisstein's World of Mathematics, Dynkin Diagram
Index entries for sequences related to Chebyshev polynomials.

Cf. A005248, A127677, A053123, A049310.
Cf. A011973, A030528, A034867, A119900.
Cf. A285072 (version with row-leading 0's and differing signs). - Eric W. Weisstein, Apr 09 2017

seq(seq((-1)^(n+m)*binomial(n+m+1,2*m+1),m=0..n),n=0..10); # Robert Israel, Oct 15 2014
# Uses function PMatrix from A357368. Adds a row above and a column to the left.
PMatrix(10, n -> -(-1)^n*n); # Peter Luschny, Oct 07 2022

T[n_, m_, d_] := If[ n == m, 2, If[n == m - 1 || n == m + 1, -1, 0]]; M[d_] := Table[T[n, m, d], {n, 1, d}, {m, 1, d}]; a = Join[M[1], Table[CoefficientList[Det[M[d] - x*IdentityMatrix[d]], x], {d, 1, 10}]]; Flatten[a] (* Roger L. Bagula, May 23 2007 *)
(* Alternative code for the matrices from MathWorld: *)
sln[n_] := 2IdentityMatrix[n] - PadLeft[PadRight[IdentityMatrix[n - 1], {n, n - 1}], {n, n}] - PadLeft[PadRight[IdentityMatrix[n - 1], {n - 1, n}], {n, n}] (* Roger L. Bagula, May 23 2007 *)

@CachedFunction
def A053122(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    return A053122(n-1,k-1)-A053122(n-2,k)-2*A053122(n-1,k)
for n in (0..9): [A053122(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

a(n, m) := 0 if n

a(n, m) = -2*a(n-1, m) + a(n-1, m-1) - a(n-2, m), a(n, -1) := 0 =: a(-1, m), a(0, 0)=1, a(n, m) := 0 if n

O.g.f. for m-th column (signed triangle): ((x/(1+x)^2)^m)/(1+x)^2.

From Jianing Song, Jun 21 2022: (Start)

T(n,k) = [x^k]f_n(x), where f_{-1}(x) = 0, f_0(x) = 1, f_n(x) = (x-2)*f_{n-1}(x) - f_{n-2}(x) for n >= 2.

f_n(x) = (((x-2+sqrt(x^2-4*x))/2)^(n+1) - ((x-2-sqrt(x^2-4*x))/2)^(n+1))/sqrt(x^2-4x).

The roots of f_n(x) are 2 + 2*cos(k*Pi/(n+1)) = 4*(cos(k*Pi/(2*n+2)))^2 for 1 <= k <= n. (End)

A110555 Triangle of partial sums of alternating binomial coefficients: T(n, k) = Sum_{j=0..k} binomial(n, j)*(-1)^j, for n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, -1, 0, 1, -2, 1, 0, 1, -3, 3, -1, 0, 1, -4, 6, -4, 1, 0, 1, -5, 10, -10, 5, -1, 0, 1, -6, 15, -20, 15, -6, 1, 0, 1, -7, 21, -35, 35, -21, 7, -1, 0, 1, -8, 28, -56, 70, -56, 28, -8, 1, 0, 1, -9, 36, -84, 126, -126, 84, -36, 9, -1, 0, 1, -10, 45, -120, 210, -252, 210, -120

Offset: 0

Reinhard Zumkeller, Jul 27 2005

			Triangle T(n, k) starts:
  [0] 1;
  [1] 1,  0;
  [2] 1, -1,  0;
  [3] 1, -2,  1,   0;
  [4] 1, -3,  3,  -1,  0;
  [5] 1, -4,  6,  -4,  1,   0;
  [6] 1, -5, 10, -10,  5,  -1,  0;
  [7] 1, -6, 15, -20, 15,  -6,  1,  0;
  [8] 1, -7, 21, -35, 35, -21,  7, -1,  0.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Ângela Mestre and José Agapito, A Family of Riordan Group Automorphisms, J. Int. Seq., Vol. 22 (2019), Article 19.8.5.
Index entries for triangles and arrays related to Pascal's triangle

T(n,1)  = -n + 1 for n>0;
T(n,2)  =  A000217(n-2) for n > 1;
T(n,3)  = -A000292(n-4) for n > 2;
T(n,4)  =  A000332(n-1) for n > 3;
T(n,5)  = -A000389(n-1) for n > 5;
T(n,6)  =  A000579(n-1) for n > 6;
T(n,7)  = -A000580(n-1) for n > 7;
T(n,8)  =  A000581(n-1) for n > 8;
T(n,9)  = -A000582(n-1) for n > 9;
T(n,10) =  A001287(n-1) for n > 10;
T(n,11) = -A001288(n-1) for n > 11;
T(n,12) =  A010965(n-1) for n > 12;
T(n,13) = -A010966(n-1) for n > 13;
T(n,14) =  A010967(n-1) for n > 14;
T(n,15) = -A010968(n-1) for n > 15;
T(n,16) =  A010969(n-1) for n > 16.
Cf. A071919 (variant), A000007 (row sums), A110556 (central terms).
Cf. A008949, A007318.

T := (n, k) -> (-1)^k * binomial(n-1, k):
seq(print(seq(T(n, k), k = 0..n)), n = 0..7); # Peter Luschny, Apr 13 2023

T[0, 0] := 1;  T[n_, n_] := 0; T[n_, k_] := (-1)^k*Binomial[n - 1, k]; Table[T[n, k], {n, 0, 20}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 31 2017 *)

concat(1, for(n=1,10, for(k=0,n, print1(if(k != n, (-1)^k*binomial(n-1,k), 0), ", ")))) \\ G. C. Greubel, Aug 31 2017

T(n, 0) = 1, T(n, n) = 0^n, T(n, k) = -T(n-1, k-1) + T(n-1, k), for 0 < k < n.
T(n, k) = binomial(n-1, k)*(-1)^k, 0 <= k < n, T(n, n) = 0^n.
T(n, n-k-1) = -T(n, k), for 0 < k < n.
T(n, k) = A071919(n, k)*(-1)^k and A071919(n, k) = abs(T(n, k)).
Triangle T(n,k), 0 <= k <= n, read by rows, given by [1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [0, -1, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 05 2005
G.f.: (1 + x*y) / (1 + x*y - x). - R. J. Mathar, Aug 11 2015

Typo in name corrected by Andrey Zabolotskiy, Feb 22 2022

Offset corrected by Peter Luschny, Apr 13 2023

A010965 a(n) = binomial(n,12).

Original entry on oeis.org

1, 13, 91, 455, 1820, 6188, 18564, 50388, 125970, 293930, 646646, 1352078, 2704156, 5200300, 9657700, 17383860, 30421755, 51895935, 86493225, 141120525, 225792840, 354817320, 548354040, 834451800, 1251677700, 1852482996, 2707475148, 3910797436, 5586853480

Offset: 12

N. J. A. Sloane

nonn

Coordination sequence for 12-dimensional cyclotomic lattice Z[zeta_13].

In this sequence only 13 is prime. - Artur Jasinski, Dec 02 2007

T. D. Noe, Table of n, a(n) for n = 12..1000
Matthias Beck and Serkan Hosten, Cyclotomic polytopes and growth series of cyclotomic lattices, arXiv:math/0508136 [math.CO], 2005-2006.
Index entries for linear recurrences with constant coefficients, signature (13, -78, 286, -715, 1287, -1716, 1716, -1287, 715, -286, 78, -13, 1).

Cf. A000581, A010966, A010967, A001787, A242091.

[Binomial(n, 12): n in [12..100]]; // Vincenzo Librandi, Apr 22 2011

seq(binomial(n,12),n=12..36); # Zerinvary Lajos, Aug 06 2008

Table[Binomial[n,12],{n,12,50}] (* Vladimir Joseph Stephan Orlovsky, Apr 22 2011 *)

for(n=12, 50, print1(binomial(n,12), ", ")) \\ G. C. Greubel, Aug 31 2017

a(n) = A110555(n+1,12). - Reinhard Zumkeller, Jul 27 2005
a(n+11) = n(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7)(n+8)(n+9)(n+10)(n+11)/12!. - Artur Jasinski, Dec 02 2007, R. J. Mathar, Jul 07 2009
G.f.: x^12/(1-x)^13. - Zerinvary Lajos, Aug 06 2008, R. J. Mathar, Jul 07 2009
From Amiram Eldar, Dec 10 2020: (Start)
Sum_{n>=12} 1/a(n) = 12/11.
Sum_{n>=12} (-1)^n/a(n) = A001787(12)*log(2) - A242091(12)/11! = 24576*log(2) - 3934820/231 = 0.9322955884... (End)

Some formulas referring to other offsets corrected by R. J. Mathar, Jul 07 2009

A128908 Riordan array (1, x/(1-x)^2).

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 3, 4, 1, 0, 4, 10, 6, 1, 0, 5, 20, 21, 8, 1, 0, 6, 35, 56, 36, 10, 1, 0, 7, 56, 126, 120, 55, 12, 1, 0, 8, 84, 252, 330, 220, 78, 14, 1, 0, 9, 120, 462, 792, 715, 364, 105, 16, 1, 0, 10, 165, 792, 1716, 2002, 1365, 560, 136, 18, 1

Offset: 0

Philippe Deléham, Apr 22 2007

Triangle T(n,k), 0 <= k <= n, read by rows given by [0,2,-1/2,1/2,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.
Row sums give A088305. - Philippe Deléham, Nov 21 2007
Column k is C(n,2k-1) for k > 0. - Philippe Deléham, Jan 20 2012
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
T is the convolution triangle of the positive integers (see A357368). - Peter Luschny, Oct 19 2022

			The triangle T(n,k) begins:
   n\k  0    1    2    3    4    5    6    7    8    9   10
   0:   1
   1:   0    1
   2:   0    2    1
   3:   0    3    4    1
   4:   0    4   10    6    1
   5:   0    5   20   21    8    1
   6:   0    6   35   56   36   10    1
   7:   0    7   56  126  120   55   12    1
   8:   0    8   84  252  330  220   78   14    1
   9:   0    9  120  462  792  715  364  105   16    1
  10:   0   10  165  792 1716 2002 1365  560  136   18    1
  ... reformatted by _Wolfdieter Lang_, Jul 31 2017
From _Peter Luschny_, Mar 06 2022: (Start)
The sequence can also be seen as a square array read by upwards antidiagonals.
   1, 1,   1,    1,    1,     1,     1,      1,      1, ...  A000012
   0, 2,   4,    6,    8,    10,    12,     14,     16, ...  A005843
   0, 3,  10,   21,   36,    55,    78,    105,    136, ...  A014105
   0, 4,  20,   56,  120,   220,   364,    560,    816, ...  A002492
   0, 5,  35,  126,  330,   715,  1365,   2380,   3876, ... (A053126)
   0, 6,  56,  252,  792,  2002,  4368,   8568,  15504, ... (A053127)
   0, 7,  84,  462, 1716,  5005, 12376,  27132,  54264, ... (A053128)
   0, 8, 120,  792, 3432, 11440, 31824,  77520, 170544, ... (A053129)
   0, 9, 165, 1287, 6435, 24310, 75582, 203490, 490314, ... (A053130)
    A27,A292, A389, A580,  A582, A1288, A10966, A10968, A165817       (End)

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened
P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv:1110.6620 [math.RT], 2014.

Cf. A002450, A007318, A034008, A053122, A078812, A084938, A088305.
Cf. Columns : A000007, A000027, A000292, A000389, A000580, A000582, A001288, A010966 ..(+2).. A011000, A017713 ..(+2).. A017763.
Cf. A000007, A001352, A008574, A054888, A084099, A084103, A163810, A357368.
Cf. A165817 (the main diagonal of the array).

# Computing the rows of the array representation:
S := proc(n,k) option remember;
if n = k then 1 elif k < 0 or k > n then 0 else
S(n-1, k-1) + 2*S(n-1, k) - S(n-2, k) fi end:
Arow := (n, len) -> seq(S(n+k-1, k-1), k = 0..len-1):
for n from 0 to 8 do Arow(n, 9) od; # Peter Luschny, Mar 06 2022
# Uses function PMatrix from A357368.
PMatrix(10, n -> n); # Peter Luschny, Oct 19 2022

With[{nmax = 10}, CoefficientList[CoefficientList[Series[(1 - x)^2/(1 - (2 + y)*x + x^2), {x, 0, nmax}, {y, 0, nmax}], x], y]] // Flatten (* G. C. Greubel, Nov 22 2017 *)

for(n=0,10, for(k=0,n, print1(if(n==0 && k==0, 1, if(k==0, 0, binomial(n+k-1,2*k-1))), ", "))) \\ G. C. Greubel, Nov 22 2017

from functools import cache
@cache
def A128908(n, k):
    if n == k: return 1
    if (k <= 0 or k > n): return 0
    return A128908(n-1, k-1) + 2*A128908(n-1, k) - A128908(n-2, k)
for n in range(10):
    print([A128908(n, k) for k in range(n+1)]) # Peter Luschny, Mar 07 2022

@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))
A128908 = lambda n,k: T(k,n)
for n in (0..10): print([A128908(n,k) for k in (0..n)]) # Peter Luschny, Mar 12 2016

T(n,0) = 0^n, T(n,k) = binomial(n+k-1, 2k-1) for k >= 1.
Sum_{k=0..n} T(n,k)*2^(n-k) = A002450(n) = (4^n-1)/3 for n>=1. - Philippe Deléham, Oct 19 2008
G.f.: (1-x)^2/(1-(2+y)*x+x^2). - Philippe Deléham, Jan 20 2012
Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A001352(n), (-1)^(n+1)*A054888(n+1), (-1)^n*A008574(n), (-1)^n*A084103(n), (-1)^n*A084099(n), A163810(n), A000007(n), A088305(n) for x = -6, -5, -4, -3, -2, -1, 0, 1 respectively. - Philippe Deléham, Jan 20 2012
Riordan array (1, x/(1-x)^2). - Philippe Deléham, Jan 20 2012

A238801 Triangle T(n,k), read by rows, given by T(n,k) = C(n+1, k+1)*(1-(k mod 2)).

Original entry on oeis.org

1, 2, 0, 3, 0, 1, 4, 0, 4, 0, 5, 0, 10, 0, 1, 6, 0, 20, 0, 6, 0, 7, 0, 35, 0, 21, 0, 1, 8, 0, 56, 0, 56, 0, 8, 0, 9, 0, 84, 0, 126, 0, 36, 0, 1, 10, 0, 120, 0, 252, 0, 120, 0, 10, 0, 11, 0, 165, 0, 462, 0, 330, 0, 55, 0, 1, 12, 0, 220, 0, 792, 0, 792, 0, 220, 0, 12, 0

Offset: 0

Philippe Deléham, Mar 05 2014

Row sums are powers of 2.

			Triangle begins:
1;
2, 0;
3, 0, 1;
4, 0, 4, 0;
5, 0, 10, 0, 1;
6, 0, 20, 0, 6, 0;
7, 0, 35, 0, 21, 0, 1;
8, 0, 56, 0, 56, 0, 8, 0;
9, 0, 84, 0, 126, 0, 36, 0, 1;
10, 0, 120, 0, 252, 0, 120, 0, 10, 0; etc.

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Cf. Columns: A000027, A000292, A000389, A000580, A000582, A001288, A010966, A010968, A010970, A010972, A010974, A010976, A010980, A010982, A010984, A010986, A010988, A010990, A010992, A010994, A010996, A010998, A011000, A017713, A017715, A017717, A017719, A017721, A017723, A017725, A017727, A017729, A017731, A017733, A017735, A017737, A017739, A017741, A017743, A017745, A017747, A017749, A017751, A017753, A017755, A017757, A017759, A017761, A017763.

Cf. A095704, A178616.

Table[Binomial[n + 1, k + 1]*(1 - Mod[k , 2]), {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Nov 22 2017 *)

T(n,k) = binomial(n+1, k+1)*(1-(k % 2));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, Nov 23 2017

G.f.: 1/((1+(y-1)*x)*(1-(y+1)*x)).
T(n,k) = 2*T(n-1,k) + T(n-2,k-2) - T(n-2,k), T(0,0) = 1, T(1,0) = 2, T(1,1) = 0, T(n,k) = 0 if k<0 or if k>n.
Sum_{k=0..n} T(n,k)*x^k = A000027(n+1), A000079(n), A015518(n+1), A003683(n+1), A079773(n+1), A051958(n+1), A080920(n+1), A053455(n), A160958(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively.

A039948 A triangle related to A000045 (Fibonacci numbers).

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 18, 12, 3, 1, 120, 72, 24, 4, 1, 960, 600, 180, 40, 5, 1, 9360, 5760, 1800, 360, 60, 6, 1, 105840, 65520, 20160, 4200, 630, 84, 7, 1, 1370880, 846720, 262080, 53760, 8400, 1008, 112, 8, 1, 19958400, 12337920, 3810240, 786240, 120960, 15120, 1512, 144, 9, 1

Offset: 0

Wolfdieter Lang

			Triangle begins :
    1;
    1,   1;
    4,   2,   1;
   18,  12,   3,  1;
  120,  72,  24,  4, 1;
  960, 600, 180, 40, 5, 1;
... - _Philippe Deléham_, Nov 08 2011

Seiichi Manyama, Rows n = 0..139, flattened
P. R. J. Asveld, Fibonacci-like differential equations with a polynomial nonhomogeneous term, Fib. Quart. 27 (1989), 303-309.

Cf. A000045, A000142, A005442, A005443, A110313 (row sums).

Diagonals include: A000027, A000217, A000292, A000332, A000389, A000579, A000580, A000581, A000582, A001287, A001288, A010965, A010966.

[(Factorial(n)/Factorial(k))*Fibonacci(n-k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 20 2022

T[n_,k_]:= (n!/k!)*Fibonacci[n-k+1];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 20 2022 *)

def A039948(n, k): return factorial(n-k)*binomial(n,k)*fibonacci(n-k+1)
flatten([[A039948(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Nov 20 2022

T(n, m) = n!*Fibonacci(n-m+1)/m!, n >= m >= 0.
T(n, 0) = A005442(n).
T(n, 1) = A005443(n).
E.g.f. for column m: x^m/(m!*(1-x-x^2)), m >= 0.
From G. C. Greubel, Nov 20 2022: (Start)
T(n, n-1) = A000027(n).
T(n, n-2) = 4*A000217(n-1), n >= 2.
T(n, n-3) = 18*A000292(n-2), n >= 3.
T(n, n-4) = 5! * A000332(n), n >= 4.
T(n, n-5) = 8 * 5! * A000389(n), n >= 5.
T(n, n-6) = 13 * 6! * A000579(n), n >= 6.
T(n, n-7) = 21 * 7! * A000580(n), n >= 7.
T(n, n-8) = 34 * 8! * A000581(n), n >= 8.
T(n, n-9) = 55 * 9! * A000582(n), n >= 9.
T(n, n-10) = 89 * 10! * A001287(n), n >= 10.
T(n, n-11) = 12 * 12! * A001288(n), n >= 11.
T(n, n-12) = 233 * 12! * A010965(n), n >= 12.
T(n, n-13) = 89 * 13! * A010966(n), n >= 13.
Sum_{k=0..n} T(n, k) = A110313(n). (End)

A101095 Fourth difference of fifth powers (A000584).

Original entry on oeis.org

1, 28, 121, 240, 360, 480, 600, 720, 840, 960, 1080, 1200, 1320, 1440, 1560, 1680, 1800, 1920, 2040, 2160, 2280, 2400, 2520, 2640, 2760, 2880, 3000, 3120, 3240, 3360, 3480, 3600, 3720, 3840, 3960, 4080, 4200, 4320, 4440, 4560, 4680, 4800, 4920, 5040, 5160, 5280

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original Name: Shells (nexus numbers) of shells of shells of shells of the power of 5.

The (Worpitzky/Euler/Pascal Cube) "MagicNKZ" algorithm is: MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n, with k>=0, n>=1, z>=0. MagicNKZ is used to generate the n-th accumulation sequence of the z-th row of the Euler Triangle (A008292). For example, MagicNKZ(3,k,0) is the 3rd row of the Euler Triangle (followed by zeros) and MagicNKZ(10,k,1) is the partial sums of the 10th row of the Euler Triangle. This sequence is MagicNKZ(5,k-1,2).

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Archive Machine link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883
Eric Weisstein, Link to section of MathWorld: Eulerian Number
Eric Weisstein, Link to section of MathWorld: Nexus number
Eric Weisstein, Link to section of MathWorld: Finite Differences
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Fourth differences of A000584, third differences of A022521, second differences of A101098, and first differences of A101096.
For other sequences based upon MagicNKZ(n,k,z):
...... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7  |  n = 8
--------------------------------------------------------------------------------------
z =  0 | A000007 | A019590 | .......  MagicNKZ(n,k,0) = T(n,k+1) from A008292  .......
z =  1 | A000012 | A040000 | A101101 | A101104 | A101100 | ....... | ....... | .......
z =  2 | A000027 | A005408 | A008458 | A101103 | thisSeq | ....... | ....... | .......
z =  3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | ....... | .......
z =  4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | ....... | .......
z =  5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181 | .......
z =  6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177 | A255182
z =  7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523 | A255178
z =  8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015 | A022524
z =  9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541 | A001016
z = 10 | A000582 | A054333 | A254469 | A254681 | A254644 | A254640 | A250212 | A000542
z = 11 | A001287 | A054334 | A254869 | A254470 | A254682 | A254645 | A254641 | A253636
z = 12 | A001288 | A057788 | ....... | A254870 | A254471 | A254683 | A254646 | A254642
z = 13 | A010965 | ....... | ....... | ....... | A254871 | A254472 | A254684 | A254647
z = 14 | A010966 | ....... | ....... | ....... | ....... | A254872 | ....... | .......
--------------------------------------------------------------------------------------
Cf. A047969.

I:=[1,28,121,240,360]; [n le 5 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, May 07 2015

MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 5, 5}, {z, 2, 2}, {k, 0, 34}]
CoefficientList[Series[(1 + 26 x + 66 x^2 + 26 x^3 + x^4)/(1 - x)^2, {x, 0, 50}], x] (* Vincenzo Librandi, May 07 2015 *)
Join[{1,28,121,240},Differences[Range[50]^5,4]] (* or *) LinearRecurrence[{2,-1},{1,28,121,240,360},50] (* Harvey P. Dale, Jun 11 2016 *)

a(n)=if(n>3, 120*n-240, 33*n^2-72*n+40) \\ Charles R Greathouse IV, Oct 11 2015

[1,28,121]+[120*(k-2) for k in range(4,36)] # Danny Rorabaugh, Apr 23 2015

a(k+1) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n; n = 5, z = 2.
For k>3, a(k) = Sum_{j=0..4} (-1)^j*binomial(4, j)*(k - j)^5 = 120*(k - 2).
a(n) = 2*a(n-1) - a(n-2), n>5. G.f.: x*(1+26*x+66*x^2+26*x^3+x^4) / (1-x)^2. - Colin Barker, Mar 01 2012

MagicNKZ material edited, Crossrefs table added, SeriesAtLevelR material removed by Danny Rorabaugh, Apr 23 2015

Name changed and keyword 'uned' removed by Danny Rorabaugh, May 06 2015

A095704 Triangle read by rows giving coefficients of the trigonometric expansion of sin(n*x).

Original entry on oeis.org

1, 2, 0, 3, 0, -1, 4, 0, -4, 0, 5, 0, -10, 0, 1, 6, 0, -20, 0, 6, 0, 7, 0, -35, 0, 21, 0, -1, 8, 0, -56, 0, 56, 0, -8, 0, 9, 0, -84, 0, 126, 0, -36, 0, 1, 10, 0, -120, 0, 252, 0, -120, 0, 10, 0, 11, 0, -165, 0, 462, 0, -330, 0, 55, 0, -1, 12, 0, -220, 0, 792, 0, -792, 0, 220, 0, -12, 0, 13, 0, -286, 0, 1287, 0

Offset: 1

Robert G. Wilson v, Jul 06 2004

			The trigonometric expansion of sin(4x) is 4*cos(x)^3*sin(x) - 4*cos(x)*sin(x)^3, so the fourth row is 4, 0, -4, 0.
Triangle begins:
1
2 0
3 0 -1
4 0 -4 0
5 0 -10 0 1
6 0 -20 0 6 0
7 0 -35 0 21 0 -1
8 0 -56 0 56 0 -8 0

Clark Kimberling, Polynomials associated with reciprocation, JIS 12 (2009) 09.3.4, section 5.

First column is A000027 = C(n, 1), third column is A000292 = C(n, 3), fifth column is A000389 = C(n, 5), seventh column is A000580 = C(n, 7), ninth column is A000582 = C(n, 9).
A001288 = C(n, 11), A010966 = C(n, 13), A010968 = C(n, 15), A010970 = C(n, 17), A010972 = C(n, 19),
A010974 = C(n, 21), A010976 = C(n, 23), A010978 = C(n, 25), A010980 = C(n, 27), A010982 = C(n, 29),
A010984 = C(n, 31), A010986 = C(n, 33), A010988 = C(n, 35), A010990 = C(n, 37), A010992 = C(n, 39),
A010994 = C(n, 41), A010996 = C(n, 43), A010998 = C(n, 45), A011000 = C(n, 47), A017713 = C(n, 49)
Another version of the triangle in A034867. Cf. A096754.
A017715 = C(n, 51), A017717 = C(n, 53), A017719 = C(n, 55), A017721 = C(n, 57), etc.

Flatten[ Table[ Plus @@ CoefficientList[ TrigExpand[ Sin[n*x]], {Sin[x], Cos[x]}], {n, 13}]]

T(n,k) = C(n+1,k+1)*sin(Pi*(k+1)/2). - Paul Barry, May 21 2006

A127157 Triangle read by rows: T(n,k) is the number of ordered trees with n edges and 2k nodes of odd degree (not outdegree; 1 <= k <= ceiling(n/2)).

Original entry on oeis.org

1, 2, 3, 2, 4, 10, 5, 30, 7, 6, 70, 56, 7, 140, 252, 30, 8, 252, 840, 330, 9, 420, 2310, 1980, 143, 10, 660, 5544, 8580, 2002, 11, 990, 12012, 30030, 15015, 728, 12, 1430, 24024, 90090, 80080, 12376, 13, 2002, 45045, 240240, 340340, 111384, 3876, 14, 2730

Offset: 1

Emeric Deutsch, Feb 27 2007

Row n has ceiling(n/2) terms.
Row sums are the Catalan numbers (A000108).
T(n,1) = n;
T(n,2) = 2*binomial(n+1, 4) = 2*A000332(n+1);
T(n,3) = 7*binomial(n+2, 7) = 7*A000580(n+2);
T(n,4) = 30*binomial(n+3, 10) = 30*A001287(n+3);
T(n,5) = 143*binomial(n+4, 13) = 143*A010966(n+4);
T(2n-1,n) = A006013(n-1).
T(n,k) is the number of ordered trees (A000108) with n edges, exactly k of whose vertices possess at least one leaf child. [David Callan, Aug 22 2014]

			Triangle starts:
  1;
  2;
  3,  2;
  4, 10;
  5, 30,  7;
  6, 70, 56;

J.-C. Aval, A. Boussicault, B. Delcroix-Oger, F. Hivert, et al., Non-ambiguous trees: new results and generalization, arXiv preprint arXiv:1511.09455 [math.CO], 2015.
Bérénice Delcroix-Oger, Florent Hivert, Patxi Laborde-Zubieta, Jean-Christophe Aval, and Adrien Boussicault, Non-ambiguous trees: new results and generalisation, hal-03165269v2 [cs.DM] [math.CO], 2021.
Tad White, Quota Trees, arXiv:2401.01462 [math.CO], 2024. See p. 20.

Cf. A000108, A000332, A000580, A001287, A010966, A006013.

T:=(n,k)->2*binomial(3*k-1,2*k)*binomial(n-1+k,3*k-2)/(3*k-1): for n from 1 to 15 do seq(T(n,k),k=1..ceil(n/2)) od;

m = 14(*rows*); G = 0; Do[G = Series[(1 + t^2 z - G^3 z^2 + G^2 z (2+z))/ (1+2z), {t, 0, m}, {z, 0, m}] // Normal // Expand, m]; Rest[ CoefficientList[#, t^2]]& /@ Rest[CoefficientList[G-1, z] ] // Flatten (* Jean-François Alcover, Jan 23 2019 *)

T(n,k) = 2*binomial(3k-1,2k)*binomial(n-1+k,3k-2)/(3k-1) (formula obtained only by inspection).

G.f.: G-1, where G=G(t,z) satisfies z^2*G^3 - z(z+2)G^2 + (1+2z)*G - t^2*z - 1 = 0.

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cp's OEIS Frontend

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

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A110555 Triangle of partial sums of alternating binomial coefficients: T(n, k) = Sum_{j=0..k} binomial(n, j)*(-1)^j, for n >= 0, 0 <= k <= n.

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A010965 a(n) = binomial(n,12).

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A128908 Riordan array (1, x/(1-x)^2).

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A238801 Triangle T(n,k), read by rows, given by T(n,k) = C(n+1, k+1)*(1-(k mod 2)).

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