A097207 -id:A097207 - cp's OEIS frontend

A029635 The (1,2)-Pascal triangle (or Lucas triangle) read by rows.

2, 1, 2, 1, 3, 2, 1, 4, 5, 2, 1, 5, 9, 7, 2, 1, 6, 14, 16, 9, 2, 1, 7, 20, 30, 25, 11, 2, 1, 8, 27, 50, 55, 36, 13, 2, 1, 9, 35, 77, 105, 91, 49, 15, 2, 1, 10, 44, 112, 182, 196, 140, 64, 17, 2, 1, 11, 54, 156, 294, 378, 336, 204, 81, 19, 2, 1, 12, 65, 210, 450, 672, 714, 540, 285, 100

Offset: 0

Mohammad K. Azarian

This is also called Vieta's array. - N. J. A. Sloane, Nov 22 2017
Dropping the first term and changing the boundary conditions to T(n,1)=n, T(n,n-1)=2 (n>=2), T(n,n)=1 yields the number of nonterminal symbols (which generate strings of length k) in a certain context-free grammar in Chomsky normal form that generates all permutations of n symbols. Summation over k (1<=k<=n) results in A003945. For the number of productions of this grammar: see A090327. Example: 1; 2, 1; 3, 2, 1; 4, 5, 2, 1; 5, 9, 7, 2, 1; 6, 14, 16, 9, 2, 1; In addition to the example of A090327 we have T(3,3)=#{S}=1, T(3,2)=#{D,E}=2 and T(3,1)=#{A,B,C}=3. - Peter R. J. Asveld, Jan 29 2004
Much as the original Pascal triangle gives the Fibonacci numbers as sums of its diagonals, this triangle gives the Lucas numbers (A000032) as sums of its diagonals; see Posamentier & Lehmann (2007). - Alonso del Arte, Apr 09 2012
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
It appears that for the infinite set of (1,N) Pascal's triangles, the binomial transform of the n-th row (n>0), followed by zeros, is equal to the n-th partial sum of (1, N, N, N, ...). Example: for the (1,2) Pascal's triangle, the binomial transform of the second row followed by zeros, i.e., of (1, 3, 2, 0, 0, 0, ...), is equal to the second partial sum of (1, 2, 2, 2, ...) = (1, 4, 9, 16, ...). - Gary W. Adamson, Aug 11 2015
Given any (1,N) Pascal triangle, let the binomial transform of the n-th row (n>1) followed by zeros be Q(x). It appears that the binomial transform of the (n-1)-th row prefaced by a zero is Q(n-1). Example: In the (1,2) Pascal triangle the binomial transform of row 3: (1, 4, 5, 2, 0, 0, 0, ...) is A000330 starting with 1: (1, 5, 14, 30, 55, 91, ...). The binomial transform of row 2 prefaced by a zero and followed by zeros, i.e., of (0, 1, 3, 2, 0, 0, 0, ...) is (0, 1, 5, 14, 30, 55, ...). - Gary W. Adamson, Sep 28 2015
It appears that in the array accompanying each (1,N) Pascal triangle (diagonals of the triangle), the binomial transform of (..., 1, N, 0, 0, 0, ...) preceded by (n-1) zeros generates the n-th row of the array (n>0). Then delete the zeros in the result. Example: in the (1,2) Pascal triangle, row 3 (1, 5, 14, 30, ...) is the binomial transform of (0, 0, 1, 2, 0, 0, 0, ...) with the resulting zeros deleted. - Gary W. Adamson, Oct 11 2015
Read as a square array (similar to the Example section Sq(m,j), but with Sq(0,0)=0 and Sq(m,j)=P(m+1,j) otherwise), P(n,k) are the multiplicities of the eigenvalues, lambda_n = n(n+k-1), of the Laplacians on the unit k-hypersphere, given by Teo (and Choi) as P(n,k) = (2n-k+1)(n+k-2)!/(n!(k-1)!). P(n,k) is also the numerator of a Dirichlet series for the Minakashisundarum-Pleijel zeta function for the sphere. Also P(n,k) is the dimension of the space of homogeneous, harmonic polynomials of degree k in n variables (Shubin, p. 169). For relations to Chebyshev polynomials and simple Lie algebras, see A034807. - Tom Copeland, Jan 10 2016
For a relation to a formulation for a universal Lie Weyl algebra for su(1,1), see page 16 of Durov et al. - Tom Copeland, Jan 15 2016

			Triangle begins:
  [0] [2]
  [1] [1, 2]
  [2] [1, 3,  2]
  [3] [1, 4,  5,  2]
  [4] [1, 5,  9,  7,   2]
  [5] [1, 6, 14, 16,   9,  2]
  [6] [1, 7, 20, 30,  25, 11,  2]
  [7] [1, 8, 27, 50,  55, 36, 13,  2]
  [8] [1, 9, 35, 77, 105, 91, 49, 15, 2]
.
Read as a square, the array begins:
  n\k| 0  1   2    3    4    5
  --------------------------------------
  0 |  2  2   2    2    2    2   A040000
  1 |  1  3   5    7    9   11   A005408
  2 |  1  4   9   16   25   36   A000290
  3 |  1  5  14   30   55   91   A000330
  4 |  1  6  20   50  105  196   A002415
  5 |  1  7  27   77  182  378   A005585
  6 |  1  8  35  112  294  672   A040977

Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers. New York: Prometheus Books (2007): 97 - 105.
M. Shubin and S. Andersson, Pseudodifferential Operators and Spectral Theory, Springer Series in Soviet Mathematics, 1987.

Vincenzo Librandi, Rows n = 0..102, flattened
P. R. J. Asveld, Generating all permutations by context-free grammars in Chomsky normal form, Theoretical Computer Science 354 (2006) 118-130.
Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Arthur T. Benjamin, The Lucas triangle recounted
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 25.
Junesang Choi, Determinants of the Laplacians on the n-dimensional unit sphere S^n , Advances in Difference Equations, 236, 2013.
N. Durov, S. Meljanac, A. Samsarov, Z. Skoda, A universal formula for representing Lie algebra generators as formal power series with coefficients in the Weyl algebra, arXiv preprint arXiv:math/0604096 [math.RT], 2006.
S. Falcon, On the Lucas triangle and its relationship with the k-Lucas numbers, Journal of Mathematical and Computational Science, 2 (2012), No. 3, 425-434. - _N. J. A. Sloane_, Sep 20 2012
Mark Feinberg, A Lucas triangle, Fibonacci Quart. 5 (1967), 486-490.
Hans-Christian Herbig, Daniel Herden, Christopher Seaton, An algebra generated by x-2, arXiv:1605.01572 [math.CO], 2016.
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
M. A. A. Obaid, S. K. Nauman, W. M. Fakieh, C. M. Ringel, The numbers of support-tilting modules for a Dynkin algebra, 2014 and J. Int. Seq. 18 (2015) 15.10.6.
Richard L. Ollerton and Anthony G. Shannon, Some properties of generalized Pascal squares and triangles, Fib. Q., 36 (1998), 98-109. [Here the initial term is 1 not 2]
C. M. Ringel, The Catalan combinatorics of the hereditary artin algebras, arXiv preprint arXiv:1502.06553 [math.RT], 2015.
Neville Robbins, The Lucas triangle revisited, Fibonacci Quarterly 43, May 2005, pp. 142-148.
Lee-Peng Teo, Zeta function of spheres and real projective spaces, arXiv:1412.0758v1 [math.NT], 2014.
Xu Wang, Xuxu Zhao, Haiyuan Yao, Structure and enumeration results of matchable Lucas cubes, arXiv:1810.07329 [math.CO], 2018. See Table 1 p. 3.
Y. Yang and J. Leida, Pascal decompositions of geometric arrays in matrices, The Fibonacci Quarterly 42.3 (2004).

Cf. A003945 (row sums), A007318, A034807, A061896, A029653 (row-reversed), A157000.
Sums along ascending antidiagonals give Lucas numbers, n>0.
Cf. A090327, A003945, A029638, A228196, A228576, A000330, A000217, A293600.

a029635 n k = a029635_tabl !! n !! k
a029635_row n = a029635_tabl !! n
a029635_tabl = [2] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,2]
-- Reinhard Zumkeller, Mar 12 2012, Feb 23 2012

T := proc(n, k) option remember;
if n = k then 2 elif k = 0 then 1 else T(n-1, k-1) + T(n-1, k) fi end:
for n from 0 to 8 do seq(T(n, k), k = 0..n) od;  # Peter Luschny, Dec 22 2024

t[0, 0] = 2; t[n_, k_] := If[k < 0 || k > n, 0, Binomial[n, k] + Binomial[n-1, k-1]]; Flatten[Table[t[n, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, May 03 2011 *)
(* The next program cogenerates A029635 and A029638. *)
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A029638  *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A029635 *)
(* Clark Kimberling, Feb 20 2012 *)
Table[Binomial[n,k]+Binomial[n-1,k-1],{n,0,20},{k,0,n}]//Flatten (* Harvey P. Dale, Feb 08 2024 *)

{T(n, k) = if( k<0 || k>n, 0, (n==0) + binomial(n, k) + binomial(n-1, k-1))}; /* Michael Somos, Jul 15 2003 */

# uses[riordan_array from A256893]
riordan_array((2-x)/(1-x), x/(1-x), 8) # Peter Luschny, Nov 09 2019

From Henry Bottomley, Apr 26 2002;  (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k).
T(n, k) = C(n, k) + C(n-1, k-1).
T(n, k) = C(n, k)*(n + k)/n.
T(n, k) = A007318(n, k) + A007318(n-1, k-1).
T(n, k) = A061896(n + k, k) but with T(0, 0) = 1 and T(1, 1) = 2.
Row sum is floor(3^2(n-1)) i.e., A003945. (End)
G.f.: 1 + (1 + x*y) / (1 - x - x*y). - Michael Somos, Jul 15 2003
G.f. for n-th row: (x+2*y)*(x+y)^(n-1).
O.g.f. for row n: (1+x)/(1-x)^(n+1). The entries in row n are the nonzero entries in column n of A053120 divided by 2^(n-1). - Peter Bala, Aug 14 2008
T(2n, n) - T(2n, n+1)= Catalan(n)= A000108(n). - Philippe Deléham, Mar 19 2009
With T(0, 0) = 1 : Triangle T(n, k), read by rows, given by [1,0,0,0,0,0,...] DELTA [2,-1,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 10 2011
With T(0, 0) = 1, as in the Example section below, this is known as Vieta's array. The LU factorization of the square array is given by Yang and Leida, equation 20. - Peter Bala, Feb 11 2012
For n > 0: T(n, k) = A097207(n-1, k), 0 <= k < n. - Reinhard Zumkeller, Mar 12 2012
For n > 0: T(n, k) = A029600(n, k) - A007318(n, k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
Riordan array ((2-x)/(1-x), x/(1-x)). - Philippe Deléham, Mar 15 2013
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + 4*x + 5*x^2/2! + 2*x^3/3!) = 1 + 5*x + 14*x^2/2! + 30*x^3/3! + 55*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
For n>=1: T(n, 0) + T(n, 1) + T(n, 2) = A000217(n+1). T(n, n-2) = (n-1)^2. - Bob Selcoe, Mar 29 2016:

More terms from David W. Wilson

a(0) changed to 2 (was 1) by Daniel Forgues, Jul 06 2010

A110813 A triangle of pyramidal numbers.

Original entry on oeis.org

1, 3, 1, 5, 4, 1, 7, 9, 5, 1, 9, 16, 14, 6, 1, 11, 25, 30, 20, 7, 1, 13, 36, 55, 50, 27, 8, 1, 15, 49, 91, 105, 77, 35, 9, 1, 17, 64, 140, 196, 182, 112, 44, 10, 1, 19, 81, 204, 336, 378, 294, 156, 54, 11, 1, 21, 100, 285, 540, 714, 672, 450, 210, 65, 12, 1, 23, 121, 385, 825

Offset: 0

Paul Barry, Aug 05 2005

Triangle A029653 less first column. In general, the product (1/(1-x),x/(1-x))*(1+m*x,x) yields the Riordan array ((1+(m-1)x)/(1-x)^2,x/(1-x)) with general term T(n,k)=(m*n-(m-1)*k+1)*C(n+1,k+1)/(n+1). This is the reversal of the (1,m)-Pascal triangle, less its first column. - Paul Barry, Mar 01 2006
The column sequences give, for k=0..10: A005408 (odd numbers), A000290 (squares), A000330, A002415, A005585, A040977, A050486, A053347, A054333, A054334, A057788.
Linked to Chebyshev polynomials by the fact that this triangle with interpolated zeros in the rows and columns is a scaled version of A053120.
Row sums are A033484. Diagonal sums are A001911(n+1) or F(n+4)-2. Factors as (1/(1-x),x/(1-x))*(1+2x,x). Inverse is A110814 or (-1)^(n-k)*A104709.
This triangle is a subtriangle of the [2,1] Pascal triangle A029653 (omit there the first column).
Subtriangle of triangles in A029653, A131084, A208510. - Philippe Deléham, Mar 02 2012
This is the iterated partial sums triangle of A005408 (odd numbers). Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given). - Wolfdieter Lang, Mar 23 2015

			The number triangle T(n, k) begins
n\k  0   1   2   3    4    5    6   7   8  9 10 11
0:   1
1:   3   1
2:   5   4   1
3:   7   9   5   1
4:   9  16  14   6    1
5:  11  25  30  20    7    1
6:  13  36  55  50   27    8    1
7:  15  49  91 105   77   35    9   1
8:  17  64 140 196  182  112   44  10   1
9:  19  81 204 336  378  294  156  54  11  1
10: 21 100 285 540  714  672  450 210  65 12  1
11: 23 121 385 825 1254 1386 1122 660 275 77 13  1
... reformatted by _Wolfdieter Lang_, Mar 23 2015
As a number square S(n, k) = T(n+k, k), rows begin
  1,   1,   1,   1,   1,   1, ...
  3,   4,   5,   6,   7,   8, ...
  5,   9,  14,  20,  27,  35, ...
  7,  16,  30,  50,  77, 112, ...
  9,  25,  55, 105, 182, 294, ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A000290, A000330, A002415, A005408, A005585, A029655, A040977, A050486, A053347, A054333, A054334, A057788.

Table[2*Binomial[n + 1, k + 1] - Binomial[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 19 2017 *)

for(n=0,10, for(k=0,n, print1(2*binomial(n+1, k+1) - binomial(n,k), ", "))) \\ G. C. Greubel, Oct 19 2017

Number triangle T(n, k) = C(n, k)*(2n-k+1)/(k+1) = 2*C(n+1, k+1) - C(n, k); Riordan array ((1+x)/(1-x)^2, x/(1-x)); As a number square read by antidiagonals, T(n, k)=C(n+k, k)(2n+k+1)/(k+1).
Equals A007318 * an infinite bidiagonal matrix with 1's in the main diagonal and 2's in the subdiagonal. - Gary W. Adamson, Dec 01 2007
Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal, all 2's in the subdiagonal and the rest zeros. - Gary W. Adamson, Dec 12 2007
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=T(1,1)=1, T(1,0)=3, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 30 2013
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(7 + 9*x + 5*x^2/2! + x^3/3!) = 7 + 16*x + 30*x^2/2! + 50*x^3/3! + 77*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014
T(n, k) = ps(1, 2; k, n-k) with ps(a, d; k, n) = sum(ps(a, d; k-1, j), j=0..n) and input ps(a, d; 0, j) = a + d*j. See the iterated partial sums comment from Mar 23 2015 above. - Wolfdieter Lang, Mar 23 2015
From Franck Maminirina Ramaharo, May 21 2018: (Start)
T(n,k) = coefficients in the expansion of ((x + 2)*(x + 1)^n - 2)/x.
T(n,k) = A135278(n,k) + A135278(n-1,k).
T(n,k) = A097207(n,n-k).
G.f.: (y + 1)/((y - 1)*(x*y + y - 1)).
E.g.f.: ((x + 2)*exp(x*y + y) - 2*exp(y))/x.
(End)

A191302 Denominators in triangle that leads to the Bernoulli numbers.

Original entry on oeis.org

1, 2, 2, 3, 2, 2, 2, 3, 15, 2, 6, 3, 2, 1, 5, 105, 2, 6, 15, 15, 2, 3, 3, 105, 105, 2, 2, 5, 7, 35, 2, 3, 3, 21, 21, 231, 2, 6, 15, 15, 21, 21, 2, 1, 5, 15, 1, 77, 15015, 2, 6, 3, 35, 15, 33, 1155

Offset: 0

Paul Curtz, May 30 2011

For the definition of the ASPEC array coefficients see the formulas; see also A029635 (Lucas triangle), A097207 and A191662 (k-dimensional square pyramidal numbers).
The antidiagonal row sums of the ASPEC array equal A042950(n) and A098011(n+3).
The coefficients of the T(n,m) array are defined in A190339. We define the coefficients of the SBD array with the aid of the T(n,n+1), see the formulas and the examples.
Multiplication of the coefficients in the rows of the ASPEC array with the coefficients in the columns of the SBD array leads to the coefficients of the BSPEC triangle, see the formulas. The BSPEC triangle can be looked upon as a spectrum for the Bernoulli numbers.
The row sums of the BSPEC triangle give the Bernoulli numbers A164555(n)/A027642(n).
For the numerators of the BSPEC triangle coefficients see A192456.

			The first few rows of the array ASPEC array:
  2, 1,  1,  1,   1,   1,    1,
  2, 3,  4,  5,   6,   7,    8,
  2, 5,  9, 14,  20,  27,   35,
  2, 7, 16, 30,  50,  77,  112,
  2, 9, 25, 55, 105, 182,  294,
The first few T(n,n+1) = T(n,n)/2 coefficients:
1/2, -1/6, 1/15, -4/105, 4/105, -16/231, 3056/15015, ...
The first few rows of the SBD array:
  1/2,   0,   0,     0
  1/2,   0,   0,     0
  1/2, -1/6,  0,     0
  1/2, -1/6,  0,     0
  1/2, -1/6, 1/15,   0
  1/2, -1/6, 1/15,   0
  1/2, -1/6, 1/15, -4/105
  1/2, -1/6, 1/15, -4/105
The first few rows of the BSPEC triangle:
  B(0) =   1   = 1/1
  B(1) =  1/2  = 1/2
  B(2) =  1/6  = 1/2 - 1/3
  B(3) =   0   = 1/2 - 1/2
  B(4) = -1/30 = 1/2 - 2/3 +  2/15
  B(5) =   0   = 1/2 - 5/6 +  1/3
  B(6) =  1/42 = 1/2 - 1/1 +  3/5  - 8/105
  B(7) =   0   = 1/2 - 7/6 + 14/15 - 4/15

Cf. A028246 (Worpitzky), A085737/A085738 (Conway-Sloane) and A051714/A051715 (Akiyama-Tanigawa) for other triangles that lead to the Bernoulli numbers. - Johannes W. Meijer, Jul 02 2011

nmax:=13: mmax:=nmax:
A164555:=proc(n): if n=1 then 1 else numer(bernoulli(n)) fi: end:
A027642:=proc(n): if n=1 then 2 else denom(bernoulli(n)) fi: end:
for m from 0 to 2*mmax do T(0,m):=A164555(m)/A027642(m) od:
for n from 1 to nmax do for m from 0 to 2*mmax do T(n,m):=T(n-1,m+1)-T(n-1,m) od: od:
seq(T(n,n+1),n=0..nmax):
for n from 0 to nmax do ASPEC(n,0):=2: for m from 1 to mmax do ASPEC(n,m):= (2*n+m)*binomial(n+m-1,m-1)/m od: od:
for n from 0 to nmax do seq(ASPEC(n,m),m=0..mmax) od:
for n from 0 to nmax do for m from 0 to 2*mmax do SBD(n,m):=0 od: od:
for m from 0 to mmax do for n from 2*m to nmax do SBD(n,m):= T(m,m+1) od: od:
for n from 0 to nmax do seq(SBD(n,m), m= 0..mmax/2) od:
for n from 0 to nmax do BSPEC(n,2) := SBD(n,2)*ASPEC(2,n-4) od:
for m from 0 to mmax do for n from 0 to nmax do BSPEC(n,m) := SBD(n,m)*ASPEC(m,n-2*m) od: od:
for n from 0 to nmax do seq(BSPEC(n,m), m=0..mmax/2) od:
seq(add(BSPEC(n, k), k=0..floor(n/2)) ,n=0..nmax):
Tx:=0:
for n from 0 to nmax do for m from 0 to floor(n/2) do a(Tx):= denom(BSPEC(n,m)): Tx:=Tx+1: od: od:
seq(a(n),n=0..Tx-1); # Johannes W. Meijer, Jul 02 2011

(* a=ASPEC, b=BSPEC *) nmax = 13; a[n_, 0] = 2; a[n_, m_] := (2n+m)*Binomial[n+m-1, m-1]/m; b[n_] := BernoulliB[n]; b[1]=1/2; bb = Table[b[n], {n, 0, nmax}]; diff = Table[ Differences[bb, n], {n, 1, nmax}]; dd = Diagonal[diff]; sbd[n_, m_] := If[n >= 2m, -dd[[m+1]], 0]; b[n_, m_] := sbd[n, m]*a[m, n-2m]; Table[b[n, m], {n, 0, nmax}, {m, 0, Floor[n/2]}] // Flatten // Denominator (* Jean-François Alcover_, Aug 09 2012 *)

ASPEC(n, 0) = 2 and ASPEC(n, m) = (2*n+m)*binomial(n+m-1, m-1)/m, n >= 0, m >= 1.
ASPEC(n, m) = ASPEC(n-1, m) + ASPEC(n, m-1), n >= 1, m >= 1, with ASPEC(n, 0) = 2, n >= 0, and ASPEC(0,m) = 1, m >= 1.
SBD(n, m) = T(m, m+1), n >= 2*m; see A190339 for the definition of the T(n, m).
BSPEC(n, m) = SBD(n, m)*ASPEC(m, n-2*m)
Sum_{k=0..floor(n/2)} BSPEC(n, k) = A164555(n)/A027642(n).

Edited, Maple program and crossrefs added by Johannes W. Meijer, Jul 02 2011

A191662 a(n) = n! / A000034(n-1).

Original entry on oeis.org

1, 1, 6, 12, 120, 360, 5040, 20160, 362880, 1814400, 39916800, 239500800, 6227020800, 43589145600, 1307674368000, 10461394944000, 355687428096000, 3201186852864000, 121645100408832000, 1216451004088320000, 51090942171709440000, 562000363888803840000

Offset: 1

Paul Curtz, Jun 10 2011

The a(n) are the denominators in the formulas of the k-dimensional square pyramidal numbers:
A005408 = (2*n+1)/1                       = 1, 3,  5,   7,   9, ... (k=1)
A000290 = (n^2)/1                         = 1, 4,  9,  16,  25, ... (k=2)
A000330 = n*(n+1)*(2*n+1)/6               = 1, 5, 14,  30,  55, ... (k=3)
A002415 = (n^2)*(n^2-1)/12                = 1, 6, 20,  50, 105, ... (k=4)
A005585 = n*(n+1)*(n+2)*(n+3)*(2*n+3)/120 = 1, 7, 27,  77, 182, ... (k=5)
A040977 = (n^2)*(n^2-1)*(n^2-4)/360       = 1, 8, 35, 112, 294, ... (k=6)
A050486 (k=7), A053347 (k=8), A054333 (k=9), A054334 (k=10), A057788 (k=11).
The first superdiagonal of this array appears in A029651. - Paul Curtz, Jul 04 2011
The general formula for the k-dimensional square pyramidal numbers is (2*n+k)*binomial(n+k-1,k-1)/k, k >= 1, n >= 0, see A097207. - Johannes W. Meijer, Jun 22 2011

Cf. A000142, A009445, A002674, A064680.
Cf. A000290, A000330, A002415, A005408, A005585, A029651, A040977, A050486, A053347, A054333, A054334, A057788.
Cf. A052612.

A191662:= proc(n): n!/A000034(n-1) end: A000034 := proc(n) op((n mod 2)+1, [1, 2]) ; end proc: seq(A191662(n),n=1..17); # Johannes W. Meijer, Jun 22 2011

Array[If[EvenQ[#],#!/2,#!]&,20] (* Harvey P. Dale, Mar 14 2014 *)

a(2*n-1) = (2*n-1)!, a(2*n) = (2*n)!/2.
a(n+1) = A064680(n+1) * a(n).
From Amiram Eldar, Jul 06 2022: (Start)
Sum_{n>=1} 1/a(n) = sinh(1) + 2*cosh(1) - 2.
Sum_{n>=1} (-1)^(n+1)/a(n) = sinh(1) - 2*cosh(1) + 2. (End)
D-finite with recurrence: a(n) - (n-1)*n*a(n-2) = 0 for n >= 3 with a(1)=a(2)=1. - Georg Fischer, Nov 25 2022
a(n) = A052612(n)/2 for n >= 1. - Alois P. Heinz, Sep 05 2023

More terms from Harvey P. Dale, Mar 14 2014

cp's OEIS Frontend

A029635 The (1,2)-Pascal triangle (or Lucas triangle) read by rows.

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A110813 A triangle of pyramidal numbers.

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A191302 Denominators in triangle that leads to the Bernoulli numbers.

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A191662 a(n) = n! / A000034(n-1).

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