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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

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Author

Mohammad K. Azarian

Keywords

Comments

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

Examples

			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

References

  • 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.

Links

Crossrefs

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.

Programs

  • Haskell
    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
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, (n==0) + binomial(n, k) + binomial(n-1, k-1))}; /* Michael Somos, Jul 15 2003 */
  • Sage
    # uses[riordan_array from A256893]
riordan_array((2-x)/(1-x), x/(1-x), 8) # Peter Luschny, Nov 09 2019

Formula

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:

Extensions

More terms from David W. Wilson
a(0) changed to 2 (was 1) by Daniel Forgues, Jul 06 2010

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