

A029635


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


63



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
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OFFSET

0,1


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,n1)=2 (n>=2), T(n,n)=1 yields the number of nonterminal symbols (which generate strings of length k) in a certain contextfree 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 closedform 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 nth row (n>0), followed by zeros, is equal to the nth 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 nth row (n>1) followed by zeros be Q(x). It appears that the binomial transform of the (n1)th row prefaced by a zero is Q(n1). 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 (n1) zeros generates the nth 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+k1), of the Laplacians on the unit khypersphere, given by Teo (and Choi) as P(n,k) = (2nk+1)(n+k2)!/(n!(k1)!). P(n,k) is also the numerator of a Dirichlet series for the MinakashisundarumPleijel 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


REFERENCES

B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5648007388. English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 25.
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

Vincenzo Librandi, Rows n = 0..102, flattened
P. R. J. Asveld, Generating all permutations by contextfree grammars in Chomsky normal form, Theoretical Computer Science 354 (2006) 118130.
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. 22212227.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Arthur T. Benjamin, The Lucas triangle recounted
Junesang Choi, Determinants of the Laplacians on the ndimensional 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 kLucas numbers, Journal of Mathematical and Computational Science, 2 (2012), No. 3, 425434.  N. J. A. Sloane, Sep 20 2012
Mark Feinberg, A Lucas triangle, Fibonacci Quart. 5 (1967), 486490.
HansChristian Herbig, Daniel Herden, Christopher Seaton, An algebra generated by x2, 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 supporttilting 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), 98109. [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. 142148.
LeePeng 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).


FORMULA

T(n,k) = T(n1, k1) + T(n1, k) = C(n, k) + C(n1, k1) = C(n, k)*(n+k)/n = A007318(n, k) + A007318(n1, k1) = A061896(n+k, k) but with T(0, 0)=1 and T(1, 1)=2. Row sum is floor[3^2(n1)] i.e., A003945.  Henry Bottomley, Apr 26 2002
G.f.: 1 + (1 + x*y) / (1  x  x*y).  Michael Somos, Jul 15 2003
G.f. for nth row: (x+2*y)*(x+y)^(n1).
O.g.f. for row n: (1+x)/(1x)^(n+1). The entries in row n are the nonzero entries in column n of A053120 divided by 2^(n1).  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(n1,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 ((2x)/(1x), x/(1x)).  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,n2) = (n1)^2.  Bob Selcoe, Mar 29 2016:


EXAMPLE

Triangle begins:
2;
1 2;
1 3 2;
1 4 5 2;
1 5 9 7 2;
...
From Peter Bala, Aug 14 2008: (Start)
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
... (End)


MATHEMATICA

t[0, 0] = 2; t[n_, k_] := If[k < 0  k > n, 0, Binomial[n, k] + Binomial[n1, k1]]; Flatten[Table[t[n, k], {n, 0, 11}, {k, 0, n}]] (* JeanFrançois 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 *)


PROG

(PARI) {T(n, k) = if( k<0  k>n, 0, (n==0) + binomial(n, k) + binomial(n1, k1))}; /* Michael Somos, Jul 15 2003 */
(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
(Sage) # uses[riordan_array from A256893]
riordan_array((2x)/(1x), x/(1x), 8) # Peter Luschny, Nov 09 2019


CROSSREFS

Cf. A007318, A034807, A061896, A029653 (rowreversed), A157000.
Sums along ascending diagonals give Lucas numbers, n>0.
Cf. A090327, A003945, A029638, A228196, A228576.
Cf. A000330, A000217, A293600.
Sequence in context: A227687 A128118 A205696 * A203301 A309853 A107456
Adjacent sequences: A029632 A029633 A029634 * A029636 A029637 A029638


KEYWORD

nonn,tabl,nice,easy


AUTHOR

Mohammad K. Azarian


EXTENSIONS

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


STATUS

approved



