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%I #201 Feb 08 2024 19:36:09
%S 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,
%T 50,55,36,13,2,1,9,35,77,105,91,49,15,2,1,10,44,112,182,196,140,64,17,
%U 2,1,11,54,156,294,378,336,204,81,19,2,1,12,65,210,450,672,714,540,285,100
%N The (1,2)-Pascal triangle (or Lucas triangle) read by rows.
%C This is also called Vieta's array. - _N. J. A. Sloane_, Nov 22 2017
%C 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
%C 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
%C For a closed-form formula for generalized Pascal's triangle see A228576. - _Boris Putievskiy_, Sep 04 2013
%C 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
%C 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
%C 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
%C 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
%C 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
%D B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8. English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 25.
%D Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers. New York: Prometheus Books (2007): 97 - 105.
%D M. Shubin and S. Andersson, Pseudodifferential Operators and Spectral Theory, Springer Series in Soviet Mathematics, 1987.
%H Vincenzo Librandi, <a href="/A029635/b029635.txt">Rows n = 0..102, flattened</a>
%H P. R. J. Asveld, <a href="http://dx.doi.org/10.1016/j.tcs.2005.11.010">Generating all permutations by context-free grammars in Chomsky normal form</a>, Theoretical Computer Science 354 (2006) 118-130.
%H Mohammad K. Azarian, <a href="http://www.m-hikari.com/ijcms/ijcms-2012/45-48-2012/azarianIJCMS45-48-2012.pdf">Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums</a>, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
%H Paul Barry, <a href="https://arxiv.org/abs/2004.04577">On a Central Transform of Integer Sequences</a>, arXiv:2004.04577 [math.CO], 2020.
%H Arthur T. Benjamin, <a href="http://www.math.hmc.edu/~benjamin/papers/LucasTriangle.pdf">The Lucas triangle recounted</a>
%H Junesang Choi, <a href="http://dx.doi.org/10.1186/1687-1847-2013-236">Determinants of the Laplacians on the n-dimensional unit sphere S^n </a>, Advances in Difference Equations, 236, 2013.
%H N. Durov, S. Meljanac, A. Samsarov, Z. Skoda, <a href="https://arxiv.org/abs/math/0604096">A universal formula for representing Lie algebra generators as formal power series with coefficients in the Weyl algebra</a>, arXiv preprint arXiv:math/0604096 [math.RT], 2006.
%H S. Falcon, <a href="http://scik.org/index.php/jmcs/article/view/102">On the Lucas triangle and its relationship with the k-Lucas numbers</a>, Journal of Mathematical and Computational Science, 2 (2012), No. 3, 425-434. - _N. J. A. Sloane_, Sep 20 2012
%H Mark Feinberg, <a href="http://www.fq.math.ca/Scanned/5-5/feinberg.pdf">A Lucas triangle</a>, Fibonacci Quart. 5 (1967), 486-490.
%H Hans-Christian Herbig, Daniel Herden, Christopher Seaton, <a href="http://arxiv.org/abs/1605.01572">An algebra generated by x-2</a>, arXiv:1605.01572 [math.CO], 2016.
%H Milan Janjić, <a href="https://arxiv.org/abs/1905.04465">On Restricted Ternary Words and Insets</a>, arXiv:1905.04465 [math.CO], 2019.
%H M. A. A. Obaid, S. K. Nauman, W. M. Fakieh, C. M. Ringel, <a href="http://www.math.uni-bielefeld.de/~ringel/opus/jeddah.pdf">The numbers of support-tilting modules for a Dynkin algebra</a>, 2014 and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Ringel/ringel22.html">J. Int. Seq. 18 (2015) 15.10.6</a>.
%H Richard L. Ollerton and Anthony G. Shannon, <a href="http://www.fq.math.ca/Scanned/36-2/ollerton.pdf">Some properties of generalized Pascal squares and triangles</a>, Fib. Q., 36 (1998), 98-109. [Here the initial term is 1 not 2]
%H C. M. Ringel, <a href="http://arxiv.org/abs/1502.06553">The Catalan combinatorics of the hereditary artin algebras</a>, arXiv preprint arXiv:1502.06553 [math.RT], 2015.
%H Neville Robbins, <a href="http://www.fq.math.ca/Papers1/43-2/paper43-2-9.pdf">The Lucas triangle revisited</a>, Fibonacci Quarterly 43, May 2005, pp. 142-148.
%H Lee-Peng Teo, <a href="http://arxiv.org/abs/1412.0758">Zeta function of spheres and real projective spaces</a>, arXiv:1412.0758v1 [math.NT], 2014.
%H Xu Wang, Xuxu Zhao, Haiyuan Yao, <a href="https://arxiv.org/abs/1810.07329">Structure and enumeration results of matchable Lucas cubes</a>, arXiv:1810.07329 [math.CO], 2018. See Table 1 p. 3.
%H Y. Yang and J. Leida, <a href="http://www.fq.math.ca/Papers1/42-3/quartyang03_2004.pdf">Pascal decompositions of geometric arrays in matrices</a>, The Fibonacci Quarterly 42.3 (2004).
%F T(n,k) = T(n-1, k-1) + T(n-1, k) = C(n, k) + C(n-1, k-1) = C(n, k)*(n+k)/n = A007318(n, k) + A007318(n-1, k-1) = 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. - _Henry Bottomley_, Apr 26 2002
%F G.f.: 1 + (1 + x*y) / (1 - x - x*y). - _Michael Somos_, Jul 15 2003
%F G.f. for n-th row: (x+2*y)*(x+y)^(n-1).
%F 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
%F T(2n,n) - T(2n,n+1)= Catalan(n)= A000108(n). - _Philippe Deléham_, Mar 19 2009
%F 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
%F 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
%F For n > 0: T(n,k) = A097207(n-1,k), 0 <= k < n. - _Reinhard Zumkeller_, Mar 12 2012
%F For n > 0: T(n,k) = A029600(n,k) - A007318(n,k), 0 <= k <= n. - _Reinhard Zumkeller_, Apr 16 2012
%F Riordan array ((2-x)/(1-x), x/(1-x)). - _Philippe Deléham_, Mar 15 2013
%F 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
%F 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:
%e Triangle begins:
%e 2;
%e 1 2;
%e 1 3 2;
%e 1 4 5 2;
%e 1 5 9 7 2;
%e ...
%e From _Peter Bala_, Aug 14 2008: (Start)
%e Read as a square, the array begins
%e ===================================
%e n\k|...0...1....2.....3.....4.....5
%e ===================================
%e 0..|...2...2....2.....2.....2.....2 A040000
%e 1..|...1...3....5.....7.....9....11 A005408
%e 2..|...1...4....9....16....25....36 A000290
%e 3..|...1...5...14....30....55....91 A000330
%e 4..|...1...6...20....50...105...196 A002415
%e 5..|...1...7...27....77...182...378 A005585
%e 6..|...1...8...35...112...294...672 A040977
%e ... (End)
%t 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 *)
%t (* The next program cogenerates A029635 and A029638. *)
%t u[1, x_] := 1; v[1, x_] := 1; z = 16;
%t u[n_, x_] := u[n - 1, x] + v[n - 1, x]
%t v[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1
%t Table[Factor[u[n, x]], {n, 1, z}]
%t Table[Factor[v[n, x]], {n, 1, z}]
%t cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
%t TableForm[cu]
%t Flatten[%] (* A029638 *)
%t Table[Expand[v[n, x]], {n, 1, z}]
%t cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
%t TableForm[cv]
%t Flatten[%] (* A029635 *)
%t (* _Clark Kimberling_, Feb 20 2012 *)
%t Table[Binomial[n,k]+Binomial[n-1,k-1],{n,0,20},{k,0,n}]//Flatten (* _Harvey P. Dale_, Feb 08 2024 *)
%o (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 */
%o (Haskell)
%o a029635 n k = a029635_tabl !! n !! k
%o a029635_row n = a029635_tabl !! n
%o a029635_tabl = [2] : iterate
%o (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,2]
%o -- _Reinhard Zumkeller_, Mar 12 2012, Feb 23 2012
%o (Sage) # uses[riordan_array from A256893]
%o riordan_array((2-x)/(1-x), x/(1-x), 8) # _Peter Luschny_, Nov 09 2019
%Y Cf. A007318, A034807, A061896, A029653 (row-reversed), A157000.
%Y Sums along ascending diagonals give Lucas numbers, n>0.
%Y Cf. A090327, A003945, A029638, A228196, A228576.
%Y Cf. A000330, A000217, A293600.
%K nonn,tabl,nice,easy
%O 0,1
%A _Mohammad K. Azarian_
%E More terms from _David W. Wilson_
%E a(0) changed to 2 (was 1) by _Daniel Forgues_, Jul 06 2010
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