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%I #120 Feb 28 2023 17:02:18
%S 1,2,1,2,3,1,2,5,4,1,2,7,9,5,1,2,9,16,14,6,1,2,11,25,30,20,7,1,2,13,
%T 36,55,50,27,8,1,2,15,49,91,105,77,35,9,1,2,17,64,140,196,182,112,44,
%U 10,1,2,19,81,204,336,378,294,156,54,11,1,2,21,100,285
%N Numbers in (2,1)-Pascal triangle (by row).
%C Reverse of A029635. Row sums are A003945. Diagonal sums are Fibonacci(n+2) = Sum_{k=0..floor(n/2)} (2n-3k)*C(n-k,n-2k)/(n-k). - _Paul Barry_, Jan 30 2005
%C Riordan array ((1+x)/(1-x), x/(1-x)). The signed triangle (-1)^(n-k)T(n,k) or ((1-x)/(1+x), x/(1+x)) is the inverse of A055248. Row sums are A003945. Diagonal sums are F(n+2). - _Paul Barry_, Feb 03 2005
%C Row sums = A003945: (1, 3, 6, 12, 24, 48, 96, ...) = (1, 3, 7, 15, 31, 63, 127, ...) - (0, 0, 1, 3, 7, 15, 31, ...); where (1, 3, 7, 15, ...) = A000225. - _Gary W. Adamson_, Apr 22 2007
%C Triangle T(n,k), read by rows, given by (2,-1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Nov 17 2011
%C A029653 is jointly generated with A208510 as an 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*v(n-1,x)+1. See the Mathematica section. - _Clark Kimberling_, Feb 28 2012
%C For a closed-form formula for arbitrary left and right borders of Pascal like triangle, see A228196. - _Boris Putievskiy_, Aug 18 2013
%C For a closed-form formula for generalized Pascal's triangle, see A228576. - _Boris Putievskiy_, Sep 04 2013
%C The n-th row polynomial is (2 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - _Peter Bala_, Feb 25 2018
%H Reinhard Zumkeller, <a href="/A029653/b029653.txt">Rows n = 0..125 of triangle, flattened</a>
%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://cs.uwaterloo.ca/journals/JIS/VOL16/Barry2/barry231.html">A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays</a>, Journal of Integer Sequences, 16 (2013), #13.5.4.
%H Hacene Belbachir and Athmane Benmezai, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL15/Belbachir/bel22.html">Expansion of Fibonacci and Lucas Polynomials: An Answer to Prodinger's Question</a>, Journal of Integer Sequences, Vol. 15 (2012), #12.7.6.
%H B. A. Bondarenko, <a href="http://www.fq.math.ca/pascal.html">Generalized Pascal Triangles and Pyramids</a> (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. 39.
%H H. Hosoya, <a href="http://dx.doi.org/10.1023/A:1019192302062">Pascal's triangle, non-adjacent numbers and D-dimensional atomic orbitals</a>, J. Math. Chemistry, vol. 23, 1998, 169-178.
%H M. Janjic and B. Petkovic, <a href="http://arxiv.org/abs/1301.4550">A Counting Function</a>, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
%H M. Janjic and B. Petkovic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Janjic/janjic45.html">A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers</a>, J. Int. Seq. 17 (2014) # 14.3.5
%H Huyile Liang, Yanni Pei, and Yi Wang, <a href="https://arxiv.org/abs/2302.11856">Analytic combinatorics of coordination numbers of cubic lattices</a>, arXiv:2302.11856 [math.CO], 2023. See p. 8.
%H Mark C. Wilson, <a href="http://emis.impa.br/EMIS/journals/DMTCS/pdfpapers/dmAD0129.pdf">Asymptotics for generalized Riordan arrays.</a> International Conference on Analysis of Algorithms DMTCS proc. AD. Vol. 323. 2005.
%F T(n, k) = C(n-2, k-1) + C(n-2, k) + C(n-1, k-1) + C(n-1, k) except for n=0.
%F G.f.: (1 + x + y + xy)/(1 - y - xy). - _Ralf Stephan_, May 17 2004
%F T(n, k) = (2n-k)*binomial(n, n-k)/n, n, k > 0. - _Paul Barry_, Jan 30 2005
%F Sum_{k=0..n} T(n, k)*x^k gives A003945-A003954 for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - _Philippe Deléham_, Jul 10 2005
%F T(n, k) = C(n-1, k) + C(n, k). - _Philippe Deléham_, Jul 10 2005
%F Equals A097806 * A007318, i.e., the pairwise operator * Pascal's Triangle as infinite lower triangular matrices. - _Gary W. Adamson_, Apr 22 2007
%F From _Peter Bala_, Dec 27 2014: (Start)
%F exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(2 + 5*x + 4*x^2/2! + x^3/3!) = 2 + 7*x + 16*x^2/2! + 30*x^3/3! + 50*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).
%F Let M denote the lower unit triangular array with 1's on the main diagonal and 1's everywhere else below the main diagonal except for the first column which consists of the sequence [1,2,2,2,...]. For k = 0,1,2,... define M(k) to be the lower unit triangular block array
%F /I_k 0\
%F \ 0 M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section. (End)
%e The triangle T(n,k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 ...
%e 0: 1
%e 1: 2 1
%e 2: 2 3 1
%e 3: 2 5 4 1
%e 4: 2 7 9 5 1
%e 5: 2 9 16 14 6 1
%e 6: 2 11 25 30 20 7 1
%e 7: 2 13 36 55 50 27 8 1
%e 8: 2 15 49 91 105 77 35 9 1
%e 9: 2 17 64 140 196 182 112 44 10 1
%e 10: 2 19 81 204 336 378 294 156 54 11 1
%e ... Reformatted. - _Wolfdieter Lang_, Jan 09 2015
%e With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins
%e /1 \/1 \/1 \ /1 \
%e |2 1 ||0 1 ||0 1 | |2 1 |
%e |2 1 1 ||0 2 1 ||0 0 1 |... = |2 3 1 |
%e |2 1 1 1 ||0 2 1 1 ||0 0 2 1 | |2 5 4 1 |
%e |2 1 1 1 1||0 2 1 1 1 ||0 0 2 1 1| |2 7 9 5 1|
%e |... ||... ||... | |... |
%e - _Peter Bala_, Dec 27 2014
%p A029653 := proc(n,k)
%p if n = 0 then
%p 1;
%p else
%p binomial(n-1, k)+binomial(n, k)
%p fi
%p end proc: # _R. J. Mathar_, Jun 30 2013
%t u[1, x_] := 1; v[1, x_] := 1; z = 16;
%t u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
%t v[n_, x_] := u[n - 1, x] + x*v[n - 1, x] + 1;
%t Table[Expand[u[n, x]], {n, 1, z/2}]
%t Table[Expand[v[n, x]], {n, 1, z/2}]
%t cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
%t TableForm[cu]
%t Flatten[%] (* A208510 *)
%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[%] (* A029653 *)
%t (* _Clark Kimberling_, Feb 28 2012 *)
%o (Haskell)
%o a029653 n k = a029653_tabl !! n !! k
%o a029653_row n = a029653_tabl !! n
%o a029653_tabl = [1] : iterate
%o (\xs -> zipWith (+) ([0] ++ xs) (xs ++ [0])) [2, 1]
%o -- _Reinhard Zumkeller_, Dec 16 2013
%o (Python)
%o from sympy import Poly
%o from sympy.abc import x
%o def u(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x)
%o def v(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x) + 1
%o def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
%o for n in range(1, 13): print(a(n)) # _Indranil Ghosh_, May 27 2017
%Y (d, 1) Pascal triangles: A007318(d=1), A093560(3), A093561(4), A093562(5), A093563(6), A093564(7), A093565(8), A093644(9), A093645(10).
%Y Cf. A003945, A208510, A228196, A228576.
%Y Cf. A078812, A106195.
%K nonn,tabl
%O 0,2
%A _Mohammad K. Azarian_
%E More terms from _James A. Sellers_
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