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 A029653 Numbers in (2,1)-Pascal triangle (by row). 57
 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, 36, 55, 50, 27, 8, 1, 2, 15, 49, 91, 105, 77, 35, 9, 1, 2, 17, 64, 140, 196, 182, 112, 44, 10, 1, 2, 19, 81, 204, 336, 378, 294, 156, 54, 11, 1, 2, 21, 100, 285 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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 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 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 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 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 For a closed-form formula for arbitrary left and right borders of Pascal like triangle, see A228196. - Boris Putievskiy, Aug 18 2013 For a closed-form formula for generalized Pascal's triangle, see A228576. - Boris Putievskiy, Sep 04 2013 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 LINKS Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened 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. P. Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, Journal of Integer Sequences, 16 (2013), #13.5.4. Hacene Belbachir and Athmane Benmezai, Expansion of Fibonacci and Lucas Polynomials: An Answer to Prodinger's Question, Journal of Integer Sequences, Vol. 15 (2012), #12.7.6. 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. 39. H. Hosoya, Pascal's triangle, non-adjacent numbers and D-dimensional atomic orbitals, J. Math. Chemistry, vol. 23, 1998, 169-178. M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From N. J. A. Sloane, Feb 13 2013 M. Janjic, B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5 Mark C. Wilson, Asymptotics for generalized Riordan arrays. International Conference on Analysis of Algorithms DMTCS proc. AD. Vol. 323. 2005. FORMULA 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. G.f.: (1 + x + y + xy)/(1 - y - xy). - Ralf Stephan, May 17 2004 T(n, k) = (2n-k)*binomial(n, n-k)/n, n, k > 0. - Paul Barry, Jan 30 2005 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 T(n, k) = C(n-1, k) + C(n, k). - Philippe Deléham, Jul 10 2005 Equals A097806 * A007318, i.e., the pairwise operator * Pascal's Triangle as infinite lower triangular matrices. - Gary W. Adamson, Apr 22 2007 From Peter Bala, Dec 27 2014: (Start) 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) ). 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 /I_k 0\ \ 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) EXAMPLE The triangle T(n,k) begins: n\k 0  1  2   3   4   5   6   7  8  9 10 ... 0:  1 1:  2  1 2:  2  3  1 3:  2  5  4   1 4:  2  7  9   5   1 5:  2  9 16  14   6   1 6:  2 11 25  30  20   7   1 7:  2 13 36  55  50  27   8   1 8:  2 15 49  91 105  77  35   9  1 9:  2 17 64 140 196 182 112  44 10  1 10: 2 19 81 204 336 378 294 156 54 11  1 ... Reformatted. - Wolfdieter Lang, Jan 09 2015 With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins /1        \/1         \/1        \      /1        \ |2 1      ||0 1       ||0 1      |      |2 1      | |2 1 1    ||0 2 1     ||0 0 1    |... = |2 3 1    | |2 1 1 1  ||0 2 1 1   ||0 0 2 1  |      |2 5 4 1  | |2 1 1 1 1||0 2 1 1 1 ||0 0 2 1 1|      |2 7 9 5 1| |...      ||...       ||...      |      |...      | - Peter Bala, Dec 27 2014 MAPLE A029653 :=  proc(n, k) if n = 0 then   1; else   binomial(n-1, k)+binomial(n, k) fi end proc: # R. J. Mathar, Jun 30 2013 MATHEMATICA u[1, x_] := 1; v[1, x_] := 1; z = 16; u[n_, x_] := u[n - 1, x] + x*v[n - 1, x]; v[n_, x_] := u[n - 1, x] + x*v[n - 1, x] + 1; Table[Expand[u[n, x]], {n, 1, z/2}] Table[Expand[v[n, x]], {n, 1, z/2}] cu = Table[CoefficientList[u[n, x], x], {n, 1, z}]; TableForm[cu] Flatten[%]  (* A208510 *) Table[Expand[v[n, x]], {n, 1, z}] cv = Table[CoefficientList[v[n, x], x], {n, 1, z}]; TableForm[cv] Flatten[%]  (* A029653 *) (* Clark Kimberling, Feb 28 2012 *) PROG (Haskell) a029653 n k = a029653_tabl !! n !! k a029653_row n = a029653_tabl !! n a029653_tabl = [1] : iterate                (\xs -> zipWith (+) ([0] ++ xs) (xs ++ [0])) [2, 1] -- Reinhard Zumkeller, Dec 16 2013 (Python) from sympy import Poly from sympy.abc import x def u(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x) def v(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x) + 1 def a(n): return Poly(v(n, x), x).all_coeffs()[::-1] for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 27 2017 CROSSREFS (d, 1) Pascal triangles: A007318(d=1), A093560(3), A093561(4), A093562(5), A093563(6), A093564(7), A093565(8), A093644(9), A093645(10). Cf. A003945, A208510, A228196, A228576. Cf. A078812, A106195. Sequence in context: A065158 A181842 A209564 * A067763 A263683 A087730 Adjacent sequences:  A029650 A029651 A029652 * A029654 A029655 A029656 KEYWORD nonn,tabl AUTHOR EXTENSIONS More terms from James A. Sellers STATUS approved

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Last modified November 23 21:51 EST 2020. Contains 338603 sequences. (Running on oeis4.)