login
A masked Pascal triangle.
12

%I #116 Sep 13 2024 03:30:55

%S 1,0,1,1,0,1,0,3,0,1,1,0,6,0,1,0,5,0,10,0,1,1,0,15,0,15,0,1,0,7,0,35,

%T 0,21,0,1,1,0,28,0,70,0,28,0,1,0,9,0,84,0,126,0,36,0,1,1,0,45,0,210,0,

%U 210,0,45,0,1,0,11,0,165,0,462,0,330,0,55,0,1,1,0,66,0,495,0,924

%N A masked Pascal triangle.

%C Row sums are A011782. Diagonal sums are F(n+1)*(1+(-1)^n)/2 (aerated version of A001519). Product by Pascal's triangle A007318 is A119468. Schur product of (1/(1-x),x/(1-x)) and (1/(1-x^2),x).

%C Exponential Riordan array (cosh(x),x). Inverse is (sech(x),x) or A119879. - _Paul Barry_, May 26 2006

%C Rows give coefficients of polynomials p_n(x) = Sum_{k=0..n} (k+1 mod 2)*binomial(n,k)*x^(n-k) having e.g.f. exp(x*t)*cosh(t)= 1*(t^0/0!) + x*(t^1/1!) + (1+x^2)*(t^2/2!) + ... - _Peter Luschny_, Jul 14 2009

%C Inverse of the coefficient matrix of the Swiss-Knife polynomials in ascending order of x^i (reversed and aerated rows of A153641). - _Peter Luschny_, Jul 16 2012

%C Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array

%C /I_k 0\

%C \ 0 M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*... is equal to A136630 but with the first row and column omitted. - _Peter Bala_, Jul 28 2014

%C The row polynomials SKv(n,x) = [(x+1)^n + (x-1)^n]/2 , with e.g.f. cosh(t)*exp(xt), are the umbral compositional inverses of the row polynomials of A119879 (basically the Swiss Knife polynomials SK(n,x) of A153641); i.e., umbrally SKv(n,SK(.,x)) = x^n = SK(n,SKv(.,x)). Therefore, this entry's matrix and A119879 are an inverse pair. Both sequences of polynomials are Appell sequences, i.e., d/dx P(n,x) = n * P(n-1,x) and (P(.,x)+y)^n = P(n,x+y). In particular, (SKv(.,0)+x)^n = SKv(n,x), reflecting that the first column has the e.g.f. cosh(t). The raising operator is R = x + tanh(d/dx); i.e., R SKv(n,x) = SKv(n+1,x). The coefficients of this operator are basically the signed and aerated zag numbers A000182, which can be expressed as normalized Bernoulli numbers. The triangle is formed by multiplying the n-th diagonal of the lower triangular Pascal matrix by the Taylor series coefficient a(n) of cosh(x). More relations for this type of triangle and its inverse are given by the formalism of A133314. - _Tom Copeland_, Sep 05 2015

%C The signed version of this matrix has the e.g.f. cos(t) e^{xt}, generating Appell polynomials that have only real, simple zeros and whose extrema are maxima above the x-axis and minima below and situated above and below the zeros of the next lower degree polynomial. The bivariate versions appear on p. 27 of Dimitrov and Rusev in conditions for entire functions that are cosine transforms of a class of functions to have only real zeros. - _Tom Copeland_, May 21 2020

%C The n-th row of the triangle is obtained by multiplying by 2^(n-1) the elements of the first row of the limit as k approaches infinity of the stochastic matrix P^(2k-1) where P is the stochastic matrix associated with the Ehrenfest model with n balls. The elements of a stochastic matrix P give the probabilities of arriving in a state j given the previous state i. In particular the sum of every row of the matrix must be 1, and so the sum of the terms of the n-th row of this triangle is 2^(n-1). Furthermore, by the properties of Markov chains, we can interpret P^(2k-1) as the (2k-1)-step transition matrix of the Ehrenfest model and its limit exists and it is again a stochastic matrix. The rows of the triangle divided by 2^(n-1) are the even rows (second, fourth, ...) and the odd rows (first, third, ...) of the limit matrix P^(2k-1). - _Luca Onnis_, Oct 29 2023

%D Paul and Tatjana Ehrenfest, Über zwei bekannte Einwände gegen das Boltzmannsche H-Theorem, Physikalische Zeitschrift, vol. 8 (1907), pp. 311-314.

%H Reinhard Zumkeller, <a href="/A119467/b119467.txt">Rows n = 0..125 of table, flattened</a>

%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Barry1/barry97r2.html">Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms</a>, J. Int. Seq. 14 (2011) # 11.2.2, example 28.

%H Paul Barry, <a href="https://arxiv.org/abs/2101.06713">On the inversion of Riordan arrays</a>, arXiv:2101.06713 [math.CO], 2021.

%H Tom Copeland, <a href="https://tcjpn.wordpress.com/2020/07/11/skipping-over-dimensions-juggling-zeros-in-the-matrix/">Skipping over Dimensions, Juggling Zeros in the Matrix</a>, 2020.

%H D. Dimitrov and P. Rusev, <a href="https://www.dcce.ibilce.unesp.br/~dimitrov/ProfRusev/EJA_Paper_24_02_11/Dim_Rus_main.pdf">Zeros of entire Fourier transforms</a>, East Journal on Approximations, Vol. 17, No. 1, p. 1-108, 2011.

%H Miguel Méndez and Rafael Sánchez, <a href="https://arxiv.org/abs/1707.00336">On the combinatorics of Riordan arrays and Sheffer polynomials: monoids, operads and monops</a>, arXiv:1707.00336 [math.CO], 2017, Section 4.3, Example 4.

%H Miguel A. Méndez and Rafael Sánchez Lamoneda, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v25i3p25">Monops, Monoids and Operads: The Combinatorics of Sheffer Polynomials</a>, The Electronic Journal of Combinatorics 25(3) (2018), #P3.25.

%H Luca Onnis, <a href="/A119467/a119467.gif">Animation of the Ehrenfest model</a>.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Ehrenfest_model">Ehrenfest model</a>.

%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>

%F G.f.: (1-x*y)/(1-2*x*y-x^2+x^2*y^2);

%F T(n,k) = C(n,k)*(1+(-1)^(n-k))/2;

%F Column k has g.f. (1/(1-x^2))*(x/(1-x^2))^k*Sum_{j=0..k+1} binomial(k+1,j)*sin((j+1)*Pi/2)^2*x^j.

%F Column k has e.g.f. cosh(x)*x^k/k!. - _Paul Barry_, May 26 2006

%F Let Pascal's triangle, A007318 = P; then this triangle = (1/2) * (P + 1/P). Also A131047 = (1/2) * (P - 1/P). - _Gary W. Adamson_, Jun 12 2007

%F Equals A007318 - A131047 since the zeros of the triangle are masks for the terms of A131047. Thus A119467 + A131047 = Pascal's triangle. - _Gary W. Adamson_, Jun 12 2007

%F T(n,k) = (A007318(n,k) + A130595(n,k))/2, 0<=k<=n. - _Reinhard Zumkeller_, Mar 23 2014

%e Triangle begins

%e 1,

%e 0, 1,

%e 1, 0, 1,

%e 0, 3, 0, 1,

%e 1, 0, 6, 0, 1,

%e 0, 5, 0, 10, 0, 1,

%e 1, 0, 15, 0, 15, 0, 1,

%e 0, 7, 0, 35, 0, 21, 0, 1,

%e 1, 0, 28, 0, 70, 0, 28, 0, 1,

%e 0, 9, 0, 84, 0, 126, 0, 36, 0, 1,

%e 1, 0, 45, 0, 210, 0, 210, 0, 45, 0, 1

%e p[0](x) = 1

%e p[1](x) = x

%e p[2](x) = 1 + x^2

%e p[3](x) = 3*x + x^3

%e p[4](x) = 1 + 6*x^2 + x^4

%e p[5](x) = 5*x + 10*x^3 + x^5

%e Connection with A136630: With the arrays M(k) as defined in the Comments section, the infinite product M(0)*M(1)*M(2)*... begins

%e /1 \/1 \/1 \ /1 \

%e |0 1 ||0 1 ||0 1 | |0 1 |

%e |1 0 1 ||0 0 1 ||0 0 1 |... = |1 0 1 |

%e |0 3 0 1 ||0 1 0 1 ||0 0 0 1 | |0 4 0 1 |

%e |1 0 6 0 1||0 0 3 0 1||0 0 1 0 1| |1 0 10 0 1|

%e |... ||... ||... | |... |

%e - _Peter Bala_, Jul 28 2014

%p # Polynomials: p_n(x)

%p p := proc(n,x) local k, pow; pow := (n,k) -> `if`(n=0 and k=0,1,n^k);

%p add((k+1 mod 2)*binomial(n,k)*pow(x,n-k),k=0..n) end;

%p # Coefficients: a(n)

%p seq(print(seq(coeff(i!*coeff(series(exp(x*t)*cosh(t),t,16),t,i),x,n),n=0..i)),i=0..8); # _Peter Luschny_, Jul 14 2009

%t Table[Binomial[n, k] (1 + (-1)^(n - k))/2, {n, 0, 12}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Sep 06 2015 *)

%t n = 15; "n-th row"

%t mat = Table[Table[0, {j, 1, n + 1}], {i, 1, n + 1}];

%t mat[[1, 2]] = 1;

%t mat[[n + 1, n]] = 1;

%t For[i = 2, i <= n, i++, mat[[i, i - 1]] = (i - 1)/n ];

%t For[i = 2, i <= n, i++, mat[[i, i + 1]] = (n - i + 1)/n];

%t mat // MatrixForm;

%t P2 = Dot[mat, mat];

%t R1 = Simplify[

%t Eigenvectors[Transpose[P2]][[1]]/

%t Total[Eigenvectors[Transpose[P2]][[1]]]]

%t R2 = Table[Dot[R1, Transpose[mat][[k]]], {k, 1, n + 1}]

%t odd = R2*2^(n - 1) (* __Luca Onnis_ *)

%o (Sage)

%o @CachedFunction

%o def A119467_poly(n):

%o R = PolynomialRing(ZZ, 'x')

%o x = R.gen()

%o return R.one() if n==0 else R.sum(binomial(n,k)*x^(n-k) for k in range(0,n+1,2))

%o def A119467_row(n):

%o return list(A119467_poly(n))

%o for n in (0..10) : print(A119467_row(n)) # _Peter Luschny_, Jul 16 2012

%o (Haskell)

%o a119467 n k = a119467_tabl !! n !! k

%o a119467_row n = a119467_tabl !! n

%o a119467_tabl = map (map (flip div 2)) $

%o zipWith (zipWith (+)) a007318_tabl a130595_tabl

%o -- _Reinhard Zumkeller_, Mar 23 2014

%o (Magma) /* As triangle */ [[Binomial(n, k)*(1 + (-1)^(n - k))/2: k in [0..n]]: n in [0.. 15]]; // _Vincenzo Librandi_, Sep 26 2015

%Y Cf. A131047, A153641, A162590.

%Y From _Peter Luschny_, Jul 14 2009: (Start)

%Y Cf. A034839, A162590.

%Y p[n](k), n=0,1,...

%Y k= 0: 1, 0, 1, 0, 1, 0, ... A128174

%Y k= 1: 1, 1, 2, 4, 8, 16, ... A011782

%Y k= 2: 1, 2, 5, 14, 41, 122, ... A007051

%Y k= 3: 1, 3, 10, 36, 136, ... A007582

%Y k= 4: 1, 4, 17, 76, 353, ... A081186

%Y k= 5: 1, 5, 26, 140, 776, ... A081187

%Y k= 6: 1, 6, 37, 234, 1513, ... A081188

%Y k= 7: 1, 7, 50, 364, 2696, ... A081189

%Y k= 8: 1, 8, 65, 536, 4481, ... A081190

%Y k= 9: 1, 9, 82, 756, 7048, ... A060531

%Y k=10: 1, 10, 101, 1030, ... A081192

%Y p[n](k), k=0,1,...

%Y p[0]: 1,1,1,1,1,1, ....... A000012

%Y p[1]: 0,1,2,3,4,5, ....... A001477

%Y p[2]: 1,2,5,10,17,26, .... A002522

%Y p[3]: 0,4,14,36,76,140, .. A079908 (End)

%Y Cf. A000182, A133314, A153641.

%K easy,nonn,tabl,look

%O 0,8

%A _Paul Barry_, May 21 2006

%E Edited by _N. J. A. Sloane_, Jul 14 2009