cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A119467 A masked Pascal triangle.

Original entry on oeis.org

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, 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, 210, 0, 45, 0, 1, 0, 11, 0, 165, 0, 462, 0, 330, 0, 55, 0, 1, 1, 0, 66, 0, 495, 0, 924
Offset: 0

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Author

Paul Barry, May 21 2006

Keywords

Comments

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).
Exponential Riordan array (cosh(x),x). Inverse is (sech(x),x) or A119879. - Paul Barry, May 26 2006
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
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
Call this array M and 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. 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
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
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
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

Examples

			Triangle begins
  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,   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, 210,  0, 45, 0, 1
p[0](x) = 1
p[1](x) = x
p[2](x) = 1 + x^2
p[3](x) = 3*x + x^3
p[4](x) = 1 + 6*x^2 + x^4
p[5](x) = 5*x + 10*x^3 + x^5
Connection with A136630: With the arrays M(k) as defined in the Comments section, the infinite product M(0)*M(1)*M(2)*... begins
/1        \/1        \/1        \      /1         \
|0 1      ||0 1      ||0 1      |      |0 1       |
|1 0 1    ||0 0 1    ||0 0 1    |... = |1 0  1    |
|0 3 0 1  ||0 1 0 1  ||0 0 0 1  |      |0 4  0 1  |
|1 0 6 0 1||0 0 3 0 1||0 0 1 0 1|      |1 0 10 0 1|
|...      ||...      ||...      |      |...       |
- _Peter Bala_, Jul 28 2014

References

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

Links

Crossrefs

Cf. A131047, A153641, A162590.
From Peter Luschny, Jul 14 2009: (Start)
Cf. A034839, A162590.
p[n](k), n=0,1,...
k= 0: 1, 0, 1, 0, 1, 0, ... A128174
k= 1: 1, 1, 2, 4, 8, 16, ... A011782
k= 2: 1, 2, 5, 14, 41, 122, ... A007051
k= 3: 1, 3, 10, 36, 136, ... A007582
k= 4: 1, 4, 17, 76, 353, ... A081186
k= 5: 1, 5, 26, 140, 776, ... A081187
k= 6: 1, 6, 37, 234, 1513, ... A081188
k= 7: 1, 7, 50, 364, 2696, ... A081189
k= 8: 1, 8, 65, 536, 4481, ... A081190
k= 9: 1, 9, 82, 756, 7048, ... A060531
k=10: 1, 10, 101, 1030, ... A081192
p[n](k), k=0,1,...
p[0]: 1,1,1,1,1,1, ....... A000012
p[1]: 0,1,2,3,4,5, ....... A001477
p[2]: 1,2,5,10,17,26, .... A002522
p[3]: 0,4,14,36,76,140, .. A079908 (End)
Cf. A000182, A133314, A153641.

Programs

  • Haskell
    a119467 n k = a119467_tabl !! n !! k
a119467_row n = a119467_tabl !! n
a119467_tabl = map (map (flip div 2)) $
               zipWith (zipWith (+)) a007318_tabl a130595_tabl
-- Reinhard Zumkeller, Mar 23 2014
  • Magma
    /* As triangle */ [[Binomial(n, k)*(1 + (-1)^(n - k))/2: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 26 2015
  • Maple
    # Polynomials: p_n(x)
p := proc(n,x) local k, pow; pow := (n,k) -> `if`(n=0 and k=0,1,n^k);
add((k+1 mod 2)*binomial(n,k)*pow(x,n-k),k=0..n) end;
# Coefficients: a(n)
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
  • Mathematica
    Table[Binomial[n, k] (1 + (-1)^(n - k))/2, {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Sep 06 2015 *)
n = 15; "n-th row"
mat = Table[Table[0, {j, 1, n + 1}], {i, 1, n + 1}];
mat[[1, 2]] = 1;
mat[[n + 1, n]] = 1;
For[i = 2, i <= n, i++, mat[[i, i - 1]] = (i - 1)/n ];
For[i = 2, i <= n, i++, mat[[i, i + 1]] = (n - i + 1)/n];
mat // MatrixForm;
P2 = Dot[mat, mat];
R1 = Simplify[
  Eigenvectors[Transpose[P2]][[1]]/
   Total[Eigenvectors[Transpose[P2]][[1]]]]
R2 = Table[Dot[R1, Transpose[mat][[k]]], {k, 1, n + 1}]
odd = R2*2^(n - 1) (* _Luca Onnis *)
  • Sage
    @CachedFunction
def A119467_poly(n):
    R = PolynomialRing(ZZ, 'x')
    x = R.gen()
    return R.one() if n==0 else R.sum(binomial(n,k)*x^(n-k) for k in range(0,n+1,2))
def A119467_row(n):
    return list(A119467_poly(n))
for n in (0..10) : print(A119467_row(n)) # Peter Luschny, Jul 16 2012

Formula

G.f.: (1-x*y)/(1-2*x*y-x^2+x^2*y^2);
T(n,k) = C(n,k)*(1+(-1)^(n-k))/2;
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.
Column k has e.g.f. cosh(x)*x^k/k!. - Paul Barry, May 26 2006
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
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
T(n,k) = (A007318(n,k) + A130595(n,k))/2, 0<=k<=n. - Reinhard Zumkeller, Mar 23 2014

Extensions

Edited by N. J. A. Sloane, Jul 14 2009

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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