OFFSET
0,5
COMMENTS
Comment from Peter Bala (Dec 06 2011): "Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of (I-t*M)^-1 (apart from the initial 1) lists the row polynomials for" A196776(n,k), which gives the number of ordered partitions of an n set into k odd-sized blocks. - Peter Luschny, Dec 06 2011
The n-th row of the triangle is formed 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 probability 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 in the n-th row of this triangle is 2^(n-1). Furthermore, by the properties of Markov chains, we can interpret P^(2k) as the (2k)-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). - Luca Onnis, Oct 29 2023
REFERENCES
Paul and Tatjana Ehrenfest, Über zwei bekannte Einwände gegen das Boltzmannsche H-Theorem, Physikalische Zeitschrift, vol. 8 (1907), pp. 311-314.
LINKS
Luca Onnis, Animation of the Ehrenfest model.
Wikipedia, Ehrenfest model.
FORMULA
p_n(x) = Sum_{k=0..n} (k mod 2)*binomial(n,k)*x^(n-k).
E.g.f.: exp(x*t)/csch(t) = 0*(t^0/0!) + 1*(t^1/1!) + (2*x)*(t^2/2!) + (3*x^2+1)*(t^3/3!) + ...
The 'co'-polynomials with generating function exp(x*t)*sech(t) are the Swiss-Knife polynomials (A153641).
EXAMPLE
Triangle begins:
0
1, 0
0, 2, 0
1, 0, 3, 0
0, 4, 0, 4, 0
1, 0, 10, 0, 5, 0
0, 6, 0, 20, 0, 6, 0
1, 0, 21, 0, 35, 0, 7, 0
...
p[0](x) = 0;
p[1](x) = 1
p[2](x) = 2*x
p[3](x) = 3*x^2 + 1
p[4](x) = 4*x^3 + 4*x
p[5](x) = 5*x^4 + 10*x^2 + 1
p[6](x) = 6*x^5 + 20*x^3 + 6*x
p[7](x) = 7*x^6 + 35*x^4 + 21*x^2 + 1
p[8](x) = 8*x^7 + 56*x^5 + 56*x^3 + 8*x
.
Cf. the triangle of odd-numbered terms in rows of Pascal's triangle (A034867).
p[n] (k), n=0,1,...
k=2: 0, 1, 4, 13, 40, 121, ... A003462
k=3: 0, 1, 6, 28, 120, 496, ... A006516
k=4: 0, 1, 8, 49, 272, 1441, ... A005059
k=10: 0, 1, 20, 301, 4040, 51001, ... ......., (A016190)
.
p[n] (k), k=0,1,...
p[0]: 0, 0, 0, 0, 0, 0, ... A000004
p[1]: 1, 1, 1, 1, 1, 1, ... A000012
p[2]: 0, 2, 4, 6, 8, 10, ... A005843
p[3]: 1, 4, 13, 28, 49, 76, ... A056107
p[4]: 0, 8, 40, 120, 272, 520, ... A105374
p[5]: 1, 16, 121, 496, 1441, 3376, ...
p[6]: 0, 32, 364, 2016, 7448, 21280, ...
MAPLE
# Polynomials: p_n(x)
p := proc(n, x) local k;
pow := (n, k) -> `if`(n=0 and k=0, 1, n^k);
add((k 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)/csch(t), t, 16), t, i), x, n), n=0..i)), i=0..8);
MATHEMATICA
p[n_, x_] := Sum[Binomial[n, 2*k-1]*x^(n-2*k+1), {k, 0, n+2}]; row[n_] := CoefficientList[p[n, x], x] // Append[#, 0]&; Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Jun 28 2013 *)
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}]
even = R1*2^(n - 1) (* Luca Onnis, Oct 29 2023_ *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Peter Luschny, Jul 07 2009
STATUS
approved