Blogrolling On III

Note that if we always set $x = 1$, $J(x,y)$ has a non-zero imaginary part when $-1 < y < 3$. If $y = 2$, and thus $z = 2x$, the coefficients take the form

$C_{i} = 2^i S_i$

for the Schroeder numbers $S_i$. Then $J$ is a cubed root of unity $\omega$, and the generating function corresponds to the rule

$J = 1 + 2J + J^2$

Can we prove a law $J^4 = J$? Similarly, if $y = 3$ and $J = -1$ the rule would be $J = 1 + 3J + J^2$. Can we interpret these rules in terms of trees? Let’s write the first rule as $J = 1 + J + J + J^2$. This looks a lot like the Motzkin rule, except for the extra factor of $J$. What if we distinguished left and right branches for Motzkin trees? That is, take a full binary rooted tree template and count whole trees with 0, 1 or 2 branches that may be fitted to the template. Then the desired rule works by differentiating left and right unary branches from the root. Instead of a fivefold bijection of the set of Motzkin trees, there is now a fourfold mapping, and the series

$1 + 4 + 24 + 176 + 1440 + \cdots = \omega$

Motzkin numbers $M_i$ are also given in terms of trinomial coefficients $T(n,1)$. The $T(n,0)$ coefficients go back to Euler. These are the number of permutations of $n$ ternary symbols (-1, 0, or 1) which sum to 0. A general $T(n,k)$ is the number of permutations of these symbols that sum to $k$.

5 Responses so far »

  1. 1

    Matt Noonan said,

    Sure enough, the substitution rule J = 1 + 2J + J^2 results in an isomorphism J = J^4! It is essentially the same as in the Motzkin case: expand until you get to a J^5 term, then start collapsing terms with the lowest power of J. You seem to need an extra expansion at one point to “borrow” some powers of J. I want to think about this more after I get all these calculus tests graded…

  2. 2

    Kea said,

    Hi Matt! Good to see you here – and thanks for working through it. I’d also love to spend more time on this, because I’m sure its related to the physics we do here – but then there are so many things to do …..

  3. 3

    Doug said,

    Hi Kea,

    The series 1 4 24 176 1440 can be found – intermixed – among a larger sequence of Sloane’s A036912 ‘Maximum inverse of phi(n) increases’ [author David W. Wilson].

    http://www.research.att.com/~njas/sequences/?q=1+4+24+176+1440&sort=0&fmt=0&language=english

    [see the list and graph as well]

    There does not appear to be any exact Sloane match for the shorter series.

    I do not know if this is significant.

  4. 4

    Doug said,

    Hi Kea,

    The list of my August 08, 2007 12:29 AM comment, [back] correlates:

    1 —-> 1
    4 —-> 3
    24 —> 9
    176 –> 22
    1440 -> 43

    There does not appear to be any exact Sloane match for 1,3,9,22,43.

    There are, however, 2914 various listings which include 1 3 9 22 43, but not necessarily in that order and many have duplicates of the individual numbers.

    The first 1-10 of 2914 examples at

    http://www.research.att.com/~njas/sequences/?q=1+3+9+22+43&sort=0&fmt=0&language=english

    1 – Array read by antidiagonals, generated by the matrix M = [1,1,1;1,N,1;1,1,1];.

    2 – Triangular array T read by rows: T(i,0)=T(i,2i)=1 for i >= 0, T(i,1)=T(i,2i-1)=[ i/2 ] for i >= 1, and for i >= 2 and 2<=j<=2i-2, T(i,j)=T(i-1,j-2)+T(i-1,j-1)+T(i-1,j) if i+j is odd, T(i,j)=T(i-1,j-2)+T(i-1,j) if i+j is even.
    3 – Continued fraction expansion of Pi^e.
    4 – Continued fraction for Li(2). 5 – Inverse permutation to A084491.
    6 – Inverse permutation to A084495.
    7 – Inverse permutation to A084499.
    8 – Continued fraction for cube root of 13.
    9 – 3^n mod 89.

    10- a(1) = 1, a(2) = 2, and a(n) = smallest number not included earlier that divides the sum of two previous terms.

    This is interesting, relating one sequence to 2914 sequences, but I do not know what it means.

    I did look at 11-30 [and the last three pages 290-292] but not beyond.

  5. 5

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