5
$\begingroup$

Wright's expansion of

$$ (1-z)^b\exp[A/(1-z)^c], \text{for } A > 0, 0<c<1\tag{1} $$

is, in the words of the late, great Mark Kac "well known to those that know it well". (See, for example, p563 of Flajolet and Sedgewick, VIII.7).

In particular, the asymptotic form of the n^th coefficient behaves as

$$ \text{const. } n^a \cdot exp(A(c+1)N^c),\tag{2} $$

where $N=(n/Ac)^{1/(c+1)}$, and "$a$" depends on $b$ and $c$.

So, in particular, the sign of the exponential is manifestly positive.

In recent numerical experiments, I have found situations involving Dyck paths under compression, self-avoiding walks and bridges under compression, and certain types of pattern avoiding permutations, where the asymptotic form of the coefficients in the o.g.f. is apparently of the form

$$ \text{const. }\mu^n. \mu_1^{n^\sigma}. n^g, \text{ where in all cases } \mu_1 < 1\tag{3} $$

To state the obvious, $\mu_1 < 1$ means that the sign of the exponential is negative, if I rewrite the variable $\mu_1$ as $\exp(t)$, so the conditions of Wright's expansion are not satisfied.

My question is, what closed form expression generates this form (3) of coefficients, in the sense that (1) generates (2)?

(There is the obvious result that the o.g.f. of the reciprocal of the coefficients, behaving as (3), looks like (2) and so the og of the reciprocal series can be said to be generated by (1), which is partially useful for my purposes, but I'd like to know if the actual ogf has a simple closed form expression?)

(Incidentally, making $A<0$ in (1) gives a series with periodic sign changes, which is definitely not how my series behave).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.