Combinatorics of geometrically distributed random variables: New q-tangent and q-secant numbers

Up-down permutations are counted by tangent resp. secant numbers. Considering words instead, where the letters are produced by independent geometric distributions, there are several ways of introducing this concept; in the limit they all coincide with the classical version. In this way, we get some new q-tangent and q-secant functions. Some of them also have nice continued fraction expansions; in one particular case, we could not find a proof for it. Divisibility results a la Andrews/Foata/Gessel are also discussed.


Introduction
For odd n, the number of them is given by n![z n ] tan z, and for even n by n![z n ] sec z.One finds that in many textbooks, e. g. [8].
Instead of speaking about exponential generating functions, we prefer to think of the coefficients of tan z and sec z as probabilities.
Parts of this research were conducted while the author was a visitor of the Technical University of Graz where he was supported by the start project Y96-MAT.

Recursions
We introduce the functions where the coefficient of u i in it is the probability that a word of length n satisfies the ≤≥≤≥ . . .condition and ends with the letter i.Also, we define which drops the technical condition about the last letter.
Furthermore, we introduce the generating functions Quantities like F ≤> (z, u) etc. are defined in an obvious way.
For the instance of secant numbers, we define similar quantities, but use the letters S, σ, G, g instead of T, τ, F, f .
Obviously we only get nonzero contributions for odd n in the tangent case and for even n in the secant case.
The reason to operate with a variable u that controls the last letter is the technique of "adding a new slice," that was applied with success in [6] and, more recently, in [12].
Theorem 1.The functions T ∇△ 2n+1 (u) satisfy the following recurrences: • (2.5) • • • (2.9) Proof.Since the technique is the same for all the instances, it is enough to discuss e. g. the "≥≤" case.Adding a new slice means adding a pair (k, j) with 1 ≤ k ≤ i, j ≥ k, replacing u i by 1 and providing the factor u j .But which explains the recursion.The starting value is just Theorem 2. The numbers τ ∇△ 2n+1 have the generating functions f ∇△ (z) = tan q (z) = sin q (z)/ cos q (z): Proof.The proofs of the first 3 relations are very similar, and we only sketch the first instance.Summing up we find Iterating that we find for f (z) = f ≥≤ (z): from which the announced formula follows by solving for f (z).
The 3 others are trickier, because of a term T ∇∆ 2n−1 (q).Again, let us discuss one case.Observe that , because one more "up" step should replace u i by k≥i pq k−1 = q i−1 .Now the generating function g ≤≥ (z) of the quantities S ≤≥ 2n (1) (upcoming) is obtained independently, whence we get Now iteration as usual derives the desired result.
Theorem 3. The functions S ∇△ 2n (u) satisfy the following recurrences: (2.17) (2.18) (2.19) Proof.The proof works as in the easy cases of the tangent recursions and is omitted.
For the starting value, we must consider the first pair of numbers.
Theorem 4. The numbers σ ∇△ 2n have the generating functions g ∇△ (z) = 1/ cos q (z): Proof.The proofs are quite similar as before; however, iteration must be done for the function G ∇∆ (z, u) − 1, and 1 must be added at the end.
3. Jackson's q-sine and q-cosine functions Jackson in [11] has introduced the functions and proved the relation sin q (z) sin 1/q (z) + cos q (z) cos 1/q (z) = 1 (3.2) Since we have here several q-sine and q-cosine functions, we call them a q-sine-cosine pair, if relation (3.2) holds.
Theorem 5.For the functions and exactly the 12 pairs in Table 1 are q-sine-cosine pairs: Proof.The desired relation gives us more and more restrictions when we look at the coefficients of z 2n .By a tedious search that will not be reported here we find these 12 possibilities, and all others can be excluded.The proof that this indeed works is very similar for all of them, so we give just one, namely the instance (1, 0, 1, 0).
Note the following expansions: So we must prove that for n ≥ 1 or, reversing the order of summation in the second sum, We rewrite this again as Therefore we have to prove that We use the formula (10.0.9) in [2], see also [1], (1 + q j z).
The desired result now follows by replacing n by 2n and plugging in z = −q −n .
Theorem 6.The 6 tan q (z) functions in Theorem 2 all involve q-sine-cosine pairs.
Remark.Replacing q by 1/q in the q-sine-cosine pairs and rewriting everything again in the q-notation means replacing the vector (A, B, C, D) of exponents by(2 This reduces the 12 pairs to 6 pairs.

continued fractions
Some of the 12 tangent functions have nice continued fraction expansions.
The two tangent functions coincide, which is classical, since Jackson [11] has shown that for his functions sin q z cos 1/q z − sin 1/q z cos q z = 0 holds.
Proof.For the proof by induction we must do the following: Set a n = [2n − 1] q q 1−n and We must show that By the induction we only have to show that [z 2n−1 ] p n (z) cos q z − q n (z) sin q z = 0.
This finishes the proof.
Proof.The proof follows the same lines; this time the polynomials (continuants) are (1 + q i ) and .
Hence we have to prove that ; from here on we can use the previous proof.
An alternative proof is by noting that tan (0,1,0,1) and using the previous result.
Although we do not need it, we also mention the dual formula A similar computation gives the result (n ≥ 1) Now we write tan q z = sinq z cos 1/q z cosq z cos 1/q z and thus cos q z cos 1/q z n≥0 T 2n+1 (q) [2n + 1] q !z 2n+1 = sin q z cos 1/q z.
The new q-secant numbers do not enjoy any divisibility results that are worthwhile to report; for the classical ones, see [3].
Plugging in iz(1−q) for z and taking real parts would result in the q-cosine with factor q n 2 .To get the corresponding q-sine, replace z by izq(1 − q), take the imaginary part and multiply by q 1/4 .We consider that merely to be a curiosity, not being of much help.