DERIVATIVE POLYNOMIALS, EULER POLYNOMIALS,

AND ASSOCIATED INTEGER SEQUENCES

Michael E. Hoffman

Mathematics Department U.S. Naval Academy Annapolis, MD 21402 meh@nadn.navy.mil Submitted: February 16, 1999; Accepted: April 2, 1999

Abstract. Let P_{n} and Q_{n} be the polynomials obtained by repeated differentiation of the tangent and secant functions respectively. From the exponential generating functions of these polynomials we develop relations among their values, which are then applied to various numerical sequences which occur as values of the P_{n} and Q_{n}. For example, P_{n}(0) and Q_{n}(0) are respectively the nth tangent and secant numbers, while P_{n}(0) + Q_{n}(0) is the nth André number. The André numbers, along with the numbers Q_{n}(1) and P_{n}(1) − Q_{n}(1), are the Springer numbers of root systems of types A_{n}, B_{n}, and D_{n} respectively, or alternatively (following V. I. Arnol’d) count the number of “snakes” of these types. We prove this for the latter two cases using combinatorial arguments. We relate the values of P_{n} and Q_{n} at \sqrt{3} to certain “generalized Euler and class numbers” of D. Shanks, which have a combinatorial interpretation in terms of 3-signed permutations as defined by R. Ehrenborg and M. A. Readdy. Finally, we express the values of Euler polynomials at any rational argument in terms of P_{n} and Q_{n}, and from this deduce formulas for Springer and Shanks numbers in terms of Euler polynomials.

1. Introduction. Consider the sequences P_{n} and Q_{n} of “derivative polynomials” defined by

\frac{d^{n}}{dx^{n}} tan x = P_{n}(tan x) and \frac{d^{n}}{dx^{n}} sec x = Q_{n}(tan x) sec x

for integer n ≥ 0. As shown in [12], their exponential generating functions

P (u, t) = ∑_{n=0}^{∞} P_{n}(u)\frac{t^{n}}{n!} and Q(u, t) = ∑_{n=0}^{∞} Q_{n}(u)\frac{t^{n}}{n!}

are given by the explicit formulas

(1)  P (u, t) = \frac{sin t + u cos t}{cos t − u sin t} and Q(u, t) = \frac{1}{cos t − u sin t}.

1991 Mathematics Subject Classification. 11B83, 11B68, 05A15. Key words and phrases. Tangent numbers, secant numbers, André numbers, Springer numbers, snakes,

generalized Euler and class numbers, Euler polynomials.

Typeset by AMS-TEX

In §2 we obtain from identities in these generating functions some useful relations among values of the polynomials (Theorem 2.2 below), and recall from [12] a result (Theorem 2.3) relating the polynomials to series of reciprocal powers. In §3 we apply results of §2 to the computation of P_{n}(u) and Q_{n}(u) for u = 0, 1, \sqrt{3}, and 1/\sqrt{3}, in the process obtaining several integer sequences studied by Glaisher [9,10,11].

In §4 we give a combinatorial interpretation of the values of the derivative polynomials at 0 and 1. These values give the Springer numbers of the irreducible root systems A_{n}, B_{n}, and D_{n} [19], which also count the corresponding types of “snakes” as defined in [3]. (Snakes for the root system A_{n−1} are alternating permutations of {1, 2, …, n}, whose study dates back to André [2].) This follows from comparison of equations (1) with the generating functions found in [19], but we also give combinatorial proofs using snakes (Theorems 4.2 and 4.3). Results from §3 then give identities for the Springer numbers (e.g., Proposition 4.4).

In §5 we recall the definition of the the “generalized Euler and class numbers” of Shanks [17]. These are arrays of positive integers c_{a,n} and d_{a,n}. The first two “rows” (i.e., the c_{a,n} and d_{a,n} with a = 1, 2) are the Springer numbers of the preceding paragraph; we show that the third row of Shanks’s numbers are given by the values P_{2n}(\sqrt{3}) and Q_{2n−1}(\sqrt{3}) of the derivative polynomials. We also give a combinatorial interpretation to the numbers c_{3,n} and d_{3,n} in terms of 3-signed alternating permutations as defined by Ehrenborg and Readdy [8].

In §6 we consider the Euler polynomials E_{n}(x), defined by

(2)  \frac{2e^{tx}}{e^{t} + 1} = ∑_{n=0}^{∞} E_{n}(x)\frac{t^{n}}{n!} .

Euler polynomials appear in many classical results (see Chapter 23 of [1]). In [6], the values of these polynomials at rational arguments were expressed in terms of the Hurwitz zeta function. Here we give explicit formulas for the Euler polynomials at rational arguments in terms of the polynomials P_{n} and Q_{n} (Theorem 6.1), and use them together with the computations of §3 to find E_{n}(p/q) for 0 ≤ p ≤ q and q = 2, 3, 4 and 6. We also write the Springer and Shanks numbers in terms of values of the Euler polynomials (Theorem 6.2).

2. Derivative polynomials. From the chain rule it follows that the polynomials P_{n} satisfy P_{0}(u) = u and P_{n+1}(u) = (u^{2} + 1)P′_{n}(u), n ≥ 0, and similarly Q_{0}(u) = 1 and Q_{n+1}(u) = (u^{2} + 1)Q′_{n}(u) + uQ_{n}(u), n ≥ 0. The following result is then clear by induction on n.

Theorem 2.1. Let n ≥ 0. Then P_{n}(u) is a polynomial of degree n + 1 consisting of even powers with positive integral coefficients when n is odd and of odd powers with positive integral coefficients when n is even; and Q_{n}(u) is a polynomial of degree n consisting of even powers with positive integral coefficients when n is even and of odd powers with positive integral coefficients when n is odd.

In particular, for n ≥ 0 we have

P_{n}(−u) = (−1)^{n+1}P_{n}(u) and Q_{n}(−u) = (−1)^{n}Q_{n}(u).

The key properties of the P_{n} and Q_{n} come from two sorts of identities in their corresponding generating functions P and Q. First, there are the composition relations

(3)  P (P (u, t), s) = P (u, t + s) and Q(P (u, t), s)Q(u, t) = Q(u, t + s);

these follow from the representations

P (u, t) = tan(tan^{−1} u + t) and Q(u, t) = \frac{sec(tan^{−1} u + t)}{sec(tan^{−1} u)},

equivalent to equations (1) above. Second, there is the functional equation

(4)  P (u, t) = P (\frac{u^{2} − 1}{2u}, 2t) + \frac{u^{2} + 1}{2u}Q (\frac{u^{2} − 1}{2u}, 2t) ,

which follows from equations (1) and the half-angle formula for tangent (cf. Theorem 3.1 of [12]). The composition relations (3) imply the following relations among values of the polynomials P_{n} and Q_{n}.

Theorem 2.2. For nonnegative integers n,

(i)  P_{n}(P (u, s)) − tan s ∑_{k=0}^{n} \binom{n}{k}P_{k}(P (u, s))P_{n−k}(u) = P_{n}(u) + δ_{0n} tan s;

(ii)  Q_{n}(P (u, s)) − tan s ∑_{k=0}^{n} \binom{n}{k}Q_{k}(P (u, s))P_{n−k}(u) = (1 − u tan s)Q_{n}(u).

Proof. The composition relation for P gives

P (P (u, s), t) = P (u, s + t) = P (P (u, t), s) = \frac{sin s + cos sP (u, t)}{cos s − sin sP (u, t)},

or cos sP (P (u, s), t) − sin sP (P (u, s), t)P (u, t) = sin s + cos sP (u, t). Take the coefficient of t^{n}/n! and divide by cos s to get (i). The proof of (ii) proceeds similarly, using the composition relation for Q.

In [12] contour integration and expansion into power series were used to obtain closed forms in terms of the P_{n} and Q_{n} for certain series. We need the following definitions. Call a function ψ : Z → C periodic mod q if ψ(0) = 0 and ψ(n + q) = ψ(n) for all n ∈ Z, and alternating mod q if ψ(0) = 0 and ψ(n + q) = −ψ(n) for all n ∈ Z. If ψ is periodic or alternating mod q, we call it even if ψ(q − j) = ψ(j) for 0 < j < q, and odd if ψ(q − j) = −ψ(j) for 0 < j < q. From [12] we have the following result.

Theorem 2.3. Let n ≥ 0 be an integer. If ψ is periodic mod q, then

∑_{j=1}^{∞} \frac{ψ(j)}{j^{n+1}} = \frac{π^{n+1}}{2q^{n+1}n!} ∑_{p=1}^{q−1} ψ(p)P_{n}(cot \frac{pπ}{q})

provided n and ψ have opposite parity. If ψ is alternating mod q, then

∑_{j=1}^{∞} \frac{ψ(j)}{j^{n+1}} = \frac{π^{n+1}}{2q^{n+1}n!} ∑_{p=1}^{q−1} ψ(p) csc \frac{pπ}{q} Q_{n}(cot \frac{pπ}{q})

provided n and ψ have the same parity.

3. Particular values of derivative polynomials. By setting u = 0 in equations (1) we see that P_{n}(0) and Q_{n}(0) are respectively the tangent and secant numbers, i.e. the coefficients of t^{n}/n! in the Maclaurin series of tan t and sec t. We shall take these numbers as known. They can be computed from the Euler-Bernoulli triangle as discussed in [4] and [3]; see also [13]. In this section we show how to compute P_{n}(u) and Q_{n}(u) for u = 1, \sqrt{3}, and 1/\sqrt{3} from the tangent and secant numbers.

Set u = 1 and consider the coefficient of t^{n}/n! in the functional equation (4) to get P_{n}(1):

(5)  P_{n}(1) = 2^{n}(P_{n}(0) + Q_{n}(0)) = \{ \begin{matrix}2^{n}Q_{n}(0), & n even, \\ 2^{n}P_{n}(0), & n odd. \end{matrix}

Now compute Q_{n}(1) using the following result.

Theorem 3.1. For integers n ≥ 0,

Q_{n}(1) = − sin \frac{nπ}{2} + ∑_{2k≤n} \binom{n}{2k}(−1)^{k}P_{n−2k}(1).

Proof. Set u = 1 in Theorem 2.2(ii), and then let s → +∞i so that tan s → i and P (1, s) → i. This gives

(1 − i)Q_{n}(1) = Q_{n}(i) − i ∑_{k=0}^{n} \binom{n}{k}Q_{k}(i)P_{n−k}(1).

Now Q(i, t) = e^{it}, so Q_{n}(i) = i^{n}. Thus, we have

(1 − i)Q_{n}(1) = i^{n} − i ∑_{k=0}^{n} \binom{n}{k}i^{k}P_{n−k}(1).

Take the imaginary part to get the conclusion.

Remarks. 1. This result, together with the corresponding one obtained by taking the real part in the proof, is equivalent to the computation of the Q_{n}(1) from the numbers P_{n}(1) via Seidel matrices as described in [7].

2. Since the Q_{n}(1) turn out to be the Springer numbers b_{n} of the next section, they can be computed via the pair of “boustrophedonic” triangles L(b) and R(b) described in [3] (see also [14]); in fact these triangles are equivalent to Seidel matrices as is explained in [7].

3. The numbers Q_{n}(1) were extensively studied by Glaisher [9,11], who wrote P_{n} for Q_{2n}(1) and Q_{n} for Q_{2n−1}(1).

Set u = −1/\sqrt{3} in equation (4) and examine the coefficient of t^{n}/n! to get

(2^{n} + (−1)^{n})P_{n}(\frac{1}{\sqrt{3}}) = \frac{2^{n+1}}{\sqrt{3}} Q_{n}(\frac{1}{\sqrt{3}});

then combine this with the equation obtained by setting u = \sqrt{3} to get

P_{n}(\sqrt{3}) = (2^{n+1} + (−1)^{n})P_{n}(\frac{1}{\sqrt{3}}).

In view of these equations, to find P_{n}(\frac{1}{\sqrt{3}}) and Q_{n}(\frac{1}{\sqrt{3}}) it is enough to find P_{n}(\sqrt{3}). Our next two results give P_{n}(\sqrt{3}) and Q_{n}(\sqrt{3}).

Theorem 3.2. i. If n is odd, P_{n}(\sqrt{3}) = \frac{1}{2}(3^{n+1} − 1)P_{n}(0). ii. If n is even, Q_{n}(\sqrt{3}) = \frac{1}{4}(3^{n+1} + 1)Q_{n}(0).

Proof. Note first that cos 3t = cos t(2 cos 2t − 1), by the addition formula for cosine and the double-angle formulas for sine and cosine. Then

P (\sqrt{3}, t) = \frac{sin t + \sqrt{3} cos t}{cos t − \sqrt{3} sin t} = \frac{\sqrt{3} + 2 sin 2t}{2 cos 2t − 1} = \frac{(\sqrt{3} + 2 sin 2t) cos t}{cos 3t},

while on the other hand

3P (0, 3t) − P (0, t) = \frac{3(sin 2t cos t + sin t cos 2t) − (2 cos 2t − 1) sin t}{cos 3t}

= \frac{3 sin 2t cos t + 2 cos^{2} t sin t}{cos 3t} = \frac{4 sin 2t cos t}{cos 3t}.

Thus

(6)  P (\sqrt{3}, t) − \frac{1}{2}(3P (0, 3t) − P (0, t)) = \frac{\sqrt{3} cos t}{cos 3t},

and (i) follows from consideration of the coefficient of t^{n}/n!, n odd (Note the right-hand side is an even function). A similar argument proves the identity

(7)  Q(\sqrt{3}, t) − \frac{1}{4}(3Q(0, 3t) + Q(0, t)) = \frac{\sqrt{3}}{2} \frac{sin 2t}{cos 3t},

from which (ii) follows upon the observation that the right-hand side is an odd function.

Theorem 3.3. i. If n > 0 is even, P_{n}(\sqrt{3}) can be computed from the tangent numbers P_{k}(0) via

P_{n}(\sqrt{3}) = \frac{\sqrt{3}}{2} ∑_{k odd} \binom{n}{k}(3^{k+1} − 1)P_{k}(0)P_{n−k}(0).

ii. If n is odd, Q_{n}(\sqrt{3}) can be computed from the tangent numbers P_{k}(0) and the secant numbers Q_{k}(0) via

Q_{n}(\sqrt{3}) = \frac{\sqrt{3}}{8} ∑_{k odd} \binom{n}{k}(3^{k+1} − 1)P_{k}(0)Q_{n−k}(0).

Proof. For (i), set s = \frac{π}{3} and u = 0 in Theorem 2.2(i) to get

P_{n}(\sqrt{3}) − \sqrt{3} ∑_{k=0}^{n} \binom{n}{k}P_{k}(\sqrt{3})P_{n−k}(0) = P_{n}(0) + \sqrt{3}δ_{0n}.

Now suppose n > 0 is even; then this reduces to

P_{n}(\sqrt{3}) = \sqrt{3} ∑_{k odd} \binom{n}{k}P_{k}(\sqrt{3})P_{n−k}(0) = \sqrt{3} ∑_{k odd} \binom{n}{k}\frac{3^{k+1} − 1}{2}P_{k}(0)P_{n−k}(0),

where we have used Theorem 3.2(i) in the last step. For (ii), proceed similarly after setting s = \frac{π}{3} and u = −\sqrt{3} in Theorem 2.2(ii).

Remarks. 1. Comparing equation (6) to the equation at the beginning of §24 in [10], we see the numbers H_{n} of [9,10] are given by H_{n} = \sqrt{3}P_{2n}(\sqrt{3})/2^{2n+1}.

2. Similarly, equation (7) shows that the numbers T_{n} of [9] are Q_{2n−1}(\sqrt{3})/\sqrt{3}.

3. In Theorem 5.1 below we show that the numbers P_{2n}(\sqrt{3}) and Q_{2n−1}(\sqrt{3}) are closely related to certain generalized Euler and class numbers as defined in [17].

4. Root systems, values of derivative polynomials at 0 and 1, and the combinatorics of snakes. Let V be a real vector space, R a root system in V, and W the Weyl group of R (for definitions see [5]). Fix a set S of simple roots for R: then any α ∈ R is either a positive or negative linear combination of elements of S; we write α > 0 in the first case and α < 0 in the second. For I ⊂ S, denote by σ(I, S) the number of elements w ∈ W such that wα > 0 for α ∈ I and wα < 0 for α ∈ S − I. Let M(R) be the maximum value of σ(I, S). T. A. Springer [19] computed the quantity M(R) for all irreducible root systems R. Setting aside the exceptional root systems, his results are as follows.

1. If R is of type A_{n}, n ≥ 1, then M(R) = a_{n} satisfies

1 + t + ∑_{n≥2} \frac{a_{n−1}}{n!} t^{n} = tan t + sec t.

2. If R is of type B_{n} or C_{n}, n ≥ 2, then M(R) = b_{n} satisfies

1 + t + ∑_{n≥2} \frac{b_{n}}{n!} t^{n} = \frac{cos t + sin t}{cos 2t}.

3. If R is of type D_{n}, n ≥ 3, then M(R) = d_{n} satisfies

t + \frac{1}{2}t^{2} + ∑_{n≥3} \frac{d_{n}}{n!} t^{n} = \frac{1 + sin 2t − cos t − sin t}{cos 2t}.

We shall call M(R) the Springer number of the root system R.

Proposition 4.1. The Springer numbers a_{n}, b_{n}, and d_{n} as defined above are given by a_{n} = P_{n+1}(0) + Q_{n+1}(0), b_{n} = Q_{n}(1), and d_{n} = P_{n}(1) − Q_{n}(1).

Proof. This amounts to writing the the generating functions above in terms of P and Q. For example, the formula for a_{n} follows from observing that P (0, t)+Q(0, t) = tan t+sec t. Similarly,

Q(1, t) = \frac{1}{cos t − sin t} = \frac{cos t + sin t}{cos 2t}

and

P (1, t) − Q(1, t) = \frac{sin t + cos t − 1}{cos t − sin t} = \frac{1 + sin 2t − cos t − sin t}{cos 2t}.

By describing Springer numbers geometrically in terms of Weyl chambers, Arnol’d [3] showed that the numbers a_{n}, b_{n}, and d_{n} can be thought of as counting various types of snakes (updown sequences). The formal definitions are as follows.

Definition. A snake of type A_{n} is a sequence (x_{0}, x_{1}, …, x_{n}) of integers such that x_{0} < x_{1} > x_{2} < ⋯ x_{n} and {x_{0}, x_{1}, …, x_{n}} = {0, 1, …, n}. A snake of type B_{n} is a sequence (x_{1}, x_{2}, …, x_{n}) of integers such that 0 < x_{1} > x_{2} < ⋯ x_{n} and {|x_{1}|, |x_{2}|, …, |x_{n}|} = {1, 2, …, n}. A snake of type D_{n} is a sequence (x_{1}, …, x_{n}) of integers such that −x_{2} < x_{1} < x_{2} > x_{3} < ⋯ x_{n} and {|x_{1}|, |x_{2}|, …, |x_{n}|} = {0, 1, …, n − 1}.

We shall write A_{n} for the set of snakes of type A_{n} and so forth. The geometric argument of [3] shows that card A_{n}, card B_{n}, and card D_{n} are a_{n}, b_{n}, and d_{n} respectively. On the other hand, it is possible to prove that these cardinalities are given by the formulas of Proposition 4.1 using combinatorial arguments about snakes. This is done for A_{n} in [2] and [3] (see the remark following Theorem 13): we do it here for B_{n} and D_{n}. For this purpose, it is convenient to introduce another type of snake from [3]: an integer sequence (x_{1}, …, x_{n}) such that x_{1} < x_{2} > x_{3} < ⋯ x_{n} and {|x_{1}|, …, |x_{n}|} = {1, …, n} is called a snake of type β_{n}. Let β_{n} denote the set of snakes of type β_{n}.

Theorem 4.2. Let b(t) = ∑_{n≥0} card B_{n}t^{n}/n! and β(t) = ∑_{n≥0} card β_{n}t^{n}/n!. Then b(t) = Q(1, t) and β(t) = P (1, t) (so card B_{n} = Q_{n}(1) and card β_{n} = P_{n}(1)).

Proof. We prove the formula for β(t) first. Given (x_{1}, …, x_{n+1}) ∈ β_{n+1}, let r be the unique element of {0, …, n} with |x_{r+1}| = n + 1. Then the sets {|x_{1}|, …, |x_{r}|} and {|x_{r+2}|, …, |x_{n+1}|} partition {1, …, n}. The sequence (x_{1}, …, x_{r}) can be shrunk into a snake of type β_{r} by applying the order-preserving bijection of {|x_{1}|, …, |x_{r}|} onto {1, …, r}; similarly ((−1)^{r+1}x_{r+2}, …, (−1)^{r+1}x_{n+1}) gives a snake of type β_{n−r}. Conversely, given r ∈ {0, …, n} and a partition of {1, …, n} into an r-set and an (n − r)-set, together with elements of β_{r} and β_{n−r}, we can construct a β_{n+1}-snake in a unique way. Hence

card β_{n+1} = ∑_{r=0}^{n} \binom{n}{r} card β_{r} card β_{n−r} + δ_{n0}.

from which follows β′(t) = β(t)^{2} + 1. (The Kronecker delta term reflects the fact that there are two β_{1}-snakes, (1) and (−1).) The unique solution of this differential equation satisfying the initial condition β(0) = card β_{0} = 1 (β_{0} consists of the empty snake) is

β(t) = tan(t + \frac{π}{4} ) = \frac{tan t + 1}{1 − tan t} = P (1, t).

Now suppose (x_{1}, …, x_{n+1}) ∈ B_{n+1}, with r ∈ {0, …, n} such that |x_{r+1}| = n + 1. Again the sequences (x_{1}, …, x_{r}) and (x_{r+2}, …, x_{n+1}) consist of integers whose absolute values partition {1, …, n}. The sequence (x_{1}, …, x_{r}) can be shrunk into a B_{r}-snake, since x_{1} > 0; but the shrinkage of the sequence ((−1)^{r}x_{r+2}, …, (−1)^{r}x_{n+1}) is a snake of type β_{n−r}. Hence

card B_{n+1} = ∑_{r=0}^{n} \binom{n}{r} card B_{r} card β_{n−r}

and we have b′(t) = b(t)β(t). Using b(0) = 1 and our formula for β(t), this gives

b(t) = \frac{1}{\sqrt{2}} sec(t + \frac{π}{4} ) = Q(1, t).

Theorem 4.3. card β_{n} = card B_{n} + card D_{n} (so card D_{n} = P_{n}(1) − Q_{n}(1)).

Proof. First note that we have a partition β_{n} = β_{n}^{−}∪β_{n}^{+}, where β_{n}^{−} and β_{n}^{+} are respectively the sets of β_{n}-snakes that start with a negative integer and with a positive integer. We shall define bijections f : β_{n}^{−} → B_{n} and g : β_{n}^{+} → D_{n}. Let f(x_{1}, …, x_{n}) = (−x_{1}, …, −x_{n}): it is easy to see that f is a bijection of β_{n}^{−} onto B_{n}. For g, let (x_{1}, …, x_{n}) ∈ β_{n}^{+}, with r ∈ {1, …, n} such that |x_{r}| = 1. Then g(x_{1}, …, x_{n}) = (x_{r}x̃_{1}, x̃_{2}, …, x̃_{n}), where x̃_{i} = (sgn x_{i})(|x_{i}| − 1). The reader may verify that the image of g is in D_{n}, and in fact that g has an inverse given by

g^{−1}(y_{1}, …, y_{n}) = \{ \begin{matrix}(1, ŷ_{2}, …, ŷ_{n}), & if r = 1; \\ (|ŷ_{1}|, ŷ_{2}, …, ŷ_{r−1}, sgn y_{1}, ŷ_{r+1}, …, ŷ_{n}), & otherwise; \end{matrix}

for (y_{1}, …, y_{n}) ∈ D_{n} with y_{r} = 0, and ŷ_{i} = (sgn y_{i})(|y_{i}| + 1) for i ≠ r.

We can use the machinery of previous sections to obtain relations among the Springer numbers (and card β_{n}). For example, equation (5) above implies card β_{n} = 2^{n}a_{n−1}, which has a simple combinatorial interpretation in terms of snakes (cf. Theorem 24 of [3]). Other relations, like the following, appear to be new.

Proposition 4.4. For positive integers n,

b_{n} = \frac{(−1)^{n} + 1}{2}a_{n−1} + ∑_{k odd} \binom{n}{k}a_{k−1}b_{n−k}

and

d_{n} = (−1)^{n−1}a_{n−1} + ∑_{k odd} \binom{n}{k}a_{k−1}d_{n−k}.

Proof. Set s = \frac{π}{4} and u = 0 in Theorem 2.2: then the first identity follows from part (ii) of the theorem, and the second upon subtracting part (ii) from part (i).

Remark. The formula for b_{n} can be given a combinatorial interpretation. Let Ā_{m} be the set of sequences (x_{0}, …, x_{m}) such that {x_{0}, …, x_{m}} = {0, 1, …, m} and x_{0} > x_{1} < x_{2} > ⋯ x_{m}: evidently Ā_{m} is in 1-1 correspondence with A_{m} via (x_{0}, …, x_{m}) → (m−x_{0}, …, m− x_{m}). Now suppose (x_{1}, …, x_{n}) ∈ B_{n}. If all the x_{i} are positive, then (x_{1} − 1, …, x_{n} − 1) ∈ Ā_{n−1}. Otherwise, there is a smallest k ∈ {1, …, n} with x_{k} < 0, and it follows from the definition of B_{n} that k must be even. Then (x_{1}, …, x_{k−1}) can be shrunk into an element of Ā_{k−2}, and (−x_{k}, −x_{k+1}, …, −x_{n}) can be shrunk into an element of B_{n−k+1}. Since there are \binom{n}{k−1} ways to choose \{x_{1}, …, x_{k−1}\} ⊂ \{1, …, n\}, we have

b_{n} = a_{n−1} + ∑_{2≤k≤n even} \binom{n}{k − 1}a_{k−2}b_{n−k+1} = a_{n−1} + ∑_{k≤n−1 odd} \binom{n}{k}a_{k−1}b_{n−k},

which is equivalent to the first identity above.

5. Values of derivative polynomials at \sqrt{3}, generalized Euler and class numbers, and 3-signed permutations. In [17] Shanks defined positive integers c_{a,n} (for integer a ≥ 1 and n ≥ 0) and d_{a,n} (for integer a, n ≥ 1) by

L_{a}(2n + 1) = K_{a}\sqrt{a}\frac{c_{a,n}}{(2n)!} (\frac{π}{2a})^{2n+1} and L_{−a}(2n) = K_{a}\sqrt{a}\frac{d_{a,n}}{(2n − 1)!} (\frac{π}{2a})^{2n} ,

where K_{a} = \frac{1}{2} if a = 1 and 1 otherwise, and

L_{a}(s) = ∑_{k=0}^{∞} (\frac{−a}{2k + 1}) (2k + 1)^{−s};

here (−a/(2k + 1)) is the Jacobi symbol. As noted in [17], the numbers c_{1,n} are just the secant numbers Q_{2n}(0), and the d_{1,n} are the tangent numbers P_{2n−1}(0). Comparison of the tables in [17] and those of [3] reveals that the numbers c_{2,n} and d_{2,n} are Springer numbers: in fact c_{2,n} = Q_{2n}(1) = b_{2n} and d_{2,n} = Q_{2n−1}(1) = b_{2n−1}. This can be proved using the recurrences given in [17] together with the generating function for the Q_{n}(1): see Proposition 6.3 of [16], where b_{n} is denoted E_{n}^{±}. Our next result gives the third row of Shanks’s numbers in terms of the numbers P_{2n}(\sqrt{3}) and Q_{2n−1}(\sqrt{3}) discussed in Theorem 3.3 above.

Theorem 5.1. i. For n ≥ 0, c_{3,n} = \frac{1}{\sqrt{3}}P_{2n}(\sqrt{3}). ii. For n ≥ 1, d_{3,n} = \frac{2}{\sqrt{3}}Q_{2n−1}(\sqrt{3}).

Proof. By substituting into the first of equations (19) of [17] the constants corresponding to the expression for L_{3}(s) in equations (19) of [18], we have

(8)  ∑_{n=0}^{∞} w^{2n} \frac{c_{3,n}}{(2n)!} = \frac{cos(3w(1 − 4/3))}{cos 3w} = \frac{cos w}{cos 3w} ,

and comparison with equation (6) above proves (i). Similarly, substitute into the second of equations (19) of [17] the constants from the expression for L_{−3}(s) in equations (19) of [18] to get

(9)  ∑_{n=1}^{∞} w^{2n−1} \frac{d_{3,n}}{(2n − 1)!} = \frac{sin(3w(1 − 4/12))}{cos 3w} = \frac{sin 2w}{cos 3w} ,

which on comparison with equation (7) gives (ii).

The notion of an alternating permutation of {1, 2, …, n} is generalized in [8] to a “Λalternating augmented r-signed permutation” of {1, 2, …, n} for any pair (p, r) of positive integers with p ≤ r. The cases (1, 1), (1, 2) and (2, 2) correspond respectively to the A_{n−1}-snakes, B_{n}-snakes, and β_{n}-snakes of the previous section. Here we give a combinatorial interpretation of the numbers c_{3,n} and d_{3,n} using the case (p, r) = (2, 3). Let S = \{nω^{m}| n, m nonnegative integers\}, where ω = e^{\frac{2πi}{3}} , with the linear order

ω^{2} < 2ω^{2} < 3ω^{2} < ⋯ < 0 < ω < 2ω < 3ω < ⋯ < 1 < 2 < 3 < ⋯

Define an ER_{n}-snake to be a sequence (x_{1}, x_{2}, …, x_{n}) of elements of S such that 0 < x_{1} > x_{2} < ⋯ x_{n} and {|x_{1}|, |x_{2}|, …, |x_{n}|} = {1, 2, …, n}. Let ER_{n} be the set of ER_{n}-snakes, so e.g., ER_{0} consists of the empty snake, ER_{1} = {(1), (ω)}, and

ER_{2} = {(2, 1), (2, ω), (2, ω^{2}), (2ω, ω), (2ω, ω^{2}), (1, 2ω), (1, 2ω^{2}), (ω, 2ω^{2})}.

Theorem 5.2. For n ≥ 0, c_{3,n} = card ER_{2n}; for n ≥ 1, d_{3,n} = card ER_{2n−1}.

Proof. In the terminology of [8], ER_{n}-snakes are Λ-alternating augmented 3-signed permutations of {1, 2, …, n} corresponding to p = 2. By Proposition 7.2 of [8], we have

∑_{n=0}^{∞} \frac{card ER_{n}}{n!} x^{n} = \frac{sin 2x + cos x}{cos 3x}

and the conclusion follows by comparison with equations (8) and (9) above.

6. Euler Polynomials. In this section we give explicit formulas for the values of the Euler polynomials at rational numbers in terms of the P_{n} and Q_{n}. The Euler polynomials E_{n}(x) are defined by equation (2) above; the Euler numbers are E_{n} = 2^{n}E_{n}(\frac{1}{2}). In view of the translation formula for Euler polynomials (23.1.7 of [1]), it suffices to give formulas for rational arguments between 0 and 1.

Theorem 6.1. If n, p and q are nonnegative integers with 0 ≤ p ≤ q (0 < p < q if n = 0) and q ≥ 2 even, then

E_{n}(\frac{p}{q}) = \frac{2}{q^{n+1}} ∑_{k=0}^{q/2−1} sin(\frac{(2k+1)πp}{q} − \frac{nπ}{2} ) csc \frac{(2k+1)π}{q} Q_{n}(cot \frac{(2k+1)π}{q})

if p is odd, and

E_{n}(\frac{p}{q}) = \frac{2}{q^{n+1}} ∑_{k=0}^{q/2−1} sin(\frac{(2k+1)πp}{q} − \frac{nπ}{2} )P_{n}(cot \frac{(2k+1)π}{q})

if p is even.

Proof. We start with the Fourier series

E_{n}(\frac{p}{q}) = \frac{4 · n!}{π^{n+1}} ∑_{k=0}^{∞} \frac{sin(\frac{(2k+1)πp}{q} − \frac{nπ}{2} )}{(2k + 1)^{n+1}}

([1], 23.1.16). Consider

ψ(j) = \{ \begin{matrix}sin(\frac{jπp}{q} − \frac{nπ}{2} ), & j odd, \\ 0, & j even. \end{matrix}

Then ψ(j + q) = (−1)^{p}ψ(j), so ψ is alternating mod q if p is odd and periodic mod q if p is even. Further, ψ(q − j) = (−1)^{p+n+1}ψ(j), so ψ has parity (as an alternating or periodic function mod q) opposite that of p + n. Thus, since

E_{n}(\frac{p}{q}) = \frac{4 · n!}{π^{n+1}} ∑_{j=1}^{∞} \frac{ψ(j)}{j^{n+1}}

the conclusion follows from Theorem 2.3.

Remark. By a proof similar to that above, one can obtain a formula for rational values B_{n}(p/q) of the nth Bernoulli polynomial in terms of P_{n−1}. This is essentially Theorem C of [20].

Examples. Set q = 2 to get

(10)  E_{n}(\frac{1}{2}) = \frac{1}{2^{n}} sin \frac{(1 − n)π}{2} Q_{n}(0) = \{ \begin{matrix}0, & n odd \\ (−1)^{\frac{n}{2}} 2^{−n}Q_{n}(0), & n even. \end{matrix}

Hence (by Theorem 2.1) Q_{n}(0) = |E_{n}|. We have also

(11)  E_{n}(0) = −E_{n}(1) = \frac{1}{2^{n}} sin \frac{(−n)π}{2} P_{n}(0) = \{ \begin{matrix}0, & n even \\ (−1)^{\frac{n+1}{2}} 2^{−n}P_{n}(0), & n odd, \end{matrix}

for n ≥ 1; this can be written in terms of Bernoulli numbers (see [1], 23.1.20) as −2(2^{n+1} − 1)B_{n+1}/(n + 1).

Taking q = 4, we obtain

(12)  E_{n}(\frac{1}{4}) = (−1)^{n}E_{n}(\frac{3}{4}) = \{ \begin{matrix}(−1)^{\frac{n}{2}} 4^{−n}Q_{n}(1), & n even, \\ (−1)^{\frac{n+1}{2}} 4^{−n}Q_{n}(1), & n odd. \end{matrix}

Finally, let q = 6 to get

(13)  E_{n}(\frac{1}{6}) = (−1)^{n}E_{n}(\frac{5}{6}) = \{ \begin{matrix}2(−1)^{\frac{n}{2}} 6^{−n−1}[2Q_{n}(\sqrt{3}) + Q_{n}(0)], & n even, \\ 4(−1)^{\frac{n+1}{2}} 6^{−n−1}\sqrt{3}Q_{n}(\sqrt{3}), & n odd, \end{matrix}

and

(14)  E_{n}(\frac{1}{3}) = (−1)^{n}E_{n}(\frac{2}{3}) = \{ \begin{matrix}2(−1)^{\frac{n}{2}} 6^{−n−1}\sqrt{3}P_{n}(\sqrt{3}), & n even, \\ 2(−1)^{\frac{n+1}{2}} 6^{−n−1}[P_{n}(\sqrt{3}) − P_{n}(0)], & n odd. \end{matrix}

If n is even, we can use Theorem 3.2(ii) and equation (13) to write

E_{n}(\frac{1}{6}) = E_{n}(\frac{5}{6}) = \frac{(−1)^{\frac{n}{2}} (3^{n} + 1)}{2 · 6^{n}}Q_{n}(0) = \frac{3^{n} + 1}{2 · 6^{n}}E_{n}

(cf. [15], Ch. 2, eqn. (46)). If n is odd, we have

E_{n}(\frac{1}{3}) = −E_{n}(\frac{2}{3}) = \frac{(−1)^{\frac{n+1}{2}} (3^{n} − 1)}{2 · 6^{n}}P_{n}(0) = −\frac{(3^{n} − 1)(2^{n+1} − 1)}{3^{n}(n + 1)}B_{n+1}

using Theorem 3.2(i) and equation (14) (cf. [15], Ch. 2, eqn. (45); [1], 23.1.22).

We close this section by expressing the Springer and Shanks numbers in terms of the Euler polynomials.

Theorem 6.2. For integers n ≥ 1,

(i)  a_{n−1} = 2^{n}|E_{n}(\frac{(−1)^{n} + 1}{4})|;
(ii)  b_{n} = 4^{n}|E_{n}(\frac{1}{4})|;
(iii)  d_{n} = 4^{n}|E_{n}(\frac{(−1)^{n} + 1}{4}) − E_{n}(\frac{1}{4})|;
(iv)  c_{3,n} = 6^{2n}|E_{2n}(\frac{1}{3})|;
(v)  d_{3,n} = 6^{2n−1}|E_{2n−1}(\frac{1}{6})|.

Proof. For (i), note first that a_{n−1} is Q_{n}(0) for n even, and P_{n}(0) for n odd; then use equations (10) and (11). For (ii), use equation (12). For (iii), observe from equation (5) that d_{n} is 2^{n}Q_{n}(0) − Q_{n}(1) for n even, and 2^{n}P_{n}(0) − Q_{n}(1) for n odd; then use equation (12) together with equations (10) and (11) respectively. For (iv) and (v), use Theorem 5.1 together with equations (13) and (14).

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