Abstract Given two strictly decreasing sequences of natural numbers of the same length

[r′_{s}, ..., r′_{1}] and [r_{s}, ..., r_{1}], we study conditions such that the fraction (∏_{1}^{s} p^{r′_{i}} − 1) / (∏_{1}^{s} p^{r_{i}} − 1)

is integral for all primes p. We also evaluate this fraction for particular sequences.

Arithmetic between sequences of natural numbers and cyclotomic polynomials

Nondas E. Kechagias Department of Mathematics University of Ioannina Ioannina, 45110 Greece

                       nkechag@cc.uoi.gr

1 Introduction

It is well known that (_{r}^{k}) = \frac{k...(r+1)}{(k−r)...1} is integral for any pair of natural numbers k and r such that k > r. The fractions \frac{9·8·7}{4·3·2} and \frac{9·8·7·6}{4·3·2·1} are both integral but the second one admits a property which the first one does not. Namely, there exists a number, it is 2 in this case, which divides two members of the sequence (4, 3, 2) and one of (9, 8, 7). We call the second pair of sequences admissible.

The reason we are interested in that property between pairs of decreasing sequences of natural numbers is the following application: Let p be a prime and consider the following fraction

\begin{matrix}(p^{9} − 1) & (p^{8} − 1) & (p^{7} − 1) \\ (p^{4} − 1) & (p^{3} − 1) & (p^{2} − 1) \end{matrix} = \frac{(p^{6} + p^{3} + 1) (p^{4} + 1) (p^{6} + ... + 1)}{p + 1}.

This fraction is not integral, if p = 7. But the following fraction is integral for any prime p

\begin{matrix}(p^{9} − 1) & (p^{8} − 1) & (p^{7} − 1) & (p^{6} − 1) \\ (p^{4} − 1) & (p^{3} − 1) & (p^{2} − 1) & (p − 1) \end{matrix}.

We would like to study conditions between sequences [r′_{s}, ..., r′_{1}] and [r_{s}, ..., r_{1}] such that the fraction

∏_{t=1}^{s} \frac{p^{r′_{t}} − 1}{p^{r_{t}} − 1}

is integral for any prime p.

Theorem 7 X[_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] is integral if and only if [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] is admissible.

Our second task is to evaluate such a fraction, i.e. to provide formulas independent of the value of p. In section three we provide such formulas for particular families of sequences.

We are able to note one application of this approach. Let V, an n-dimensional 𝔽_{p}-vector space, p a prime number. Let GL (n, 𝔽_{p}) (the general linear group) act naturally on V and G a parabolic subgroup of type (t, k). It is well known that the rings of invariants, S^{G} and S^{GL_{n}}, of the symmetric algebra S of V are polynomial algebras. Moreover, S^{G} is a finite free module over S^{GL_{n}} ([1]). To construct such a basis it is important to evaluate the following fraction

[GL (k + t, 𝔽_{p}) : G] = ∏_{t=1}^{s} \frac{p^{k+t} − 1}{p^{t} − 1}.

2 Admissibility

In this section, we study pairs of sequences of natural numbers whose divisors satisfy certain properties described in the following definition.

Notation Throughout this paper, p denotes a prime number and s, r_{1}, r_{2}, ..., r_{s} and r′_{1}, r′_{2}, ..., r′_{s} are natural numbers such that 1 ≤ t ≤ s and r_{t+1} > r_{t} ≥ 1, r′_{t+1} > r′_{t} ≥ 1.

Definition 1) Let the symbol [_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] stand for two sequences consisting of s natural

numbers (r′_{s}, ..., r′_{1}) and (r_{s}, ..., r_{1}). [_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] is called integral, if the fraction \frac{∏ r′_{i}}{∏ r_{i}} is integral.

2) Given a pair of sequences as above and n a natural number greater than 0 define

d(n) := ∑_{i=1}^{s} d_{i} (n) , where d_{i}(n) = \{ ; \begin{matrix}1, n|r_{i} \\ 0, n ∤ r_{i} \end{matrix} .

Then [_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] is called admissible, if d(n) ≤ d′(n) for every n.

Lemma 1. If [_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] is admissible, then [_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] is also integral.

Example a) [_{6,5,4,3,2,1}^{13,12,11,10,9,8}] is admissible: d(6) = 1, d(5) = 1, d(4) = 1, d(3) = 2 and d(2) = 3; d′(13) = 1, d′(12) = 1, d′(11) = 1, d′(10) = 1, d′(9) = 1, d′(8) = 1, d′(6) = 1, d′(5) = 1, d′(4) = 2, d′(3) = 2, d′(2) = 3.

b) [_{12,10,5,4,2}^{30,28,23,22,20}] is integral but it is not admissible:

i) (20 × 22 × 23 × 28 × 30)/(2 × 4 × 5 × 10 × 12) = 1771; ii) d(12) = 1 > d′(12) = 0.

Definition Given a sequence as above and n a natural number greater than 1 define

b(n) := ∑_{i=1}^{s} b_{i} where n^{b_{i}}||r_{i}.

Proposition 2. If [_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] is admissible, then b(n) ≤ b′(n) for every n.

The following example is a counter example to the contrary.

Example 1) [_{9,8,4}^{19,18,16}] is integral and b(n) ≤ b′(n) for every n; although [_{9,8,4}^{19,18,16}] is not admissible: d(4) = 2 > d′(4) = 1 but b (4) = 2 ≤ b′ (4) = 2.

2) [_{8,6,4}^{18,16,14}] is integral and b(n) ≤ b′(n) for every n; although [_{8,6,4}^{18,16,14}] is not admissible.

Proposition 3. [_{r,...,1}^{k+r,...,k+1}] is admissible for k and r ≥ 1.

Proof. Using double induction and the well known formula (_{r}^{k+r}) = (_{r}^{k+r−1}) + (_{r−1}^{k+r−1}), the integrality of (_{r}^{k+r}) is proved. We assume that [_{r−1,...,1}^{k+r−1,...,k+1}] is admissible. If r is a prime number or divides (k + r), then the statement follows. Otherwise, let \frac{r}{q} be a divisor of r. Then \frac{r}{q} is also a divisor of m\frac{r}{q} for 1 ≤ m ≤ q. We need to show that there exist at least q natural numbers k + r′ between k + 1 and k + r which are divisible by \frac{r}{q}. Let k = x\frac{r}{q} + l with 0 ≤ l ≤ \frac{r}{q} − 1. Then k + r = (x + q)\frac{r}{q} + l and the statement follows.

Remark [_{4,3,2,1}^{9,8,7,6}] is admissible but [_{4,3,2}^{9,8,7}] is not.

Lemma 4. If [_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] is admissible and m is a multiple of lcm(r_{s}, ..., r_{1}), then [_{r_{s},...,r_{1}}^{m+r′_{s},...,m+r′_{1}}] is also admissible.

Remark The admissibility of both [_{6,...,1}^{13,...,8}] and [_{6,...,1}^{6,...,1}] does not imply that 7 is a multiple of 60.

Lemma 5. If [_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] is admissible and σ is a permutation of (k_{s}r_{s}, ..., k_{1}r_{1}) such that σ (k_{s}r_{s}, ..., k_{1}r_{1}) is in decreasing order, then [_{r_{s},...,r_{1}}^{σ(k_{s}r_{s},...,k_{1}r_{1})}] is also admissible.

Lemma 6. If [_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] is admissible and m is a divisor of gcd(r_{s}, ..., r_{1}), then [_{r_{s}/m,...,r_{1}/m}^{r′_{s}/m,...,r′_{1}/m}] is also admissible.

Definition 1) Given [_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] and a prime number p, define

X[{r′_{s}, ..., r′_{1} \atop r_{s}, ..., r_{1}}] := ∏_{t=1}^{s} \frac{p^{r′_{t}} − 1}{p^{r_{t}} − 1}.

2) Given [_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}], X[_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] is called integral if X[_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] is integral for every prime p.

Theorem 7. X[_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] is integral if and only if [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] is admissible.

Proof. It is known that x^{k} − 1 = ∏_{n|k} C_{n} ([2]), where C_{n} stands for the n-th cyclotomic

polynomial. We recall that C_{n}(x) = ∏_{q=1,gcd(q,n)=1}^{n}(x − ζ^{q}) and ζ is a primitive root of unity

over 𝔽_{p}. Thus X[_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] = ∏_{t=1}^{s} \frac{x^{k+r_{t}}−1}{x^{r_{t}}−1} = ∏_{t=1}^{s} \frac{∏_{n|(k+r_{t})} C_{n}}{∏_{n′|r_{t}} C_{n′}} and the statement follows because

of the definition of admissibility. Please note that d (1) = d′ (1).

Example [_{14,12,10,6,4}^{298,296,294,290,288}] is admissible and hence X[_{14,12,10,6,4}^{298,296,294,290,288}] is integral.

3 Partition of sequences - Applications

Let [_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] be admissible. We would like to decompose the given pair of sequences in terms of more sequences of smaller length such that they are still admissible. We shall use such a decomposition to evaluate X[_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}]. We are mainly interested in expressing X[_{r,...,1}^{k+r,...,k+1}] in terms of r variables, A_{r_{t}}, defined bellow.

We begin by defining a peculiar division in [_{r,...,1}^{k+r,...,k+1}] as follows:

Definition 1) Given [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] we let r_{t} ”divide” k + r_{l}, r_{t}”\”k + r_{l}, if \frac{k+r_{l}}{r_{t}} is integral, l ≤ t and l is maximal with this property.

2) Given [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}], let I_{exact} := {r_{t} | r_{t}|k}.

Now we define a partition of I := {r_{s}, ..., r_{1}} according to the given ”division” as follows: For each r_{t(0)} ∈ I_{exact} let I_{r_{t(0)}} be the union of all subsequences {r_{t(0)}, r_{t(0)+s_{1}}, ..., r_{t(0)+s_{l}}} such

that r_{t(0)+s_{i+1}}”\”k + r_{t(0)+s_{i}} and s_{l} is as large as possible. Now I = ⨆_{r_{t(0)}∈I_{exact}} I_{r_{t(0)}}.

Example a) [_{11,...,1}^{11+12,...,1+12}], I_{exact} = {6, 4, 3, 2, 1}, I_{6} = {9, 6}, I_{4} = {11, 10, 8, 4}, I_{3} = {5, 3}, I_{2} = {7, 2} and I_{1} = {1}. [_{9,6}^{21,18}], [_{11,10,8,4}^{23,22,20,16}], [_{5,3}^{17,15}], [_{7,2}^{19,14}] are all admissible.

b) [_{11,...,1}^{29,...,19}], I_{exact} = {9, 6, 3, 2, 1}, I_{9} = {9}, I_{6} = {8, 6}, I_{3} = {7, 3}, I_{2} = {11, 4, 2} ∪ {10, 2} ∪ {5, 2} and I_{1} = {1}. [_{9}^{27}], [_{8,6}^{26,24}], [_{7,3}^{25,21}] are admissible sequences but [_{11,10,5,4,2}^{29,28,23,22,20}] is not even integral.

Definition Given [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}], let ⨆_{r_{t(0)}∈I_{exact}} I_{r_{t(0)}} be the corresponding partition of I. For

each element r_{t_{l−i}} of I_{r_{t(0)}} = {r_{t(0)}, r_{t(0)+s_{1}}, ..., r_{t(0)+s_{l}}} and a prime p, we define

A_{r_{t(0)+s_{i}}} := p^{(r_{t(0)+s_{i}}−r_{t(0)+s_{i−1}})} \frac{p^{(k+r_{t(0)+s_{i−1}})} − 1}{p^{(r_{t(0)+s_{i}})} − 1} and A_{r_{t(0)}} := \frac{p^{(k+r_{t(0)})} − p^{(r_{t(0)})}}{p^{(r_{t(0)})} − 1} .

Here r_{t(0)+s_{i}}\r_{t(0)+s_{i−1}} and s_{i−1} is minimal with this property.

Remark i) A_{i} expresses the same natural number in X[_{r}^{k+r}] and also in X[_{r−1}^{k+r−1}], for i = 1, ..., r − 1.

Proposition 8. If r′_{i} = k_{i}r_{i} for i = 1, ..., s, then X[_{r_{s},...,r_{1}}^{r′_{s},...,r′_{1}}] = ∏_{i=1}^{s} (A_{r_{i}} + 1). Here A_{r_{i}} :=

\frac{p^{k_{i}r_{i}}−p^{r_{i}}}{p^{r_{i}}−1} .

Lemma 9. Let [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] such that r_{t}|k + r_{t−1}, r_{1}|k and for each divisor n of gcd(r_{1}, r_{2}), n|k + r_{s} or ∃t : n ∤ r_{t} and n|k + r_{t−1}. Then [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] is admissible.

Remark k in the last lemma must satisfy the following conditions: k = l′ lcm(r_{1}, ...r_{s}) + l lcm(r_{1}, r_{2}) − r_{1} such that l is a non-negative integer and l′ is a positive integer and l lcm(r_{1}, r_{2}) − r_{1} + r_{i−1} ≡ 0 mod r_{i}. If r_{s} = s, then k = l lcm(2, ..., s) + 1.

The following example demonstrates the need of the extra condition given in last lemma. Example [_{9,8,4}^{19,18,16}] is integral although [_{9,8,4}^{19,18,16}] is not admissible.

Proposition 10. Let [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] such that the r_{t}’s are as in the last lemma. Then

X[_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] = ∑_{i=0}^{s} ∏_{t=0}^{i} A_{r_{t}}. Here A_{r_{i}} = p^{r_{i}−r_{i−1}} \frac{p^{k+r_{i−1}}−1}{p^{r_{i}}−1}, A_{r_{i}} = p^{r_{1}} \frac{p^{k}−1}{p^{r_{1}}−1} and A_{0} = 1.

Proof. We use induction on s. 1 + A_{r_{1}} = \frac{p^{k+r_{1}}−1}{p^{r_{1}}−1}. For the general step we use the following

decomposition: X[_{r_{s−1},...,r_{1}}^{k+r_{s−1},...,k+r_{1}}] + ∏_{t=0}^{s} A_{r_{t}} = ∏_{t=1}^{s} \frac{p^{k+r_{t}}−1}{p^{r_{t}}−1} ⇔ ∏_{t=0}^{s} A_{r_{t}} = X[_{r_{s−1},...,r_{1}}^{k+r_{s−1},...,k+r_{1}}]p^{r_{s}} \frac{p^{k}−1}{p^{rs}−1}

⇔ p^{r_{s}} \frac{p^{k}−1}{p^{r_{1}}−1} ∏_{t=2}^{s−1} \frac{p^{k+r_{t−1}}−1}{p^{r_{t}}−1} = ∏_{t=1}^{s} \frac{p^{k+r_{t}}−1}{p^{r_{t}}−1}.

Example X[_{11,...,1}^{23,...,13}] = X[_{1}^{13}]X[_{7,2}^{19,14}]X[_{5,3}^{17,15}]X[_{11,10,8,4}^{23,22,20,16}]X[_{9,6}^{21,18}]. Thus

X[{23, ..., 13 \atop 11, ..., 1}] = (1 + A_{1}) (1 + A_{2} + A_{2}A_{7}) (1 + A_{3} + A_{3}A_{5})
(1 + A_{4} + A_{4}A_{8} + A_{4}A_{8}A_{10} + A_{4}A_{8}A_{10}A_{11}) (1 + A_{6} + A_{6}A_{9}) .

Lemma 11. Let [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] such that: r_{t}|k + r_{t−1} for 2 ≤ t ≤ s − 1 and r_{s}, r_{1}|k + r_{1}; for each divisor n of \frac{gcd(r_{1},r_{s})}{gcd(r_{1},r_{2},r_{s})} or \frac{gcd(r_{2},r_{s})}{gcd(r_{1},r_{2},r_{s})}, n|k + r_{s}, or k + r_{s−1}, or ∃t : n ∤ r_{t} and n|k + r_{t−1}; for each divisor n of gcd(r_{1}, r_{2}, r_{s}), either (n divides both k + r_{s} and k + r_{s−1}), ( n divides one of k + r_{s} and k + r_{s−1}, and ∃t : n ∤ r_{t}, r_{t′} and n|k + r_{t−1}), or ( ∃t and t′ : n ∤ r_{t}, r_{t′} and n|k + r_{t−1}, k + r_{t′−1}). Then [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] is admissible.

Proof. Only the divisors of r_{1}, r_{2}, and r_{s} should be considered i.e. four cases:

i) n|r_{s} and n ∤ r_{1}, r_{2}; ii) n|r_{s}, r_{1} and n ∤ r_{2}; iii) n|r_{s}, r_{2} and n ∤ r_{1}; iv) n|r_{s},r_{1}, r_{2}. Condition r_{s}|k + r_{1} implies 1 ≤ d (n) ≤ d′ (n) and covers case i). Conditions n|\frac{gcd(r_{1},r_{s})}{gcd(r_{1},r_{2},r_{s})} and either (n|k + r_{s} or k + r_{s−1}), or (∃t : n ∤ r_{t} and n|k + r_{t−1}) imply 2 ≤ d (n) ≤ d′ (n). Now case ii) follows. Case iii) is similar. The last condition expressed in lemma imply that 3 ≤ d (n) ≤ d′ (n). Hence case iv) follows.

Remark a) k as in the last lemma must satisfy the following conditions:

k = l′ lcm(r_{1}, ...r_{s}) + l lcm(r_{1}, r_{2}, r_{s}) − r_{1}

such that l is a positive integer and l′ is a non-negative integer and

l′ lcm(r_{1}, r_{2}, r_{s}) − r_{1} + r_{i−1} ≡ 0 mod r_{i}.

b) If r_{s} = s, then k = l lcm(2, ..., s) + l′ lcm(2, ..., s − 1) + 1 and l′ lcm(2, ..., s − 1) ≡ (s − 2) mod s. Thus s is a prime number or s = 4.

Example [_{4,...,1}^{11,...,8}] and [_{7,...,1}^{182,...,176}] satisfy conditions described above.

Proposition 12. Let [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] such that: r_{t}|k + r_{t−1} for 2 ≤ t ≤ s − 1, r_{s},r_{1}|k + r_{1}, r_{s} − r_{s−1} = r_{2} − r_{1}, and [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] be admissible. Then

X[{k + r_{s}, ..., k + r_{1} \atop r_{s}, ..., r_{1}}] = ∑_{t=0}^{s−1} ∏_{i=0}^{t} A_{r_{i}} + A_{r_{s}} ∏_{i=1,i≠2}^{s−1} A_{r_{i}}(A_{r_{2}} − p^{r_{2}−r_{1}}) +

p^{r_{s}−r_{s−1}} ∏_{i=1}^{s−1} A_{r_{i}} = ∑_{t=0}^{s−1} ∏_{i=0}^{t} A_{r_{i}} + ∏_{i=1,i≠2}^{s} A_{r_{i}}(A_{r_{2}} + \frac{p^{r_{s}−r_{1}} − p^{r_{2}−r_{1}}}{p^{r_{2}} − 1} ).

Here A_{0} = 1, A_{r_{1}} = p^{r_{1}} \frac{p^{k}−1}{p^{r_{1}}−1}, A_{r_{i}} = p^{r_{i}−r_{i−1}} \frac{p^{k+r_{i−1}}−1}{p^{r_{i}}−1}, for 2 ≤ i ≤ s − 1, and A_{r_{s}} =

p^{r_{s}−r_{1}} \frac{p^{k+r_{1}}−1}{p^{rs}−1}.

Proof. We use Proposition 10 on ∑_{t=0}^{s−1} ∏_{i=0}^{t} A_{r_{i}}. By direct computation we show that

A_{r_{s}} ∏_{i=1,i≠2}^{s−1} A_{r_{i}}(A_{r_{2}} − p^{r_{2}−r_{1}}) + p^{r_{s}−r_{s−1}} ∏_{i=1}^{s−1} A_{r_{i}} = X[_{r_{s−1},...,r_{1}}^{k+r_{s−1},...,k+r_{1}}]\frac{p^{k+rs}−p^{rs}}{p^{rs}−1} .

Example X[_{7,...,1}^{182,...,176}] = X[_{7}^{182}]X[_{6,5}^{181,180}]X[_{4,...,1}^{179,...,176}] =

(1 + A_{7}) (1 + A_{5} + A_{5}A_{6}) (1 + A_{1} + A_{1}A_{2} + A_{1}A_{2}A_{3} + A_{1}A_{3}A_{4} (A_{2} − p) + pA_{1}A_{2}A_{3}) .

Lemma 13. Let [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] such that r_{1}, r_{s−1}|k + r_{1}, r_{t}|k + r_{t−1} for 2 ≤ t ≤ s − 2 and r_{s}|k + r_{s−2}; for each divisor n of \frac{gcd(r_{1},r_{s−1})}{gcd(r_{1},r_{2},r_{s−1})} or \frac{gcd(r_{2},r_{s−1})}{gcd(r_{1},r_{2},r_{s−1})}, n|k + r_{s}, or k + r_{s−1}, or ∃t : n ∤ r_{t} and n|k + r_{t−1}; for each divisor n of gcd(r_{1}, r_{2}, r_{s−1}), either (n divides both k + r_{s} and k + r_{s−1}), (n divides one of k + r_{s} and k + r_{s−1}, and ∃t : n ∤ r_{t} and n|k + r_{t−1}), or (∃t and t′ : n ∤ r_{t}, r_{t′} and n|k + r_{t−1}, k + r_{t′−1}). Then [_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] is admissible.

Proof. The proof is identical as in the last lemma.

Remark a) k as in the last lemma must satisfy the following conditions: k = l′ lcm(r_{1}, ..., r_{s})+l lcm(r_{1}, r_{2}, r_{s−1})−r_{1} where l is a positive integer and l′ is a non-negative integer and l lcm(r_{1}, r_{2}, r_{s−1}) − r_{1} + r_{t−1} ≡ mod r_{t}, l lcm(r_{1}, r_{2}, r_{s−1}) − r_{1} + r_{s−2} ≡ mod r_{s}.

b) If r_{s} = s, then k = l lcm(2, ..., s) + l′ lcm(2, ..., s − 2) + 1, and l′ lcm(2, ..., s − 2) + 2 ≡ 0 mod(s − 1) and l′ lcm(2, ..., s − 2) + s − 1 ≡ 0 mod s. Thus s = 5.

Example [_{14,12,10,6,4}^{298,296,294,290,288}] and [_{5,...,1}^{12,...,8}] satisfy conditions described above.

Proposition 14. Let [_{r_{3},r_{2},r_{1}}^{k+r_{3},k+r_{2},k+r_{1}}] be admissible such that r_{1}, r_{2}, r_{3}|k + r_{1}. Then

X[{k + r_{3}, k + r_{2}, k + r_{1} \atop r_{3}, r_{2}, r_{1}}] = 1 + A_{r_{1}} + A_{r_{1}}A_{r_{2}} + A_{r_{1}}A_{r_{2}}A_{r_{3}} +

+A_{r_{1}}A_{r_{3}} \frac{p^{r_{2}−r_{1}} − 1}{p^{r_{2}} − 1}

Here A_{r_{1}} = p^{r_{1}} \frac{p^{k}−1}{p^{r_{1}}−1}, A_{r_{2}} = p^{r_{2}−r_{1}} \frac{p^{k+r_{1}}−1}{p^{r_{2}}−1}, A_{r_{3}} = p^{r_{3}−r_{1}} \frac{p^{k+r_{1}}−1}{p^{r_{3}}−1}, and A_{0} = 1.

Proof. We show that X[_{r_{2},r_{1}}^{k+r_{2},k+r_{1}}]\frac{p^{k+r_{3}}−p^{r_{3}}}{p^{r3}−1} = A_{r_{1}}A_{r_{3}}(A_{r_{2}} + \frac{p^{r_{2}−r_{1}}−1}{p^{r_{2}}−1}) by direct computation.

Corollary 15. Let [_{r_{3},r_{2},r_{1}}^{k+r_{3},k+r_{2},k+r_{1}}] be admissible such that r_{1}, r_{2}, r_{3}|k + r_{1} and r_{3} = r_{2} + r_{1}. Then

X[{k + r_{3}, k + r_{2}, k + r_{1} \atop r_{3}, r_{2}, r_{1}}] = 1 + A_{r_{1}} + A_{r_{1}}A_{r_{2}} + A_{r_{1}}A_{r_{2}}A_{r_{3}} + A_{r_{1}}A_{r_{3}} −
−A_{r_{1}}p^{r_{3}−r_{1}}A_{r_{2}} + A_{r_{1}}p^{r_{3}−r_{1}}A_{r_{3}}.

Corollary 16. Let [_{r_{3},r_{2},r_{1}}^{k+r_{3},k+r_{2},k+r_{1}}] be admissible such that r_{1}, r_{2}, r_{3}|k + r_{1} and r_{3} + r_{1} = 2r_{2}. Then

X[{k + r_{3}, k + r_{2}, k + r_{1} \atop r_{3}, r_{2}, r_{1}}] = 1 + A_{r_{1}} + A_{r_{1}}A_{r_{2}} + A_{r_{1}}A_{r_{2}}A_{r_{3}} + A_{r_{1}}p^{r_{3}−r_{2}}A_{r_{2}} − A_{r_{1}}p^{r_{2}−r_{1}}A_{r_{3}}.

Example X[_{8,5,3}^{125,122,120}] = 1 + A_{3} + A_{3}A_{5} + A_{3}A_{5}A_{8} + A_{3}A_{8} − A_{3}p^{5}A_{5} + A_{3}p^{5}A_{8}.

Example X[_{7,...,1}^{39,...,33}] = X[_{1}^{33}]X[_{2}^{34}]X[_{7,5,3}^{39,37,35}]X[_{6,4}^{38,36}].

Proposition 17. Let [_{r_{5},...,r_{1}}^{k+r_{5},...,k+r_{1}}] such that r_{5} = r_{1} + r_{4}, r_{1}, r_{2}, r_{4}|k + r_{1}, r_{3}|k + r_{2}, r_{5}|k + r_{3}, and [_{r_{5},...,r_{1}}^{k+r_{5},...,k+r_{1}}] be admissible. Then

X[{k + r_{5}, ..., k + r_{1} \atop r_{5}, ..., r_{1}}] = ∑_{t=0}^{3} ∏_{i=0}^{t} A_{r_{i}} + A_{r_{1}}A_{r_{2}}A_{r_{3}}A_{r_{5}} +
A_{r_{1}}A_{r_{3}}A_{r_{5}} [p^{r_{2}−2r_{1}} + A_{r_{4}}A_{r_{2}} + p^{r_{4}+r_{2}−r_{5}}(A_{r_{2}} − p^{k})]

If in addition r_{2} = 2r_{1} and r_{3} = 3r_{1}, then

X[{k + r_{5}, ..., k + r_{1} \atop r_{5}, ..., r_{1}}] = ∑_{t=0}^{3} ∏_{i=0}^{t} A_{r_{i}} + A_{r_{1}}A_{r_{3}}[(A_{r_{2}} − p^{r_{2}−r_{1}})A_{r_{4}} + p^{r_{4}−r_{3}}A_{r_{2}}] +
A_{r_{1}}A_{r_{3}}A_{r_{4}} [p^{r_{2}−2r_{1}} + p^{r_{2}−r_{1}}A_{r_{2}} + A_{r_{5}}(A_{r_{2}} − p^{r_{2}−r_{1}})] .

Here A_{0} = 1, A_{r_{1}} = p^{r_{1}} \frac{p^{k}−1}{p^{r_{1}}−1}, A_{r_{i}} = p^{r_{i}−r_{i−1}} \frac{p^{k+r_{i−1}}−1}{p^{r_{i}}−1}, for 2 ≤ i ≤ 3, A_{r_{4}} = p^{r_{4}−r_{1}} \frac{p^{k+r_{1}}−1}{p^{r_{4}}−1} and

A_{r_{5}} = p^{r_{5}−r_{3}} \frac{p^{k+r_{3}}−1}{p^{r_{5}}−1}.

Proof. We use Proposition 10 on ∑_{t=0}^{3} ∏_{i=0}^{t} A_{r_{i}} + A_{r_{1}}A_{r_{2}}A_{r_{3}}A_{r_{5}}. By direct computation we show

that X[_{r_{5},r_{3},r_{2},r_{1}}^{k+r_{5},k+r_{3},k+r_{2},k+r_{1}}]\frac{p^{k+r_{4}}−p^{r_{4}}}{p^{r_{4}}−1} = A_{r_{1}}A_{r_{3}}A_{r_{5}} [p^{r_{2}−2r_{1}} + A_{r_{4}}A_{r_{2}} + p^{r_{4}+r_{2}−r_{5}}(A_{r_{2}} − p^{k})]. For the second statement we use Proposition 12 on

∑_{t=0}^{3} ∏_{i=0}^{t} A_{r_{i}} + A_{r_{1}}A_{r_{3}}[(A_{r_{2}} − p^{r_{2}−r_{1}})A_{r_{4}} + p^{r_{4}−r_{3}}A_{r_{2}}]. By direct computation we show that

X[_{r_{4},r_{3},r_{2},r_{1}}^{k+r_{4},k+r_{3},k+r_{2},k+r_{1}}]\frac{p^{k+r_{5}}−p^{r_{5}}}{p^{r_{5}}−1} = A_{r_{1}}A_{r_{3}}A_{r_{4}} [p^{r_{2}−2r_{1}} + p^{r_{2}−r_{1}}A_{r_{2}} + A_{r_{5}}(A_{r_{2}} − p^{r_{2}−r_{1}})].

Example X[_{5,...,1}^{12,...,8}] =

∑_{t=0}^{3} ∏_{i=0}^{t} A_{i} + A_{1}A_{3}[(A_{2} − p)A_{4} + pA_{2}] + A_{1}A_{3}A_{4} [1 + pA_{2} + A_{5}(A_{2} − p)].

Lemma 18. Let [_{r_{4},...,r_{1}}^{k+r_{4},...,k+r_{1}}] such that r_{4} = r_{1} +r_{2}, r_{4}|k+r_{2} and 2r_{3} = 2r_{1} +r_{4}, r_{1}, r_{2}, r_{3}|k+ r_{1}. Then [_{r_{4},...,r_{1}}^{k+r_{4},...,k+r_{1}}] is admissible.

Remark In this case, k = l′ lcm(r_{1}, r_{2}, r_{3}, r_{4}) + l lcm(r_{1}, r_{2}, r_{3}) − r_{1} where l and l′ are non-negative integers and l lcm(r_{1}, r_{2}, r_{3}) − r_{1} + r_{2} ≡ mod r_{4}.

Example [_{10,9,6,4}^{294,293,290,288}] and [_{8,7,5,3}^{635,634,632,630}] are both admissible.

Proposition 19. Let [_{r_{4},...,r_{1}}^{k+r_{4},...,k+r_{1}}] as in the last lemma. Then

X[{k + r_{4}, ..., k + r_{1} \atop r_{4}, ..., r_{1}}] = 1 + A_{r_{1}} + A_{r_{1}}A_{r_{2}} + A_{r_{1}}A_{r_{2}}A_{r_{4}} +
A_{r_{1}}A_{r_{4}}(A_{r_{2}}A_{r_{3}} + p^{r_{3}−r_{1}}(A_{r_{3}} − p^{k}) + p^{r_{4}−r_{3}}).

Here A_{r_{1}} = p^{r_{1}} \frac{p^{k}−1}{p^{r_{1}}−1}, A_{r_{2}} = p^{r_{2}−r_{1}} \frac{p^{k+r_{1}}−1}{p^{r_{2}}−1}, A_{r_{3}} = p^{r_{3}−r_{1}} \frac{p^{k+r_{1}}−1}{p^{r_{3}}−1}, A_{r_{4}} = p^{r_{4}−r_{2}} \frac{p^{k+r_{2}}−1}{p^{r_{4}}−1} and A_{0} = 1.

Proof. We show that X[_{r_{4},r_{2},r_{1}}^{k+r_{4},k+r_{2},k+r_{1}}]\frac{p^{k+r_{3}}−p^{r_{3}}}{p^{r3}−1} = A_{r_{1}}A_{r_{4}}(A_{r_{2}}A_{r_{3}} + p^{r_{3}−r_{1}}(A_{r_{3}} − p^{k}) + p^{r_{4}−r_{3}}) by direct computation.

Example X[_{10,9,6,4}^{294,293,290,288}] = 1+A_{4} +A_{4}A_{6} +A_{4}A_{6}A_{10}+ A_{4}A_{10}(A_{6}A_{9} +p^{5}(A_{9} −p^{284})+p)

We close this work by posing a problem: Does there exist an algorithm which decomposes X[_{r_{s},...,r_{1}}^{k+r_{s},...,k+r_{1}}] in terms of our A_{r_{t}} ’s?

References

[1] Nondas E. Kechagias, Invariants of parabolic subgroups as modules over the Dickson algebra. In preparation.

[2] R. Lidl and H. Niederreiter, Encycl. of Mathematics, Cambridge University Press. Vol. 20, (1983).

2000 Mathematics Subject Classification: 11P81, 11P83. Keywords: Fractions of Cyclotomic polynomials.