The 2-adic valuation of generalized Fibonacci sequences with an application to certain Diophantine equations

TL;DR

This paper investigates the 2-adic valuation of generalized Fibonacci sequences, derives bounds for solutions to related Diophantine equations, and explores connections with p-regular sequences, advancing understanding of these number sequences.

Contribution

It provides explicit formulas for the 2-adic valuation of 2k-nacci sequences and applies these to solve specific Diophantine equations, extending to a broader family of sequences.

Findings

Explicit 2-adic valuation formulas for 2k-nacci sequences

Bounds on factorials expressed as products of these sequences

Potential links between these sequences and p-regular sequences

Abstract

In this paper we focus on finding all the factorials expressible as a product of a fixed number of $2k$-nacci numbers with $k \geq 2$. We derive the 2-adic valuation of the $2k$-nacci sequence and use it to establish bounds on the solutions of the initial equation. In addition, we specify a more general family of sequences, for which we can perform a similar procedure. We also investigate a possible connection of these results with $p$-regular sequences.

Tables

	1	2	3	4	5	6	7	8	9	10
2	11	19	27	35	43	51	59	67	75	84
3	13	22	31	40	50	59	68	77	87	96
4	11	19	28	36	44	52	60	68	76	84
5	14	25	35	46	56	67	77	88	98	109

Equations

$$t_n=\{\begin{array}{ll}0& \text{ for }n=0,\\ 1& \text{ for }1\le n\le r-1,\\ \sum _{i=1}^r{t}_{n-i}& \text{ for }n\ge r-1.\end{array}$$

$$m!=\prod _{i=1}^d{t}_{n_i}$$

$${\mathcal{N}}_p(a)=\left\{\{s_{p^{l}n+j}{\}}_{n\ge 0}:l\ge 0,0\le j<p^l\right\}$$

$$\frac{m}{p-1}-\lfloor \frac{\log m}{\log p}\rfloor -1\le {\nu }_p(m!)\le \frac{m-1}{p-1}.$$

$$\log s_n=\Omega (n).$$

$${\nu }_p(s_n)=O\left(n^C\right)$$

$$m!=\prod _{i=1}^d{s}_{n_i}$$

$$\frac{1}{2}m\le m-\lfloor {\log }_{2}m\rfloor -1\le {\nu }_2(m!),$$

$${\nu }_2\left(\prod _{i=1}^d{s}_{n_i}\right)=\sum _{i=1}^d{\nu }_2(s_{n_i})\le K_1\sum _{i=1}^d{n}_i^C\le d{K}_1{\left(\underset{1\le i\le d}{\max }{n}_i\right)}^C.$$

$$m\le 2d{K}_1{\left(\underset{1\le i\le d}{\max }{n}_i\right)}^C.$$

$${\log }_2\left(\prod _{i=1}^d{s}_{n_i}\right)=\sum _{i=1}^d{\log }_2{s}_{n_i}\ge K_2\sum _{i=1}^d{n}_i\ge K_2\underset{1\le i\le d}{\max }{n}_i.$$

$$m!<{\left(\frac{m}{2}\right)}^m,$$

$$m{\log }_2\frac{m}{2}>K_2\underset{1\le i\le d}{\max }{n}_i.$$

$${\log }_2\frac{m}{2}>K{m}^{\frac{1}{C}-1},$$

$${\nu }_2(t_n)=\{\begin{array}{ll}0& \text{ for }n\equiv 1,2,...,2k(\mathrm{mod}2k+1),\\ 1& \text{ for }n\equiv 2k+1(\mathrm{mod}2(2k+1)),\\ {\nu }_2(n)+{\nu }_2(k-1)+2& \text{ for }n\equiv 0(\mathrm{mod}2(2k+1)).\end{array}$$

$$t_n\ge {\varphi }^{n-r-1},$$

$${\nu }_2(t_n)\le {\nu }_2(n)+{\nu }_2(k-1)+2\le {\log }_2(n)+{\nu }_2(k-1)+2\le \sqrt{n},$$

$$\log t_n\ge (n-2k-1)\log \varphi \ge \frac{1}{2}n\log \varphi ,$$

$$m-\lfloor {\log }_{2}m\rfloor -1$$

$$=d[{\nu }_2(k-1)+2]+{\nu }_2\left(\prod _{i=1}^d{n}_i\right)$$

$$\le d[{\nu }_2(k-1)+2]+{\log }_2\left(\prod _{i=1}^d{n}_i\right).$$

$$\left[\sum _{i=1}^d{n}_i-d(2k-1)\right]{\log }_2{\varphi }_k\le \sum _{i=1}^d{\log }_2(t_{n_i})={\log }_2(m!)<m({\log }_{2}m-1),$$

$$m-\lfloor {\log }_{2}m\rfloor -1-d[{\nu }_2(k-1)+2]<d{\log }_2\left[\frac{m}{d{\log }_2{\varphi }_k}({\log }_{2}m-1)+(2k-1)\right].$$

$$T_n=\left[\begin{array}{c}{t}_n\\ t_{n+1}\\ ⋮\\ t_{n+2k-1}\end{array}\right],B_n=\left[\begin{array}{cccc}{t}_n& t_{n+1}& \dots & t_{n+2k-1}\\ t_{n+1}& t_{n+2}& \dots & t_{n+2k}\\ ⋮& ⋮& & ⋮\\ t_{n+2k-1}& t_{n+2k}& \dots & t_{n+4k-2}\end{array}\right],$$

$$C=\left[\begin{array}{cccccc}0& 1& 0& \dots & 0& 0\\ ⋮& 0& 1& \ddots & & 0\\ & ⋮& \ddots & \ddots & \ddots & ⋮\\ & & & \ddots & 1& 0\\ 0& 0& 0& \cdots & 0& 1\\ 1& 1& 1& \dots & 1& 1\end{array}\right],$$

$$C^n{T}_w$$

$$C^n{B}_w$$

$$t_{n+w}=T_n^T{B}_0^{-1}{T}_w.$$

$$\det B_0\equiv \det \left[\begin{array}{cccccc}0& 1& 1& \dots & & 1\\ 1& 1& \dots & & & 1\\ 1& ⋮& & & \text{udots}& 0\\ ⋮& & & \text{udots}& \text{udots}& 1\\ & & \text{udots}& \text{udots}& \text{udots}& ⋮\\ 1& 1& 0& 1& \cdots & 1\end{array}\right](\mathrm{mod}2),$$

$$T_{n+w}=C^n{T}_w=B_n{B}_0^{-1}{T}_w.$$

$$T_{2^l(2k+1)}\equiv T_0(\mathrm{mod}{2}^{l+1}).$$

$$T_{2^{l_0}(2k+1)}\equiv 2^{l_0+1}A+T_0(\mathrm{mod}{2}^{l_0+{\nu }_2(k-1)+3}),$$

Full text

1. Introduction

For a fixed integer $r\geq 2$ define the generalized Fibonacci ($r$-nacci) sequence $\{t_{n}\}_{n\geq 0}$ as follows:

[TABLE]

Notice, that for $r=2$ we obtain the usual Fibonacci sequence, which has already been studied extensively. In this paper we will mostly restrict ourselves to the case of even $r\geq 4$ and write $r=2k$ for some $k\geq 2$. The main motivation for our considerations is to completely solve the equation

[TABLE]

in positive integers $m,n_{1},...,n_{d}$.

For $p$ prime define the $p$-adic valuation of a non-zero integer $s$ as $\nu_{p}(s)=\max\{l\geq 0:p^{l}|s\}$ and $\nu_{p}(0)=\infty$. Equation (2) for the case of $r=3$ and $d=1$ was solved by Lengyel and Marques in [5] by means of computing $\nu_{2}(t_{n})$ and then applying this result to obtain an effective upper bound on $m$ and $n_{1}$. In this paper we will follow a similar procedure for $2k$-nacci sequences with $k\geq 2$.

To begin with, in Theorem 2 we specify a more general family of integer sequences $\{s_{n}\}_{n\geq 0}$ for which we are able to solve equation (2) and show a general procedure to achieve this goal. Informally speaking, we need the term $s_{n}$ to grow at least exponentially and $\nu_{p}(s_{n})$ – at most polynomially with an exponent less than 1, for some $p$ prime.

Theorem 3 provides a simple expression for $\nu_{2}(t_{n})$ when $r=2k\geq 4$ and the subsequent corollary shows that the sequence $\{t_{n}\}_{n\geq 0}$ satisfies the conditions of Theorem 2. We then find all the solutions of equation (2) for small values of $k$ and $d$.

We also briefly discuss how our results are related to $p$-regular sequences. Recall that a sequence $\{s_{n}\}_{n\geq 0}$ with rational values is $p$-regular iff its $p$-kernel

[TABLE]

is contained in a finitely generated $\mathbb{Z}$-module. More details on regular sequences can be found in [1] and [2]. As we note later, the formula given in Theorem 3 implies $2$-regularity of $\{\nu_{2}(t_{n})\}_{n\geq 0}$. However, it turns out that exponential growth of a sequence $\{s_{n}\}_{n\geq 0}$ and $p$-regularity of $\{\nu_{p}(s_{n})\}_{n\geq 0}$ still do not guarantee that the assumptions of Theorem 2 are met. In this case we cannot determine, using the shown method, whether the equation (2) has only a finite number of solutions.

2. Main results

As we mentioned before, we start with describing a general situation in which equation (2) can be completely solved. For two sequences $\{a_{n}\}_{n\geq 0}$ and $\{b_{n}\}_{n\geq 0}$ we denote $a_{n}=O(b_{n})$ if there exists a positive constant $K$ such that $|a_{n}|\leq K|b_{n}|$ for sufficiently large $n$. Similarily, we write $a_{n}=\Omega(b_{n})$ if there exists a positive constant $K$ such that $|a_{n}|\geq K|b_{n}|$ for sufficiently large $n$. First, we give an auxiliary lemma, also used in [5], which is an easy corollary from Legendre’s formula for $\nu_{p}(m!)$.

Lemma 1**.**

For any integer $m\geq 1$ and prime $p$, we have

[TABLE]

Theorem 2**.**

Let $\{s_{n}\}_{n\geq 0}$ be a sequence of positive integers such that

[TABLE]

Let $p$ be a prime. Assume that

[TABLE]

for some constant $C<1$. Then for each fixed positive integer $d$ the equation

[TABLE]

has only a finite number of solutions in $m,n_{1},n_{2},\ldots,n_{d}$.

Proof.

We adjust the method used in [5] to a more general setting. Roughly speaking, we will show that if (5) is satisfied and we let both sides grow, then the $p$-adic valuation of the right hand side increases slower than $\nu_{p}(m!)$. For each value of $p$ we proceed in the same way, so for simplicity assume that $p=2$. By our assumptions, there exist some positive constants $K_{1},K_{2}$ and an integer $n_{0}\geq 0$ such that $\nu_{2}(s_{n})\leq K_{1}n^{C}$ and $\log_{2}s_{n}\geq K_{2}n$ for $n\geq n_{0}$.

Suppose that $n_{i}\geq n_{0}$ for $i=1,2,\ldots,d$. There is only a finite number of solutions with $m<6$ because $s_{n}$ grows at least exponentially. By Lemma 1, for $p=2$ we get

[TABLE]

where the leftmost inequality is true for $m\geq 6$. On the other hand,

[TABLE]

Hence, for $m\geq 6$

[TABLE]

We need another inequality with $n_{i}$ bounded from above by $m$. By the assumption (3)

[TABLE]

Furthermore, for $m\geq 5$ the following inequality holds:

[TABLE]

and hence

[TABLE]

Combining inequalities (7) and (9) yields

[TABLE]

where $K=K_{2}/(2dK_{1})^{1/C}$. But $C<1$ implies that the exponent in (10) is strictly greater than 0, so (10) holds only for finitely many $m$. Thus, by inequality (9) we obtain an upper bound for all $n_{i}$.

Now assume without loss of generality that $n_{i}<n_{0}$ for $i=j+1,j+2,\ldots,d$. Observe that for each fixed $n_{j+1},n_{j+2},\ldots,n_{d}$ the problem is equivalent to solving the equation (5) with $d=j$ and the left hand side divided by a positive integer constant $M$. Then in (6) and (8) we need to replace $m!$ with $m!/M$ which only changes the set of $m$ for which both of those inequalities hold. But this leads to the same conclusion as before. ∎

Remark 1**.**

Observe that the method of Theorem 2 does not work if we let $d$ be unbounded. Indeed, the constant $K$ in (10) becomes arbitrarily small as $d$ increases, so we cannot use the subsequent argument. Informally speaking, the $p$-adic valuation of the expression $\prod_{i=1}^{d}s_{n_{i}}$ might grow too fast for the method to work.

Remark 2**.**

The condition (3) is satisfied for sequences expressible in Binet form. Hence, for linear recurrence sequences, we usually need to check only the condition (4) for some $p$. One might also ask whether we can replace it with some other assumption. Shu and Yao proved in [7] a condition on a binary recurrence sequence, which guarantees $p$-regularity of $\{\nu_{p}(s_{n})\}_{n\geq 0}$, and mentioned a possible generalization to recurrences of higher order. It is known that $p$-regular sequences grow at most polynomially, which is a result by Allouche and Shallit [1]. Unfortunately, this does not give a bound on $C$ in (4) and the proof of Theorem 2 fails if $C\geq 1$. Therefore, some additional information besides regularity needs to be known about $\{\nu_{p}(s_{n})\}_{n\geq 0}$ in order to put the theorem to use.

The reasoning in Theorem 2 provides an upper bound on the solutions of the equation (5) if we are able to find the values of $C,K_{1},K_{2}$ and $n_{0}$. We will show that it is indeed the case for our sequence $\{t_{n}\}_{n\geq 0}$. We start by determining the $2$-adic valuation of each $t_{n}$. A similar characterization of $\{\nu_{2}(t_{n})\}_{n\geq 0}$ for $r=2$ is given by Lengyel in [3] and [4] (by a different method).

Theorem 3**.**

If $k\geq 2$, then the sequence $\{\nu_{2}(t_{n})\}_{n\geq 0}$ satisfies the following conditions:

[TABLE]

The proof of the theorem is presented in Section 3.

Remark 3**.**

Theorem 3 implies that $\{\nu_{2}(t_{n})\}_{n\geq 0}$ is a $2$-regular sequence. Note that this conclusion does not follow directly from the results of [7], as we consider recurrence of any even order.

Now we proceed to show that Theorem 2 can be applied to our sequence $\{t_{n}\}_{n\geq 0}$. The following lemma establishes a lower bound on $t_{n}$.

Lemma 4**.**

For all $n\geq 1$ we have

[TABLE]

where $\phi$ is the unique real root of the equation $x^{r}=x^{r-1}+...+x+1$ lying in the interval $(1,2)$.

Proof.

By lemma 3.6 in [8] there is indeed exactly one such $\phi$, which in addition lies in the interval $(2(1-2^{-r}),2)$. For $n=1,2,...,r$ the inequality (11) follows from starting conditions for $t_{n}$ and the fact that $t_{r}=2(r-1)\geq 2>\phi$. Then we proceed easily by induction. ∎

Corollary 5**.**

If $r=2k\geq 4$ then the equation (2) has only a finite number of solutions in $m,n_{1},n_{2},\ldots,n_{d}$ and this number can be effectively bounded from above.

Proof.

By theorem 3 we have

[TABLE]

where the last inequality is true for example for $n\geq 2^{2\max\{2,\nu_{2}(k-1)\}}$. By lemma 4

[TABLE]

for $n\geq 2(2k+1)$, because $\phi>1$. Hence, we can take $n_{0}=\max\{2(2k+1),2^{2\max\{2,\nu_{2}(k-1)\}}\}$ and apply the method used in Theorem 2 to find an upper bound on $m,n_{1},n_{2},\ldots,n_{d}$. ∎

We could apply Corollary 5 to find all non-trivial solutions (with $t_{n_{i}}>1$ for each $i$) of the equation (2) for given $k$ and $d$. However, using the explicit form of $\nu_{2}(t_{n})$, one can make the bounds much more precise. We will once again follow the approach shown in [5]. The computations are quite similar to those in Theorem 2, so we omit the details.

By Lemma 1 and Theorem 3, we obtain

[TABLE]

On the other hand, from Lemma 4, and inequality (8), we get for $m\geq 5$

[TABLE]

where $\phi_{k}$ is the value of $\phi$ in Lemma 4 corresponding to $r=2k$. The AM–GM inequality applied to all $n_{i}$, together with (2) and (13), yields

[TABLE]

This gives an upper bound on $m$ and, consequently, on each $n_{i}$. As an example, in the table below we give the upper bound on $m$ obtained for $2k$-nacci sequences with $2\leq k\leq 5$ and $1\leq d\leq 10$.

Using this result we find that the only non–trivial solution of the equation (2) with $2\leq k\leq 5$ and $1\leq d\leq 10$ appears in the $4$-nacci sequence and is the single term $t_{5}=3!$.

3. Proof of Theorem 3

In order to study the 2-adic valuation of $t_{n}$ it is enough to focus on $n$ divisible by $2k+1$ as all the other terms are odd. In this case we can write $n=s2^{l}(2k+1)$ for $s$ odd and $l\geq 0$. We will divide our proof into two main parts.

First, we will show by induction on $l$ and $s$ that $2k$ consecutive terms of the sequence $\{t_{n}\}_{n\geq 0}$, starting with $t_{s2^{l}(2k+1)}$, satisfy a particular system of congruences, given in Lemma 7. However, as it will turn out, this argument works only for $l\geq l_{0}$, where $l_{0}$ depends on $\nu_{2}(k-1)$. Moreover, the initial system of congruences for $l=l_{0}$ involves some constants, which need to be computed.

Therefore, we will have to employ another method for $l\leq l_{0}$. We will show how to obtain the values of $t_{n+2k+1},\ldots,t_{n+4k}$ in terms of $t_{n},\ldots,t_{n+2k-1}$ for any $n$ and then proceed by induction. As a result, we will be able to express $t_{n}$ in quite concrete form, given in Lemma 10, involving binomial coefficients weighted by powers of 2.

For simplicity, we introduce the following matrix notation:

[TABLE]

where $n\geq 0$. By $C$ we will denote the companion matrix of $t_{n}$ which has the form

[TABLE]

where the entries above the diagonal and in the bottom row are equal to 1 and all other entries are zero. It is easy to check that $CT_{n}=T_{n+1}$ and $CB_{n}=B_{n+1}$, so for any positive integers $n$ and $w$ we have

[TABLE]

First, we state an identity involving the terms of the sequence $\{t_{n}\}_{n\geq 0}$.

Lemma 6**.**

The matrix $B_{0}$ is invertible and its determinant is odd. Moreover, for all positive integers $n,w$ we have the formula

[TABLE]

Proof.

It is easy to see that $t_{n}$ is even iff $n$ is divisible by $2k+1$. Therefore,

[TABLE]

where zeros in the latter matrix appear only at positions corresponding to $t_{0}$ and $t_{2k+1}$ in $B_{0}$, that is, at $(1,1)$ and $(i,j)$ such that $i+j=2k+2$. By subtracting the first row from all the others, we easily obtain $\det B_{0}\equiv 1\pmod{2}$, which proves the first part of the statement.

From (16) and (17), we get

[TABLE]

The first coordinate gives us the formula for $t_{n+w}$. ∎

The identity (18) might seem difficult to apply without an explicit expression for $B_{0}^{-1}$. However, it plays a major role in deriving the congruence relations in the following lemma.

Lemma 7**.**

For any $l\geq 0$ the following congruence relation holds:

[TABLE]

Moreover, if a column vector $A\in\mathbb{Z}^{2k}$ satisfies

[TABLE]

where $l_{0}=\nu_{2}(k-1)+2$, then for any $l\geq l_{0}$ and $s\geq 1$ we also have

[TABLE]

Proof.

Obviously (19) is true for $l=0$. Now assume that (19) holds for some $l\geq 0$. We can write

[TABLE]

where $a_{l,j}$ are some positive integers for $j=0,1,\ldots,2k-1$. Define also $a_{l,2k},a_{l,2k+1},\ldots,a_{l,4k-2}$ by the same recurrence as $\{t_{n}\}_{n\geq 0}$. Then (21) is satisfied for $j=0,1,\ldots,4k-2$. For convenience denote by $e_{j}\in\mathbb{Z}^{2k}$ the vector with 1 on the $j$-th position (counting from 0) and 0 on the other positions, and additionally define

[TABLE]

for $j=0,1,\ldots,2k-1$. It follows from the definition of $B_{0}$ that $B_{0}^{-1}T_{j}=e_{j}$.

Fix any $0\leq i\leq 2k-1$. The formula (18) yields

[TABLE]

Therefore, using (21) we get

[TABLE]

for some rational $c_{l,i}$ such that $\det(B_{0})c_{l,i}$ is an integer. In Lemma 6, however, we showed that $\det B_{0}$ is odd which means that $c_{l,i}$ must be an integer. Thus,

[TABLE]

from which (19) follows. If we choose $l\geq l_{0}=\nu_{2}(k-1)+2$ then the term $2^{2l+2}c_{l,i}$ in (7) is reduced modulo $2^{l+\nu_{2}(k-1)+4}$. We can take $A=A_{l_{0},0}$ to complete the proof of (20) for $s=1$.

To proceed by induction on $s$ notice that the index $s2^{l}(2k+1)+i$ can be expressed as a sum of indices in the following way:

[TABLE]

We can then perform a similar computation as in (7) to get the desired result. ∎

Our specific choice of the divisor in (20) equal to $2^{l+\nu_{2}(k-1)+3}$ is based on the observation of $t_{n}$ and is indeed effective in proving the formula for $\nu_{2}(t_{n})$. The numbers $a_{l,j}$ in (21) are determined uniquely modulo $2^{\nu_{2}(k-1)+2}$. We are particularly interested in finding the value of $a_{l_{0},0}$ which will directly give us the 2-adic valuation of $t_{2^{l}(2k+1)}$ for $l\geq l_{0}=\nu_{2}(k-1)+2$, provided that $a_{l_{0},0}\leq\nu_{2}(k-1)+1$. However, as mentioned before, we need to develop another method to analyze the case when $l\leq l_{0}$.

We start with deriving a formula for expressing $t_{n+2k+1},\ldots,t_{n+4k}$ in terms of $t_{n},\ldots,t_{n+2k-1}$.

Lemma 8**.**

Define $C$ as in (15). Then

[TABLE]

Proof.

Using the identity $t_{n+2k+1}=2t_{n+2k}-t_{n}$, one can show by induction that for any $0\leq i\leq 2k-1$

[TABLE]

We know that $D=C^{2k+1}$ is the only matrix satisfying $T_{n+2k+1}=DT_{n}$ for all $n\geq 0$. Thus, for each $0\leq i\leq 2k-1$ the coefficients at $t_{n+j}$ in (23) correspond to the $i$-th row of $C^{2k+1}$ (counting from 0). ∎

We are also going to need two standard identities involving binomial coefficients. For the convenience of the reader we include the proof.

Lemma 9**.**

For all positive integers $m,w$ we have

(a)* $\sum\limits_{i=0}^{w}\binom{m+i}{m}=\binom{m+w+1}{m+1}$,*

(b)* $\sum\limits_{i=0}^{w}\binom{m+i}{m}2^{i}=(-1)^{m+1}+2^{w+1}\sum\limits_{j=0}^{m}\binom{m+w+1}{m-j}(-2)^{j}.$*

Proof.

For any fixed $m\geq 0$ the formula (a) follows easily from induction on $w$.

To prove (b) take any $w,m\geq 0$ and consider the function

[TABLE]

for $x\neq 1$. It is easy to see that the left side of (b) is equal to $f^{(m)}(x)$ evaluated at $x=2$. Applying the Leibniz formula we get

[TABLE]

Substituting $x=2$ we get the desired result. ∎

The following lemma gives us an easily computable expression for $2k$ subsequent terms $t_{n}$.

Lemma 10**.**

Define the following column vectors in $\mathbb{N}^{2k}$:

[TABLE]

for $m\geq 0$. Then for any $m\geq 1$ we have

[TABLE]

Proof.

Using the form of $C^{2k+1}$ given in Lemma 8, it is easy to see that

[TABLE]

Now fix $m\geq 0$. Applying Lemma 9 to each coordinate gives us

[TABLE]

One can check that

[TABLE]

so (24) is true for $m=1$. Now assume that (24) holds for some $m\geq 1$. Using (25) and (26) we get

[TABLE]

where $\epsilon_{m}$ is equal to $m$ modulo 2. Thus, the coefficient at $v_{0}$ is equal to $(-1)^{m}\cdot 4(k-1)$, so we can incorporate it into the sum. Finally, we obtain

[TABLE]

∎

We are now ready to prove Theorem 3.

Proof (of Theorem 3).

The term $t_{n}$ is even iff $n$ is divisible by $2k+1$, which proves that $\nu_{2}(t_{n})=0$ for $n\equiv 1,2,...,2k\pmod{2k+1}$. Observe that if $k\geq 2$, then $2\nu_{2}(k-1)+5\leq 2k+1$, so by Lemma 8 for $m\geq 1$ we have

[TABLE]

Looking at the first entry of this vector, we obtain

[TABLE]

Thus, for odd $m$ we get $t_{m(2k+1)}\equiv 2\pmod{4}$, hence $\nu_{2}(t_{n})=1$ for $n\equiv 2k+1\pmod{2(2k+1)}$.

Now let $n=s(2k+1)2^{l}$ for odd $s$ and $l\geq 1$, so that $n\equiv 0\pmod{2(2k+1)}$. We will further split the third case into two subcases, depending whether $l\leq l_{0}=\nu_{2}(k-1)+2$ or $l>l_{0}$. If $l\leq l_{0}$ then from (27) we obtain

[TABLE]

so $\nu_{2}(t_{n})=l+2+\nu_{2}(k-1)=\nu_{2}(n)+\nu_{2}(k-1)+2$.

We cannot extend the same argument to $l>l_{0}$ because we only know the congruence modulo $2^{2\nu_{2}(k-1)+5}$. However, substituting $s=1$ and $l=l_{0}$ in (28) gives us a possible value $a_{l_{0},0}=2(k-1)$, as defined in Lemma 7. Using Lemma 7 for any $l>l_{0}$, we get in the first coordinate

[TABLE]

which again yields $\nu_{2}(t_{n})=\nu_{2}(n)+\nu_{2}(k-1)+2$. ∎