# Abstract

Double sequences appear in a natural way in cases of iteratively given sequences if the iteration allows to determine besides the successors from the predecessors also the predecessors from their followers. A particular pair of double sequences is considered which appears in a parqueting-reflection process of the complex plane. While one end of each sequence is a natural number sequence, the other consists of rational numbers. The natural numbers sequences are not yet listed in OEIS Wiki. Complex versions from the double sequences are provided.

MSC 2010: 40B05; 40A05

## 1 Introduction

While sequences are known as mappings from the set of natural numbers, double (two-sided) sequences are based on the entire numbers. Although a double sequence can be rearranged in a sequence, such a rearrangement does not necessarily result in a convergent sequence even if the two ends of the double sequence are convergent (when the indices tend to + or - ). Of course, other combinations of the limit behavior are possible. Double sequences appear in a natural way in the cases of iteratively given sequences if the iteration recipe allows to determine besides the successors from the predecessors also the predecessors from their followers.

Repeatedly, iteratively given sequences appear when applying the parqueting-reflection principle to certain circular domains of the (complex) plane; see e.g. . A pair of double sequences arises when repeatedly reflecting a certain circular rectangle at its boundary parts . This pair of double sequences consists of sequences, the tails of which are, on one hand, natural numbers sequences (see ) and, on the other hand, made from rational numbers. Sequences can be extended to ones with complex numbers bearing similar properties as their origins.

## 2 Recurrence relations

With a 0 = b 0 = 1 for k , the relations

a 2 k = 3 a 2 k - 1 + b 2 k - 1 , a 2 k + 1 = a 2 k + b 2 k ,
b 2 k = a 2 k - 1 + b 2 k - 1 , b 2 k + 1 = a 2 k + 3 b 2 k

determine two real sequences. Obviously, both sets of equations may be solved for the lower indexed numbers as

2 a 2 k - 1 = a 2 k - b 2 k , 2 a 2 k = 3 a 2 k + 1 - b 2 k + 1 ,
2 b 2 k - 1 = 3 b 2 k - a 2 k , 2 b 2 k = b 2 k + 1 - a 2 k + 1 .

Hence, ( a k , b k ) is a pair of double sequences ( a k ) , ( b k ) , their first numbers being

, a - 4 = - 5 4 , a - 3 = - 1 2 , a - 2 = - 1 2 , a - 1 = 0 ,
a 0 = 1 , a 1 = 2 , a 2 = 10 , a 3 = 16 , a 4 = 76 , ,
, b - 4 = 1 4 , b - 3 = 1 , b - 2 = 1 2 , b - 1 = 1 ,
b 0 = 1 , b 1 = 4 , b 2 = 6 , b 3 = 28 , b 4 = 44 , .

## 3 Properties of the double sequences

In order to determine the particular explicit numbers in the sequences, some properties are investigated.

## Lemma 3.1.

For any k Z the relations 3 b 2 k 2 - a 2 k 2 = 2 k + 1 and b 2 k + 1 2 - 3 a 2 k + 1 = 2 2 k + 2 hold.

## Proof.

For k = 0 and k = 1 , the relations 3 b 0 2 - a 0 2 = 2 and b 1 2 - 3 a 1 2 = 2 2 are true. By the recurrence relations, for 0 < k ,

3 b 2 k 2 - a 2 k 2 = 3 ( a 2 k - 1 + b 2 k - 1 ) 2 - ( 3 a 2 k - 1 + b 2 k - 1 ) 2 = 2 ( b 2 k - 1 2 - 3 a 2 k - 1 2 ) = 2 [ ( a 2 k - 2 + 3 b 2 k - 2 ) 2 - 3 ( a 2 k - 2 + b 2 k - 2 ) 2 ] = 4 ( 3 b 2 k - 2 2 - a 2 k - 2 2 ) ,
b 2 k + 1 2 - 3 a 2 k + 1 = ( a 2 k + 3 b 2 k ) 2 - 3 ( a 2 k + b 2 k ) 2 = 2 ( 3 b 2 k 2 - a 2 k 2 ) = 2 [ 3 ( a 2 k - 1 + b 2 k - 1 ) 2 - ( 3 a 2 k - 1 + b 2 k - 1 ) 2 ] = 4 ( b 2 k - 1 2 - 3 a 2 k - 1 2 )

hold, and for k < 0 ,

3 b 2 k 2 - a 2 k 2 = 3 4 ( b 2 k + 1 - a 2 k + 1 ) 2 - 1 4 ( 3 a 2 k + 1 - b 2 k + 1 ) 2 = 1 2 ( b 2 k + 1 2 - 3 a 2 k + 1 2 ) = 1 8 [ ( 3 b 2 k + 2 - a 2 k + 2 ) 2 - 3 ( a 2 k + 2 - b 2 k + 2 ) 2 ] = 1 4 ( 3 b 2 k + 2 2 - a 2 k + 2 2 ) ,
b 2 k - 1 2 - 3 a 2 k - 1 2 = 1 4 ( 3 b 2 k - a 2 k ) 2 - 3 4 ( a 2 k - b 2 k ) 2 = 1 2 ( 3 b 2 k 2 - a 2 k 2 ) = 1 8 [ 3 ( b 2 k + 1 - a 2 k + 1 ) 2 - ( 3 a 2 k + 1 - b 2 k + 1 ) 2 ] = 1 4 ( b 2 k + 1 2 - 3 a 2 k + 1 2 ) .

## Remark 3.2.

Both formulas from the last lemma are unified as

3 1 + [ k 2 ] - [ k + 1 2 ] b k 2 - 3 [ k + 1 2 ] - [ k 2 ] a k 2 = 2 k + 1 .

For convenience, the notation m 1 = 3 will be further used.

## Lemma 3.3.

For k Z , we have m 1 b 2 k ± a 2 k = ( m 1 ± 1 ) 2 k + 1 and b 2 k + 1 ± m 1 a 2 k + 1 = ( m 1 ± 1 ) 2 k + 2 .

## Proof.

For k = 0 and k = 1 , the relations m 1 b 0 ± a 0 = m 1 ± 1 and b 1 ± m 1 a 1 = 4 ± 2 m 1 = ( m 1 ± 1 ) 2 hold. For 0 < k ,

m 1 b 2 k ± a 2 k = m 1 ( a 2 n - 1 + b 2 k - 1 ) ± ( 3 a 2 n - 1 + b 2 n - 1 ) = ( m 1 ± 1 ) ( b 2 k - 1 ± m 1 a 2 n - 1 ) = ( m 1 ± 1 ) [ ( a 2 k - 2 + 3 b 2 k - 2 ) ± m 1 ( a 2 k - 2 + b 2 k - 2 ) ] = ( m 1 ± 1 ) 2 ( m 1 b 2 n - 2 ± a 2 k - 2 ) ,
b 2 k + 1 ± m 1 a 2 k + 1 = a 2 k + 3 b 2 k ± m 1 ( a 2 k + b 2 k ) = ( m 1 ± 1 ) ( m 1 b 2 k ± a 2 k ) = ( m 1 ± 1 ) [ m 1 ( a 2 k - 1 + b 2 k - 1 ) ± ( 3 a 2 k - 1 + b 2 k - 1 ) ] = ( m 1 ± 1 ) 2 ( b 2 k - 1 ± m 1 a 2 k - 1 ) ,

and for k < 0 , taking into account ( m 1 ± 1 ) ( m 1 1 ) = 2 ,

m 1 b 2 k ± a 2 k = m 1 2 ( b 2 k + 1 - a 2 k + 1 ) ± 1 2 ( 3 a 2 k + 1 - b 2 k + 1 ) = 1 2 [ ( m 1 1 ) b 2 k + 1 - m 1 ( 1 m 1 ) a 2 k + 1 ] = m 1 1 2 ( b 2 k + 1 ± m 1 a 2 k + 1 ) = m 1 1 2 [ 3 b 2 k + 2 - a 2 k + 2 ± m 1 ( a 2 k + 2 - b 2 k + 2 ) ] = m 1 1 4 [ ( 3 m 1 ) b 2 k + 2 - ( 1 m 1 ) a 2 k + 2 ] = ( m 1 1 ) 2 4 ( m 1 b 2 k + 2 ± a 2 k + 2 ) = ( ( m 1 1 ) 2 4 ) - k ( m 1 b 0 ± a 0 ) = ( ( m 1 1 ) 2 4 ) - k ( m 1 ± 1 ) = ( ( m 1 1 ) 2 4 ) - k - 1 m 1 1 2 = ( ( m 1 1 ) 2 ) - 2 k - 1 = ( m 1 ± 1 ) 2 k + 1 ,
b 2 k - 1 ± m 1 a 2 k - 1 = 1 2 [ 3 b 2 k - a 2 k ± m 1 ( a 2 k - b 2 k ) ] = m 1 1 2 ( m 1 b 2 k ± a 2 k ) = m 1 1 4 [ m 1 ( b 2 k + 1 - a 2 k + 1 ) ± ( 3 a 2 k + 1 - b 2 k + 1 ) ] = ( m 1 1 ) 2 4 ( b 2 k + 1 ± m 1 a 2 k + 1 ) = ( ( m 1 1 ) 2 4 ) - k ( b - 1 ± a - 1 ) = ( ( m 1 1 ) 2 4 ) - k = ( 4 2 m 1 4 ) - k = ( m 1 1 2 ) - 2 k = ( m 1 ± 1 ) 2 k .

With the formulas from Lemma 3.3, the terms of the sequences are determined.

## Theorem 3.4.

For k Z , the double sequences ( a k ) and ( b k ) are given via

2 a 2 k = ( m 1 + 1 ) 2 k + 1 - ( m 1 - 1 ) 2 k + 1 , 2 m 1 a 2 k + 1 = ( m 1 + 1 ) 2 k + 2 - ( m 1 - 1 ) 2 k + 2 ,
2 m 1 b 2 k = ( m 1 + 1 ) 2 k + 1 + ( m 1 - 1 ) 2 k + 1 , 2 b 2 k + 1 = ( m 1 + 1 ) 2 k + 2 + ( m 1 - 1 ) 2 k + 2 .

## 4 Complex version

The double sequence pair ( a k , b k ) of rational numbers handled in ( 3 ) has the counterpart in ( 3 , i ) .

For k , define complex numbers as

c 2 k = ( - 1 ) k a 2 k , c 2 k + 1 = ( - 1 ) k + 1 i m 1 a 2 k + 1 , d 2 k = ( - 1 ) k + 1 i m 1 b 2 k , d 2 k + 1 = ( - 1 ) k + 1 b 2 k + 1 .

Then the recursion relations for a k and b k are reflected into ones for c k and d k as

c k = - i m 1 c k - 1 + d k - 1 , d k = - c k - 1 - i m 1 d k - 1 with c 0 = 1 , d 0 = - i m 1 ,
2 c k - 1 = i m 1 c k + d k , 2 d k - 1 = i m 1 d k - c k with c - 1 = 0 , d - 1 = 1 .

The first terms are

, c - 4 = - 5 4 , c - 3 = 1 2 i m 1 , c - 2 = 1 2 , c - 1 = 0 ,
c 0 = 1 , c 1 = - 2 i m 1 , c 2 = 10 , c 3 = - 16 i m 1 , c 4 = 76 , ,
, d - 4 = - 3 4 i m 1 , d - 3 = - 1 , d - 2 = 1 2 i m 1 , d - 1 = 1 ,
d 0 = - i m 1 , d 1 = - 4 , d 2 = 6 i m 1 , d 3 = 28 , d 4 = - 44 i m 1 , .

Their properties are

| d k | 2 - | c k | 2 = 2 k + 1 , | c k | + | d k | = ( m 1 + 1 ) k + 1 , | d k | - | c k | = ( m 1 - 1 ) k + 1 ,
c k 2 + d k 2 = ( - 2 ) k + 1 , c k d k ¯ + c k ¯ d k = 0 , - i c k d k ¯ = i c k ¯ d k = | c k d k |

for k .

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