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A pair of rational double sequences

A pair of rational double sequences 1IntroductionWhile 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 +∞{+\infty}or -∞{-\infty}).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. [1].A pair of double sequences arises when repeatedly reflecting a certain circular rectangle at its boundary parts [2].This pair of double sequences consists of sequences, the tails of which are, on one hand, natural numbers sequences (see [3]) and, on the other hand, made from rational numbers.Sequences can be extended to ones with complex numbers bearing similar properties as their origins.2Recurrence relationsWith a0=b0=1{a_{0}=b_{0}=1}for k∈ℕ{k\in\mathbb{N}}, the relationsa2⁢k=3⁢a2⁢k-1+b2⁢k-1,\displaystyle a_{2k}=3a_{2k-1}+b_{2k-1},a2⁢k+1=a2⁢k+b2⁢k,\displaystyle\hskip 10.0pta_{2k+1}=a_{2k}+b_{2k},b2⁢k=a2⁢k-1+b2⁢k-1,\displaystyle b_{2k}=a_{2k-1}+b_{2k-1},b2⁢k+1=a2⁢k+3⁢b2⁢k\displaystyle\hskip 10.0ptb_{2k+1}=a_{2k}+3b_{2k}determine two real sequences.Obviously, both sets of equations may be solved for the lower indexed numbers as2⁢a2⁢k-1=a2⁢k-b2⁢k,\displaystyle 2a_{2k-1}=a_{2k}-b_{2k},2⁢a2⁢k=3⁢a2⁢k+1-b2⁢k+1,\displaystyle\hskip 10.0pt2a_{2k}=3a_{2k+1}-b_{2k+1},2⁢b2⁢k-1=3⁢b2⁢k-a2⁢k,\displaystyle 2b_{2k-1}=3b_{2k}-a_{2k},2⁢b2⁢k=b2⁢k+1-a2⁢k+1.\displaystyle\hskip 10.0pt2b_{2k}=b_{2k+1}-a_{2k+1}.Hence, (ak,bk){(a_{k},b_{k})}is a pair of double sequences (ak){(a_{k})}, (bk){(b_{k})}, their first numbers being…,a-4=-54,\displaystyle\dots,a_{-4}=\frac{-5}{4},a-3=-12,\displaystyle\hskip 10.0pta_{-3}=\frac{-1}{2},a-2 =-12,\displaystyle a_{-2}\hskip 10.0pt=\frac{-1}{2},a-1=0,\displaystyle a_{-1}=0,a0=1,\displaystyle\hskip 10.0pta_{0}=1,a1 =2,\displaystyle a_{1}\hskip 10.0pt=2,a2=10,\displaystyle a_{2}=10,a3=16,\displaystyle\hskip 10.0pta_{3}=16,a4 =76,…,\displaystyle a_{4}\hskip 10.0pt=76,\dots,…,b-4=14,\displaystyle\dots,b_{-4}=\frac{1}{4},b-3=1,\displaystyle\hskip 10.0ptb_{-3}=1,b-2 =12,\displaystyle b_{-2}\hskip 10.0pt=\frac{1}{2},b-1=1,\displaystyle b_{-1}=1,b0=1,\displaystyle\hskip 10.0ptb_{0}=1,b1 =4,\displaystyle b_{1}\hskip 10.0pt=4,b2=6,\displaystyle b_{2}=6,b3=28,\displaystyle\hskip 10.0ptb_{3}=28,b4 =44,….\displaystyle b_{4}\hskip 10.0pt=44,\dots.3Properties of the double sequencesIn order to determine the particular explicit numbers in the sequences, some properties are investigated.Lemma 3.1.For any k∈Z{k\in\mathbb{Z}}the relations 3⁢b2⁢k2-a2⁢k2=2k+1{3b_{2k}^{2}-a_{2k}^{2}=2^{k+1}}and b2⁢k+12-3⁢a2⁢k+1=22⁢k+2{b_{2k+1}^{2}-3a_{2k+1}=2^{2k+2}}hold.Proof.For k=0{k=0}and k=1{k=1}, the relations 3⁢b02-a02=2{3b_{0}^{2}-a_{0}^{2}=2}and b12-3⁢a12=22{b_{1}^{2}-3a_{1}^{2}=2^{2}}are true.By the recurrence relations, for 0<k{0<k},3⁢b2⁢k2-a2⁢k2=3⁢(a2⁢k-1+b2⁢k-1)2-(3⁢a2⁢k-1+b2⁢k-1)2=2⁢(b2⁢k-12-3⁢a2⁢k-12)=2⁢[(a2⁢k-2+3⁢b2⁢k-2)2-3⁢(a2⁢k-2+b2⁢k-2)2]=4⁢(3⁢b2⁢k-22-a2⁢k-22),\displaystyle\begin{split}\displaystyle 3b_{2k}^{2}-a_{2k}^{2}&\displaystyle=3%(a_{2k-1}+b_{2k-1})^{2}-(3a_{2k-1}+b_{2k-1})^{2}=2(b_{2k-1}^{2}-3a_{2k-1}^{2})%\\&\displaystyle=2[(a_{2k-2}+3b_{2k-2})^{2}-3(a_{2k-2}+b_{2k-2})^{2}]=4(3b_{2k-2%}^{2}-a_{2k-2}^{2}),\end{split}b2⁢k+12-3⁢a2⁢k+1=(a2⁢k+3⁢b2⁢k)2-3⁢(a2⁢k+b2⁢k)2=2⁢(3⁢b2⁢k2-a2⁢k2)=2⁢[3⁢(a2⁢k-1+b2⁢k-1)2-(3⁢a2⁢k-1+b2⁢k-1)2]=4⁢(b2⁢k-12-3⁢a2⁢k-12)\displaystyle\begin{split}\displaystyle b_{2k+1}^{2}-3a_{2k+1}&\displaystyle=(%a_{2k}+3b_{2k})^{2}-3(a_{2k}+b_{2k})^{2}=2(3b_{2k}^{2}-a_{2k}^{2})\\&\displaystyle=2[3(a_{2k-1}+b_{2k-1})^{2}-(3a_{2k-1}+b_{2k-1})^{2}]=4(b_{2k-1}%^{2}-3a_{2k-1}^{2})\end{split}hold, and for k<0{k<0},3⁢b2⁢k2-a2⁢k2=34⁢(b2⁢k+1-a2⁢k+1)2-14⁢(3⁢a2⁢k+1-b2⁢k+1)2=12⁢(b2⁢k+12-3⁢a2⁢k+12)=18⁢[(3⁢b2⁢k+2-a2⁢k+2)2-3⁢(a2⁢k+2-b2⁢k+2)2]=14⁢(3⁢b2⁢k+22-a2⁢k+22),\displaystyle\begin{split}\displaystyle 3b_{2k}^{2}-a_{2k}^{2}&\displaystyle=%\frac{3}{4}(b_{2k+1}-a_{2k+1})^{2}-\frac{1}{4}(3a_{2k+1}-b_{2k+1})^{2}=\frac{1%}{2}(b_{2k+1}^{2}-3a_{2k+1}^{2})\\&\displaystyle=\frac{1}{8}[(3b_{2k+2}-a_{2k+2})^{2}-3(a_{2k+2}-b_{2k+2})^{2}]=%\frac{1}{4}(3b_{2k+2}^{2}-a_{2k+2}^{2}),\end{split}b2⁢k-12-3⁢a2⁢k-12=14⁢(3⁢b2⁢k-a2⁢k)2-34⁢(a2⁢k-b2⁢k)2=12⁢(3⁢b2⁢k2-a2⁢k2)=18⁢[3⁢(b2⁢k+1-a2⁢k+1)2-(3⁢a2⁢k+1-b2⁢k+1)2]=14⁢(b2⁢k+12-3⁢a2⁢k+12).∎\displaystyle\begin{split}\displaystyle b_{2k-1}^{2}-3a_{2k-1}^{2}&%\displaystyle=\frac{1}{4}(3b_{2k}-a_{2k})^{2}-\frac{3}{4}(a_{2k-b_{2k}})^{2}=%\frac{1}{2}(3b_{2k}^{2}-a_{2k}^{2})\\&\displaystyle=\frac{1}{8}[3(b_{2k+1}-a_{2k+1})^{2}-(3a_{2k+1}-b_{2k+1})^{2}]=%\frac{1}{4}(b_{2k+1}^{2}-3a_{2k+1}^{2}).\qed\end{split}Remark 3.2.Both formulas from the last lemma are unified as31+[k2]-[k+12]⁢bk2-3[k+12]-[k2]⁢ak2=2k+1.3^{1+[\frac{k}{2}]-[\frac{k+1}{2}]}b_{k}^{2}-3^{[\frac{k+1}{2}]-[\frac{k}{2}]}%a_{k}^{2}=2^{k+1}.For convenience, the notation m1=3{m_{1}=\sqrt{3}}will be further used.Lemma 3.3.For k∈Z{k\in\mathbb{Z}}, we have m1⁢b2⁢k±a2⁢k=(m1±1)2⁢k+1{m_{1}b_{2k}\pm a_{2k}=(m_{1}\pm 1)^{2k+1}}and b2⁢k+1±m1⁢a2⁢k+1=(m1±1)2⁢k+2{b_{2k+1}\pm m_{1}a_{2k+1}=(m_{1}\pm 1)^{2k+2}}.Proof.For k=0{k=0}and k=1{k=1}, the relations m1⁢b0±a0=m1±1{m_{1}b_{0}\pm a_{0}=m_{1}\pm 1}and b1±m1⁢a1=4±2⁢m1=(m1±1)2{b_{1}\pm m_{1}a_{1}=4\pm 2m_{1}=(m_{1}\pm 1)^{2}}hold.For 0<k{0<k},m1⁢b2⁢k±a2⁢k=m1⁢(a2⁢n-1+b2⁢k-1)±(3⁢a2⁢n-1+b2⁢n-1)=(m1±1)⁢(b2⁢k-1±m1⁢a2⁢n-1)=(m1±1)⁢[(a2⁢k-2+3⁢b2⁢k-2)±m1⁢(a2⁢k-2+b2⁢k-2)]=(m1±1)2⁢(m1⁢b2⁢n-2±a2⁢k-2),\displaystyle\begin{split}\displaystyle m_{1}b_{2k}\pm a_{2k}&\displaystyle=m_%{1}(a_{2n-1}+b_{2k-1})\pm(3a_{2n-1}+b_{2n-1})=(m_{1}\pm 1)(b_{2k-1}\pm m_{1}a_%{2n-1})\\&\displaystyle=(m_{1}\pm 1)[(a_{2k-2}+3b_{2k-2})\pm m_{1}(a_{2k-2}+b_{2k-2})]=%(m_{1}\pm 1)^{2}(m_{1}b_{2n-2}\pm a_{2k-2}),\end{split}b2⁢k+1±m1⁢a2⁢k+1=a2⁢k+3⁢b2⁢k±m1⁢(a2⁢k+b2⁢k)=(m1±1)⁢(m1⁢b2⁢k±a2⁢k)=(m1±1)⁢[m1⁢(a2⁢k-1+b2⁢k-1)±(3⁢a2⁢k-1+b2⁢k-1)]=(m1±1)2⁢(b2⁢k-1±m1⁢a2⁢k-1),\displaystyle\begin{split}\displaystyle b_{2k+1}\pm m_{1}a_{2k+1}&%\displaystyle=a_{2k}+3b_{2k}\pm m_{1}(a_{2k}+b_{2k})=(m_{1}\pm 1)(m_{1}b_{2k}%\pm a_{2k})\\&\displaystyle=(m_{1}\pm 1)[m_{1}(a_{2k-1}+b_{2k-1})\pm(3a_{2k-1}+b_{2k-1})]=(%m_{1}\pm 1)^{2}(b_{2k-1}\pm m_{1}a_{2k-1}),\end{split}and for k<0{k<0}, taking into account (m1±1)⁢(m1∓1)=2{(m_{1}\pm 1)(m_{1}\mp 1)=2},m1⁢b2⁢k±a2⁢k=m12⁢(b2⁢k+1-a2⁢k+1)±12⁢(3⁢a2⁢k+1-b2⁢k+1)=12⁢[(m1∓1)⁢b2⁢k+1-m1⁢(1∓m1)⁢a2⁢k+1]=m1∓12⁢(b2⁢k+1±m1⁢a2⁢k+1)=m1∓12⁢[3⁢b2⁢k+2-a2⁢k+2±m1⁢(a2⁢k+2-b2⁢k+2)]=m1∓14⁢[(3∓m1)⁢b2⁢k+2-(1∓m1)⁢a2⁢k+2]=(m1∓1)24⁢(m1⁢b2⁢k+2±a2⁢k+2)=((m1∓1)24)-k⁢(m1⁢b0±a0)=((m1∓1)24)-k⁢(m1±1)=((m1∓1)24)-k-1⁢m1∓12=((m1∓1)2)-2⁢k-1=(m1±1)2⁢k+1,\displaystyle\begin{split}\displaystyle m_{1}b_{2k}\pm a_{2k}&\displaystyle=%\frac{m_{1}}{2(b_{2k+1}-a_{2k+1})}\pm\frac{1}{2}(3a_{2k+1}-b_{2k+1})\\&\displaystyle=\frac{1}{2}[(m_{1}\mp 1)b_{2k+1}-m_{1}(1\mp m_{1})a_{2k+1}]=%\frac{m_{1}\mp 1}{2}(b_{2k+1}\pm m_{1}a_{2k+1})\\&\displaystyle=\frac{m_{1}\mp 1}{2}[3b_{2k+2}-a_{2k+2}\pm m_{1}(a_{2k+2}-b_{2k%+2})]\\&\displaystyle=\frac{m_{1}\mp 1}{4}[(3\mp m_{1})b_{2k+2}-(1\mp m_{1})a_{2k+2}]%=\frac{(m_{1}\mp 1)^{2}}{4}(m_{1}b_{2k+2}\pm a_{2k+2})\\&\displaystyle=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k}(m_{1}b_{0}\pma%_{0})=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k}(m_{1}\pm 1)\\&\displaystyle=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k-1}\frac{m_{1}%\mp 1}{2}=\biggl{(}\frac{(m_{1}\mp 1)}{2}\biggr{)}^{-2k-1}=(m_{1}\pm 1)^{2k+1}%,\end{split}b2⁢k-1±m1⁢a2⁢k-1=12⁢[3⁢b2⁢k-a2⁢k±m1⁢(a2⁢k-b2⁢k)]=m1∓12⁢(m1⁢b2⁢k±a2⁢k)=m1∓14⁢[m1⁢(b2⁢k+1-a2⁢k+1)±(3⁢a2⁢k+1-b2⁢k+1)]=(m1∓1)24⁢(b2⁢k+1±m1⁢a2⁢k+1)=((m1∓1)24)-k⁢(b-1±a-1)=((m1∓1)24)-k=(4∓2⁢m14)-k=(m1∓12)-2⁢k=(m1±1)2⁢k.∎\displaystyle\begin{split}\displaystyle b_{2k-1}\pm m_{1}a_{2k-1}&%\displaystyle=\frac{1}{2}[3b_{2k}-a_{2k}\pm m_{1}(a_{2k}-b_{2k})]=\frac{m_{1}%\mp 1}{2}(m_{1}b_{2k}\pm a_{2k})\\&\displaystyle=\frac{m_{1}\mp 1}{4}[m_{1}(b_{2k+1}-a_{2k+1})\pm(3a_{2k+1}-b_{2%k+1})]=\frac{(m_{1}\mp 1)^{2}}{4}(b_{2k+1}\pm m_{1}a_{2k+1})\\&\displaystyle=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k}(b_{-1}\pm a_{%-1})=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k}\\&\displaystyle=\biggl{(}\frac{4\mp 2m_{1}}{4}\biggr{)}^{-k}=\biggl{(}\frac{m_{%1}\mp 1}{2}\biggr{)}^{-2k}=(m_{1}\pm 1)^{2k}.\qed\end{split}With the formulas from Lemma 3.3, the terms of the sequences are determined.Theorem 3.4.For k∈Z{k\in\mathbb{Z}}, the double sequences (ak){(a_{k})}and (bk){(b_{k})}are given via2⁢a2⁢k=(m1+1)2⁢k+1-(m1-1)2⁢k+1,\displaystyle 2a_{2k}=(m_{1}+1)^{2k+1}-(m_{1}-1)^{2k+1},2⁢m1⁢a2⁢k+1=(m1+1)2⁢k+2-(m1-1)2⁢k+2,\displaystyle\hskip 10.0pt2m_{1}a_{2k+1}=(m_{1}+1)^{2k+2}-(m_{1}-1)^{2k+2},2⁢m1⁢b2⁢k=(m1+1)2⁢k+1+(m1-1)2⁢k+1,\displaystyle 2m_{1}b_{2k}=(m_{1}+1)^{2k+1}+(m_{1}-1)^{2k+1},2⁢b2⁢k+1=(m1+1)2⁢k+2+(m1-1)2⁢k+2.\displaystyle\hskip 10.0pt2b_{2k+1}=(m_{1}+1)^{2k+2}+(m_{1}-1)^{2k+2}.4Complex versionThe double sequence pair (ak,bk){(a_{k},b_{k})}of rational numbers handled in ℚ⁢(3){\mathbb{Q}(\sqrt{3})}has the counterpart in ℚ⁢(3,i){\mathbb{Q}(\sqrt{3},i)}.For k∈ℤ{k\in\mathbb{Z}}, define complex numbers asc2⁢k=(-1)k⁢a2⁢k,c2⁢k+1=(-1)k+1⁢i⁢m1⁢a2⁢k+1,d2⁢k=(-1)k+1⁢i⁢m1⁢b2⁢k,d2⁢k+1=(-1)k+1⁢b2⁢k+1.c_{2k}=(-1)^{k}a_{2k},\quad c_{2k+1}=(-1)^{k+1}im_{1}a_{2k+1},\quad d_{2k}=(-1%)^{k+1}im_{1}b_{2k},\quad d_{2k+1}=(-1)^{k+1}b_{2k+1}.Then the recursion relations for ak{a_{k}}and bk{b_{k}}are reflected into ones for ck{c_{k}}and dk{d_{k}}asck=-i⁢m1⁢ck-1+dk-1,\displaystyle c_{k}=-im_{1}c_{k-1}+d_{k-1},dk=-ck-1-i⁢m1⁢dk-1\displaystyle\hskip 10.0ptd_{k}=-c_{k-1}-im_{1}d_{k-1}with⁢c0=1,d0=-i⁢m1,\hskip 10.0pt\text{with}\ c_{0}=1,\,d_{0}=-im_{1},2⁢ck-1=i⁢m1⁢ck+dk,\displaystyle 2c_{k-1}=im_{1}c_{k}+d_{k},2⁢dk-1=i⁢m1⁢dk-ck\displaystyle\hskip 10.0pt2d_{k-1}=im_{1}d_{k}-c_{k}with⁢c-1=0,d-1=1.\hskip 10.0pt\text{with}\ c_{-1}=0,\,d_{-1}=1.The first terms are…,c-4=-54,\displaystyle\dots,c_{-4}=-\frac{5}{4},c-3=12⁢i⁢m1,\displaystyle\hskip 10.0ptc_{-3}=\frac{1}{2}im_{1},c-2 =12,\displaystyle c_{-2}\hskip 10.0pt=\frac{1}{2},c-1=0,\displaystyle c_{-1}=0,c0=1,\displaystyle\hskip 10.0ptc_{0}=1,c1 =-2im1,\displaystyle c_{1}\hskip 10.0pt=-2im_{1},c2=10,\displaystyle c_{2}=10,c3=-16⁢i⁢m1,\displaystyle\hskip 10.0ptc_{3}=-16im_{1},c4 =76,…,\displaystyle c_{4}\hskip 10.0pt=76,\dots,…,d-4=-34⁢i⁢m1,\displaystyle\dots,d_{-4}=-\frac{3}{4}im_{1},d-3=-1,\displaystyle\hskip 10.0ptd_{-3}=-1,d-2 =12im1,\displaystyle d_{-2}\hskip 10.0pt=\frac{1}{2}im_{1},d-1=1,\displaystyle d_{-1}=1,d0=-i⁢m1,\displaystyle\hskip 10.0ptd_{0}=-im_{1},d1 =-4,\displaystyle d_{1}\hskip 10.0pt=-4,d2=6⁢i⁢m1,\displaystyle d_{2}=6im_{1},d3=28,\displaystyle\hskip 10.0ptd_{3}=28,d4 =-44im1,….\displaystyle d_{4}\hskip 10.0pt=-44im_{1},\dots.Their properties are|dk|2-|ck|2=2k+1,|ck|+|dk|=(m1+1)k+1,|dk|-|ck|=(m1-1)k+1,\displaystyle\lvert d_{k}\rvert^{2}-\lvert c_{k}\rvert^{2}=2^{k+1},\quad\lvertc%_{k}\rvert+\lvert d_{k}\rvert=(m_{1}+1)^{k+1},\quad\lvert d_{k}\rvert-\lvert c%_{k}\rvert=(m_{1}-1)^{k+1},ck2+dk2=(-2)k+1,ck⁢dk¯+ck¯⁢dk=0,-i⁢ck⁢dk¯=i⁢ck¯⁢dk=|ck⁢dk|\displaystyle c_{k}^{2}+d_{k}^{2}=(-2)^{k+1},\quad c_{k}\overline{d_{k}}+%\overline{c_{k}}d_{k}=0,\quad-ic_{k}\overline{d_{k}}=i\overline{c_{k}}d_{k}=%\lvert c_{k}d_{k}\rvertfor k∈ℤ{k\in\mathbb{Z}}. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Georgian Mathematical Journal de Gruyter

A pair of rational double sequences

Georgian Mathematical Journal , Volume 29 (2): 4 – Apr 1, 2022

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de Gruyter
Copyright
© 2022 Walter de Gruyter GmbH, Berlin/Boston
ISSN
1572-9176
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1572-9176
DOI
10.1515/gmj-2021-2119
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Abstract

1IntroductionWhile 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 +∞{+\infty}or -∞{-\infty}).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. [1].A pair of double sequences arises when repeatedly reflecting a certain circular rectangle at its boundary parts [2].This pair of double sequences consists of sequences, the tails of which are, on one hand, natural numbers sequences (see [3]) and, on the other hand, made from rational numbers.Sequences can be extended to ones with complex numbers bearing similar properties as their origins.2Recurrence relationsWith a0=b0=1{a_{0}=b_{0}=1}for k∈ℕ{k\in\mathbb{N}}, the relationsa2⁢k=3⁢a2⁢k-1+b2⁢k-1,\displaystyle a_{2k}=3a_{2k-1}+b_{2k-1},a2⁢k+1=a2⁢k+b2⁢k,\displaystyle\hskip 10.0pta_{2k+1}=a_{2k}+b_{2k},b2⁢k=a2⁢k-1+b2⁢k-1,\displaystyle b_{2k}=a_{2k-1}+b_{2k-1},b2⁢k+1=a2⁢k+3⁢b2⁢k\displaystyle\hskip 10.0ptb_{2k+1}=a_{2k}+3b_{2k}determine two real sequences.Obviously, both sets of equations may be solved for the lower indexed numbers as2⁢a2⁢k-1=a2⁢k-b2⁢k,\displaystyle 2a_{2k-1}=a_{2k}-b_{2k},2⁢a2⁢k=3⁢a2⁢k+1-b2⁢k+1,\displaystyle\hskip 10.0pt2a_{2k}=3a_{2k+1}-b_{2k+1},2⁢b2⁢k-1=3⁢b2⁢k-a2⁢k,\displaystyle 2b_{2k-1}=3b_{2k}-a_{2k},2⁢b2⁢k=b2⁢k+1-a2⁢k+1.\displaystyle\hskip 10.0pt2b_{2k}=b_{2k+1}-a_{2k+1}.Hence, (ak,bk){(a_{k},b_{k})}is a pair of double sequences (ak){(a_{k})}, (bk){(b_{k})}, their first numbers being…,a-4=-54,\displaystyle\dots,a_{-4}=\frac{-5}{4},a-3=-12,\displaystyle\hskip 10.0pta_{-3}=\frac{-1}{2},a-2 =-12,\displaystyle a_{-2}\hskip 10.0pt=\frac{-1}{2},a-1=0,\displaystyle a_{-1}=0,a0=1,\displaystyle\hskip 10.0pta_{0}=1,a1 =2,\displaystyle a_{1}\hskip 10.0pt=2,a2=10,\displaystyle a_{2}=10,a3=16,\displaystyle\hskip 10.0pta_{3}=16,a4 =76,…,\displaystyle a_{4}\hskip 10.0pt=76,\dots,…,b-4=14,\displaystyle\dots,b_{-4}=\frac{1}{4},b-3=1,\displaystyle\hskip 10.0ptb_{-3}=1,b-2 =12,\displaystyle b_{-2}\hskip 10.0pt=\frac{1}{2},b-1=1,\displaystyle b_{-1}=1,b0=1,\displaystyle\hskip 10.0ptb_{0}=1,b1 =4,\displaystyle b_{1}\hskip 10.0pt=4,b2=6,\displaystyle b_{2}=6,b3=28,\displaystyle\hskip 10.0ptb_{3}=28,b4 =44,….\displaystyle b_{4}\hskip 10.0pt=44,\dots.3Properties of the double sequencesIn order to determine the particular explicit numbers in the sequences, some properties are investigated.Lemma 3.1.For any k∈Z{k\in\mathbb{Z}}the relations 3⁢b2⁢k2-a2⁢k2=2k+1{3b_{2k}^{2}-a_{2k}^{2}=2^{k+1}}and b2⁢k+12-3⁢a2⁢k+1=22⁢k+2{b_{2k+1}^{2}-3a_{2k+1}=2^{2k+2}}hold.Proof.For k=0{k=0}and k=1{k=1}, the relations 3⁢b02-a02=2{3b_{0}^{2}-a_{0}^{2}=2}and b12-3⁢a12=22{b_{1}^{2}-3a_{1}^{2}=2^{2}}are true.By the recurrence relations, for 0<k{0<k},3⁢b2⁢k2-a2⁢k2=3⁢(a2⁢k-1+b2⁢k-1)2-(3⁢a2⁢k-1+b2⁢k-1)2=2⁢(b2⁢k-12-3⁢a2⁢k-12)=2⁢[(a2⁢k-2+3⁢b2⁢k-2)2-3⁢(a2⁢k-2+b2⁢k-2)2]=4⁢(3⁢b2⁢k-22-a2⁢k-22),\displaystyle\begin{split}\displaystyle 3b_{2k}^{2}-a_{2k}^{2}&\displaystyle=3%(a_{2k-1}+b_{2k-1})^{2}-(3a_{2k-1}+b_{2k-1})^{2}=2(b_{2k-1}^{2}-3a_{2k-1}^{2})%\\&\displaystyle=2[(a_{2k-2}+3b_{2k-2})^{2}-3(a_{2k-2}+b_{2k-2})^{2}]=4(3b_{2k-2%}^{2}-a_{2k-2}^{2}),\end{split}b2⁢k+12-3⁢a2⁢k+1=(a2⁢k+3⁢b2⁢k)2-3⁢(a2⁢k+b2⁢k)2=2⁢(3⁢b2⁢k2-a2⁢k2)=2⁢[3⁢(a2⁢k-1+b2⁢k-1)2-(3⁢a2⁢k-1+b2⁢k-1)2]=4⁢(b2⁢k-12-3⁢a2⁢k-12)\displaystyle\begin{split}\displaystyle b_{2k+1}^{2}-3a_{2k+1}&\displaystyle=(%a_{2k}+3b_{2k})^{2}-3(a_{2k}+b_{2k})^{2}=2(3b_{2k}^{2}-a_{2k}^{2})\\&\displaystyle=2[3(a_{2k-1}+b_{2k-1})^{2}-(3a_{2k-1}+b_{2k-1})^{2}]=4(b_{2k-1}%^{2}-3a_{2k-1}^{2})\end{split}hold, and for k<0{k<0},3⁢b2⁢k2-a2⁢k2=34⁢(b2⁢k+1-a2⁢k+1)2-14⁢(3⁢a2⁢k+1-b2⁢k+1)2=12⁢(b2⁢k+12-3⁢a2⁢k+12)=18⁢[(3⁢b2⁢k+2-a2⁢k+2)2-3⁢(a2⁢k+2-b2⁢k+2)2]=14⁢(3⁢b2⁢k+22-a2⁢k+22),\displaystyle\begin{split}\displaystyle 3b_{2k}^{2}-a_{2k}^{2}&\displaystyle=%\frac{3}{4}(b_{2k+1}-a_{2k+1})^{2}-\frac{1}{4}(3a_{2k+1}-b_{2k+1})^{2}=\frac{1%}{2}(b_{2k+1}^{2}-3a_{2k+1}^{2})\\&\displaystyle=\frac{1}{8}[(3b_{2k+2}-a_{2k+2})^{2}-3(a_{2k+2}-b_{2k+2})^{2}]=%\frac{1}{4}(3b_{2k+2}^{2}-a_{2k+2}^{2}),\end{split}b2⁢k-12-3⁢a2⁢k-12=14⁢(3⁢b2⁢k-a2⁢k)2-34⁢(a2⁢k-b2⁢k)2=12⁢(3⁢b2⁢k2-a2⁢k2)=18⁢[3⁢(b2⁢k+1-a2⁢k+1)2-(3⁢a2⁢k+1-b2⁢k+1)2]=14⁢(b2⁢k+12-3⁢a2⁢k+12).∎\displaystyle\begin{split}\displaystyle b_{2k-1}^{2}-3a_{2k-1}^{2}&%\displaystyle=\frac{1}{4}(3b_{2k}-a_{2k})^{2}-\frac{3}{4}(a_{2k-b_{2k}})^{2}=%\frac{1}{2}(3b_{2k}^{2}-a_{2k}^{2})\\&\displaystyle=\frac{1}{8}[3(b_{2k+1}-a_{2k+1})^{2}-(3a_{2k+1}-b_{2k+1})^{2}]=%\frac{1}{4}(b_{2k+1}^{2}-3a_{2k+1}^{2}).\qed\end{split}Remark 3.2.Both formulas from the last lemma are unified as31+[k2]-[k+12]⁢bk2-3[k+12]-[k2]⁢ak2=2k+1.3^{1+[\frac{k}{2}]-[\frac{k+1}{2}]}b_{k}^{2}-3^{[\frac{k+1}{2}]-[\frac{k}{2}]}%a_{k}^{2}=2^{k+1}.For convenience, the notation m1=3{m_{1}=\sqrt{3}}will be further used.Lemma 3.3.For k∈Z{k\in\mathbb{Z}}, we have m1⁢b2⁢k±a2⁢k=(m1±1)2⁢k+1{m_{1}b_{2k}\pm a_{2k}=(m_{1}\pm 1)^{2k+1}}and b2⁢k+1±m1⁢a2⁢k+1=(m1±1)2⁢k+2{b_{2k+1}\pm m_{1}a_{2k+1}=(m_{1}\pm 1)^{2k+2}}.Proof.For k=0{k=0}and k=1{k=1}, the relations m1⁢b0±a0=m1±1{m_{1}b_{0}\pm a_{0}=m_{1}\pm 1}and b1±m1⁢a1=4±2⁢m1=(m1±1)2{b_{1}\pm m_{1}a_{1}=4\pm 2m_{1}=(m_{1}\pm 1)^{2}}hold.For 0<k{0<k},m1⁢b2⁢k±a2⁢k=m1⁢(a2⁢n-1+b2⁢k-1)±(3⁢a2⁢n-1+b2⁢n-1)=(m1±1)⁢(b2⁢k-1±m1⁢a2⁢n-1)=(m1±1)⁢[(a2⁢k-2+3⁢b2⁢k-2)±m1⁢(a2⁢k-2+b2⁢k-2)]=(m1±1)2⁢(m1⁢b2⁢n-2±a2⁢k-2),\displaystyle\begin{split}\displaystyle m_{1}b_{2k}\pm a_{2k}&\displaystyle=m_%{1}(a_{2n-1}+b_{2k-1})\pm(3a_{2n-1}+b_{2n-1})=(m_{1}\pm 1)(b_{2k-1}\pm m_{1}a_%{2n-1})\\&\displaystyle=(m_{1}\pm 1)[(a_{2k-2}+3b_{2k-2})\pm m_{1}(a_{2k-2}+b_{2k-2})]=%(m_{1}\pm 1)^{2}(m_{1}b_{2n-2}\pm a_{2k-2}),\end{split}b2⁢k+1±m1⁢a2⁢k+1=a2⁢k+3⁢b2⁢k±m1⁢(a2⁢k+b2⁢k)=(m1±1)⁢(m1⁢b2⁢k±a2⁢k)=(m1±1)⁢[m1⁢(a2⁢k-1+b2⁢k-1)±(3⁢a2⁢k-1+b2⁢k-1)]=(m1±1)2⁢(b2⁢k-1±m1⁢a2⁢k-1),\displaystyle\begin{split}\displaystyle b_{2k+1}\pm m_{1}a_{2k+1}&%\displaystyle=a_{2k}+3b_{2k}\pm m_{1}(a_{2k}+b_{2k})=(m_{1}\pm 1)(m_{1}b_{2k}%\pm a_{2k})\\&\displaystyle=(m_{1}\pm 1)[m_{1}(a_{2k-1}+b_{2k-1})\pm(3a_{2k-1}+b_{2k-1})]=(%m_{1}\pm 1)^{2}(b_{2k-1}\pm m_{1}a_{2k-1}),\end{split}and for k<0{k<0}, taking into account (m1±1)⁢(m1∓1)=2{(m_{1}\pm 1)(m_{1}\mp 1)=2},m1⁢b2⁢k±a2⁢k=m12⁢(b2⁢k+1-a2⁢k+1)±12⁢(3⁢a2⁢k+1-b2⁢k+1)=12⁢[(m1∓1)⁢b2⁢k+1-m1⁢(1∓m1)⁢a2⁢k+1]=m1∓12⁢(b2⁢k+1±m1⁢a2⁢k+1)=m1∓12⁢[3⁢b2⁢k+2-a2⁢k+2±m1⁢(a2⁢k+2-b2⁢k+2)]=m1∓14⁢[(3∓m1)⁢b2⁢k+2-(1∓m1)⁢a2⁢k+2]=(m1∓1)24⁢(m1⁢b2⁢k+2±a2⁢k+2)=((m1∓1)24)-k⁢(m1⁢b0±a0)=((m1∓1)24)-k⁢(m1±1)=((m1∓1)24)-k-1⁢m1∓12=((m1∓1)2)-2⁢k-1=(m1±1)2⁢k+1,\displaystyle\begin{split}\displaystyle m_{1}b_{2k}\pm a_{2k}&\displaystyle=%\frac{m_{1}}{2(b_{2k+1}-a_{2k+1})}\pm\frac{1}{2}(3a_{2k+1}-b_{2k+1})\\&\displaystyle=\frac{1}{2}[(m_{1}\mp 1)b_{2k+1}-m_{1}(1\mp m_{1})a_{2k+1}]=%\frac{m_{1}\mp 1}{2}(b_{2k+1}\pm m_{1}a_{2k+1})\\&\displaystyle=\frac{m_{1}\mp 1}{2}[3b_{2k+2}-a_{2k+2}\pm m_{1}(a_{2k+2}-b_{2k%+2})]\\&\displaystyle=\frac{m_{1}\mp 1}{4}[(3\mp m_{1})b_{2k+2}-(1\mp m_{1})a_{2k+2}]%=\frac{(m_{1}\mp 1)^{2}}{4}(m_{1}b_{2k+2}\pm a_{2k+2})\\&\displaystyle=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k}(m_{1}b_{0}\pma%_{0})=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k}(m_{1}\pm 1)\\&\displaystyle=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k-1}\frac{m_{1}%\mp 1}{2}=\biggl{(}\frac{(m_{1}\mp 1)}{2}\biggr{)}^{-2k-1}=(m_{1}\pm 1)^{2k+1}%,\end{split}b2⁢k-1±m1⁢a2⁢k-1=12⁢[3⁢b2⁢k-a2⁢k±m1⁢(a2⁢k-b2⁢k)]=m1∓12⁢(m1⁢b2⁢k±a2⁢k)=m1∓14⁢[m1⁢(b2⁢k+1-a2⁢k+1)±(3⁢a2⁢k+1-b2⁢k+1)]=(m1∓1)24⁢(b2⁢k+1±m1⁢a2⁢k+1)=((m1∓1)24)-k⁢(b-1±a-1)=((m1∓1)24)-k=(4∓2⁢m14)-k=(m1∓12)-2⁢k=(m1±1)2⁢k.∎\displaystyle\begin{split}\displaystyle b_{2k-1}\pm m_{1}a_{2k-1}&%\displaystyle=\frac{1}{2}[3b_{2k}-a_{2k}\pm m_{1}(a_{2k}-b_{2k})]=\frac{m_{1}%\mp 1}{2}(m_{1}b_{2k}\pm a_{2k})\\&\displaystyle=\frac{m_{1}\mp 1}{4}[m_{1}(b_{2k+1}-a_{2k+1})\pm(3a_{2k+1}-b_{2%k+1})]=\frac{(m_{1}\mp 1)^{2}}{4}(b_{2k+1}\pm m_{1}a_{2k+1})\\&\displaystyle=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k}(b_{-1}\pm a_{%-1})=\biggl{(}\frac{(m_{1}\mp 1)^{2}}{4}\biggr{)}^{-k}\\&\displaystyle=\biggl{(}\frac{4\mp 2m_{1}}{4}\biggr{)}^{-k}=\biggl{(}\frac{m_{%1}\mp 1}{2}\biggr{)}^{-2k}=(m_{1}\pm 1)^{2k}.\qed\end{split}With the formulas from Lemma 3.3, the terms of the sequences are determined.Theorem 3.4.For k∈Z{k\in\mathbb{Z}}, the double sequences (ak){(a_{k})}and (bk){(b_{k})}are given via2⁢a2⁢k=(m1+1)2⁢k+1-(m1-1)2⁢k+1,\displaystyle 2a_{2k}=(m_{1}+1)^{2k+1}-(m_{1}-1)^{2k+1},2⁢m1⁢a2⁢k+1=(m1+1)2⁢k+2-(m1-1)2⁢k+2,\displaystyle\hskip 10.0pt2m_{1}a_{2k+1}=(m_{1}+1)^{2k+2}-(m_{1}-1)^{2k+2},2⁢m1⁢b2⁢k=(m1+1)2⁢k+1+(m1-1)2⁢k+1,\displaystyle 2m_{1}b_{2k}=(m_{1}+1)^{2k+1}+(m_{1}-1)^{2k+1},2⁢b2⁢k+1=(m1+1)2⁢k+2+(m1-1)2⁢k+2.\displaystyle\hskip 10.0pt2b_{2k+1}=(m_{1}+1)^{2k+2}+(m_{1}-1)^{2k+2}.4Complex versionThe double sequence pair (ak,bk){(a_{k},b_{k})}of rational numbers handled in ℚ⁢(3){\mathbb{Q}(\sqrt{3})}has the counterpart in ℚ⁢(3,i){\mathbb{Q}(\sqrt{3},i)}.For k∈ℤ{k\in\mathbb{Z}}, define complex numbers asc2⁢k=(-1)k⁢a2⁢k,c2⁢k+1=(-1)k+1⁢i⁢m1⁢a2⁢k+1,d2⁢k=(-1)k+1⁢i⁢m1⁢b2⁢k,d2⁢k+1=(-1)k+1⁢b2⁢k+1.c_{2k}=(-1)^{k}a_{2k},\quad c_{2k+1}=(-1)^{k+1}im_{1}a_{2k+1},\quad d_{2k}=(-1%)^{k+1}im_{1}b_{2k},\quad d_{2k+1}=(-1)^{k+1}b_{2k+1}.Then the recursion relations for ak{a_{k}}and bk{b_{k}}are reflected into ones for ck{c_{k}}and dk{d_{k}}asck=-i⁢m1⁢ck-1+dk-1,\displaystyle c_{k}=-im_{1}c_{k-1}+d_{k-1},dk=-ck-1-i⁢m1⁢dk-1\displaystyle\hskip 10.0ptd_{k}=-c_{k-1}-im_{1}d_{k-1}with⁢c0=1,d0=-i⁢m1,\hskip 10.0pt\text{with}\ c_{0}=1,\,d_{0}=-im_{1},2⁢ck-1=i⁢m1⁢ck+dk,\displaystyle 2c_{k-1}=im_{1}c_{k}+d_{k},2⁢dk-1=i⁢m1⁢dk-ck\displaystyle\hskip 10.0pt2d_{k-1}=im_{1}d_{k}-c_{k}with⁢c-1=0,d-1=1.\hskip 10.0pt\text{with}\ c_{-1}=0,\,d_{-1}=1.The first terms are…,c-4=-54,\displaystyle\dots,c_{-4}=-\frac{5}{4},c-3=12⁢i⁢m1,\displaystyle\hskip 10.0ptc_{-3}=\frac{1}{2}im_{1},c-2 =12,\displaystyle c_{-2}\hskip 10.0pt=\frac{1}{2},c-1=0,\displaystyle c_{-1}=0,c0=1,\displaystyle\hskip 10.0ptc_{0}=1,c1 =-2im1,\displaystyle c_{1}\hskip 10.0pt=-2im_{1},c2=10,\displaystyle c_{2}=10,c3=-16⁢i⁢m1,\displaystyle\hskip 10.0ptc_{3}=-16im_{1},c4 =76,…,\displaystyle c_{4}\hskip 10.0pt=76,\dots,…,d-4=-34⁢i⁢m1,\displaystyle\dots,d_{-4}=-\frac{3}{4}im_{1},d-3=-1,\displaystyle\hskip 10.0ptd_{-3}=-1,d-2 =12im1,\displaystyle d_{-2}\hskip 10.0pt=\frac{1}{2}im_{1},d-1=1,\displaystyle d_{-1}=1,d0=-i⁢m1,\displaystyle\hskip 10.0ptd_{0}=-im_{1},d1 =-4,\displaystyle d_{1}\hskip 10.0pt=-4,d2=6⁢i⁢m1,\displaystyle d_{2}=6im_{1},d3=28,\displaystyle\hskip 10.0ptd_{3}=28,d4 =-44im1,….\displaystyle d_{4}\hskip 10.0pt=-44im_{1},\dots.Their properties are|dk|2-|ck|2=2k+1,|ck|+|dk|=(m1+1)k+1,|dk|-|ck|=(m1-1)k+1,\displaystyle\lvert d_{k}\rvert^{2}-\lvert c_{k}\rvert^{2}=2^{k+1},\quad\lvertc%_{k}\rvert+\lvert d_{k}\rvert=(m_{1}+1)^{k+1},\quad\lvert d_{k}\rvert-\lvert c%_{k}\rvert=(m_{1}-1)^{k+1},ck2+dk2=(-2)k+1,ck⁢dk¯+ck¯⁢dk=0,-i⁢ck⁢dk¯=i⁢ck¯⁢dk=|ck⁢dk|\displaystyle c_{k}^{2}+d_{k}^{2}=(-2)^{k+1},\quad c_{k}\overline{d_{k}}+%\overline{c_{k}}d_{k}=0,\quad-ic_{k}\overline{d_{k}}=i\overline{c_{k}}d_{k}=%\lvert c_{k}d_{k}\rvertfor k∈ℤ{k\in\mathbb{Z}}.

Journal

Georgian Mathematical Journalde Gruyter

Published: Apr 1, 2022

Keywords: Two-sided sequences; natural numbers sequences; complex sequences; 40B05; 40A05

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