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Number defined by a recursive binary sequence
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In a math column in Scientific American many years ago, I encountered a peculiar binary sequence I describe below. Unfortunately I can't find a reference on this, so I would be grateful for any pointers or references.
Let $mathbbN$ be the set of positive integers and let $T = 2^n: nin mathbbNcup 0$ denote the set of powers of $2$. Let $textm:mathbbNto Tcup0$ be defined by $nmapsto maxbig(0cup tin T: t<nbig)$.
We define $a:mathbbNto0,1$ recursively by
$a(1) = 1$, and
$a(n) = 1-a(n-textm(n))$ for $ngeq 2$.
This sequence starts by $10010110ldots$ and I recall that it has some peculiar properties such as, no non-empty finite sub-sequence occurs $3$ times in a row.
Question. Is $sum_n=1^infty 2^-na(n)$ transcendent?
nt.number-theory reference-request real-analysis transcendental-number-theory transcendence
asked 4 hours ago
Dominic van der Zypen
13.6k43173
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up vote
4
down vote
favorite
In a math column in Scientific American many years ago, I encountered a peculiar binary sequence I describe below. Unfortunately I can't find a reference on this, so I would be grateful for any pointers or references.
Let $mathbbN$ be the set of positive integers and let $T = 2^n: nin mathbbNcup 0$ denote the set of powers of $2$. Let $textm:mathbbNto Tcup0$ be defined by $nmapsto maxbig(0cup tin T: t<nbig)$.
We define $a:mathbbNto0,1$ recursively by
$a(1) = 1$, and
$a(n) = 1-a(n-textm(n))$ for $ngeq 2$.
This sequence starts by $10010110ldots$ and I recall that it has some peculiar properties such as, no non-empty finite sub-sequence occurs $3$ times in a row.
Question. Is $sum_n=1^infty 2^-na(n)$ transcendent?
nt.number-theory reference-request real-analysis transcendental-number-theory transcendence
asked 4 hours ago
Dominic van der Zypen
13.6k43173
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In a math column in Scientific American many years ago, I encountered a peculiar binary sequence I describe below. Unfortunately I can't find a reference on this, so I would be grateful for any pointers or references.
Let $mathbbN$ be the set of positive integers and let $T = 2^n: nin mathbbNcup 0$ denote the set of powers of $2$. Let $textm:mathbbNto Tcup0$ be defined by $nmapsto maxbig(0cup tin T: t<nbig)$.
We define $a:mathbbNto0,1$ recursively by
$a(1) = 1$, and
$a(n) = 1-a(n-textm(n))$ for $ngeq 2$.
This sequence starts by $10010110ldots$ and I recall that it has some peculiar properties such as, no non-empty finite sub-sequence occurs $3$ times in a row.
Question. Is $sum_n=1^infty 2^-na(n)$ transcendent?
nt.number-theory reference-request real-analysis transcendental-number-theory transcendence
asked 4 hours ago
Dominic van der Zypen
13.6k43173
In a math column in Scientific American many years ago, I encountered a peculiar binary sequence I describe below. Unfortunately I can't find a reference on this, so I would be grateful for any pointers or references.
Let $mathbbN$ be the set of positive integers and let $T = 2^n: nin mathbbNcup 0$ denote the set of powers of $2$. Let $textm:mathbbNto Tcup0$ be defined by $nmapsto maxbig(0cup tin T: t<nbig)$.
We define $a:mathbbNto0,1$ recursively by
$a(1) = 1$, and
$a(n) = 1-a(n-textm(n))$ for $ngeq 2$.
This sequence starts by $10010110ldots$ and I recall that it has some peculiar properties such as, no non-empty finite sub-sequence occurs $3$ times in a row.
Question. Is $sum_n=1^infty 2^-na(n)$ transcendent?
nt.number-theory reference-request real-analysis transcendental-number-theory transcendence
nt.number-theory reference-request real-analysis transcendental-number-theory transcendence
asked 4 hours ago
Dominic van der Zypen
13.6k43173
asked 4 hours ago
Dominic van der Zypen
13.6k43173
asked 4 hours ago
Dominic van der Zypen
13.6k43173
asked 4 hours ago
Dominic van der Zypen
13.6k43173
asked 4 hours ago
Dominic van der Zypen
13.6k43173
13.6k43173
add a comment |Â
add a comment |Â
1 Answer
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Let $t(i)_0^infty$ be the Thue-Morse sequence. (It starts $0,1,1,0,1,0,0,1,ldots$.).
I claim that your sequence is described by $a(n)=1-t(n-1)$ where $n$ is a positive integer. (It is similar to sequence A010059 in the OEIS, but its index starts at $1$ instead of at $0$.)
The proof is as follows:
For $n=1$, $a(1)=1-t(1-1)=1$. We now consider $nge 2$.
From this OEIS link, let $A_k$ denote the first $2^k$ terms of $t$; then $A_0=0$ and for $kge 0$, $A_k+1=A_k,B_k$, where $B_k$ is obtained from $A_k$ by interchanging $0$'s and $1$'s. That is, $1-t(i)=t(i-2^k)$ where $2^kle ile 2^k+1-1$. Since $mathrmm(i)$ is the largest power of $2$ less than $i$, then $mathrmm(i+1)=2^k$. Thus, $1-t(i)=t(i-mathrmm(i+1))$. Letting $i=n-1$ yields $1-t(n-1)=t(n-1-mathrmm(n))$. Letting $a(n)=1-t(n-1)$ yields $a(n)=t(n-1-mathrmm(n))=1-(1-t(n-mathrmm(n)-1))=1-a(n-mathrmm(n))$. $blacksquare$
The Prouhet-Thue-Morse constant $0.01101001ldots$ (in binary) (which is based on the Thue-Morse sequence) was shown to be transcendental by Kurt Mahler in 1929. It follows that the constant formed from your sequence is also transcendental.
answered 4 hours ago
Joel Reyes Noche
1,04831531
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Let $t(i)_0^infty$ be the Thue-Morse sequence. (It starts $0,1,1,0,1,0,0,1,ldots$.).
I claim that your sequence is described by $a(n)=1-t(n-1)$ where $n$ is a positive integer. (It is similar to sequence A010059 in the OEIS, but its index starts at $1$ instead of at $0$.)
The proof is as follows:
For $n=1$, $a(1)=1-t(1-1)=1$. We now consider $nge 2$.
From this OEIS link, let $A_k$ denote the first $2^k$ terms of $t$; then $A_0=0$ and for $kge 0$, $A_k+1=A_k,B_k$, where $B_k$ is obtained from $A_k$ by interchanging $0$'s and $1$'s. That is, $1-t(i)=t(i-2^k)$ where $2^kle ile 2^k+1-1$. Since $mathrmm(i)$ is the largest power of $2$ less than $i$, then $mathrmm(i+1)=2^k$. Thus, $1-t(i)=t(i-mathrmm(i+1))$. Letting $i=n-1$ yields $1-t(n-1)=t(n-1-mathrmm(n))$. Letting $a(n)=1-t(n-1)$ yields $a(n)=t(n-1-mathrmm(n))=1-(1-t(n-mathrmm(n)-1))=1-a(n-mathrmm(n))$. $blacksquare$
The Prouhet-Thue-Morse constant $0.01101001ldots$ (in binary) (which is based on the Thue-Morse sequence) was shown to be transcendental by Kurt Mahler in 1929. It follows that the constant formed from your sequence is also transcendental.
answered 4 hours ago
Joel Reyes Noche
1,04831531
add a comment |Â
up vote
3
down vote
Let $t(i)_0^infty$ be the Thue-Morse sequence. (It starts $0,1,1,0,1,0,0,1,ldots$.).
I claim that your sequence is described by $a(n)=1-t(n-1)$ where $n$ is a positive integer. (It is similar to sequence A010059 in the OEIS, but its index starts at $1$ instead of at $0$.)
The proof is as follows:
For $n=1$, $a(1)=1-t(1-1)=1$. We now consider $nge 2$.
From this OEIS link, let $A_k$ denote the first $2^k$ terms of $t$; then $A_0=0$ and for $kge 0$, $A_k+1=A_k,B_k$, where $B_k$ is obtained from $A_k$ by interchanging $0$'s and $1$'s. That is, $1-t(i)=t(i-2^k)$ where $2^kle ile 2^k+1-1$. Since $mathrmm(i)$ is the largest power of $2$ less than $i$, then $mathrmm(i+1)=2^k$. Thus, $1-t(i)=t(i-mathrmm(i+1))$. Letting $i=n-1$ yields $1-t(n-1)=t(n-1-mathrmm(n))$. Letting $a(n)=1-t(n-1)$ yields $a(n)=t(n-1-mathrmm(n))=1-(1-t(n-mathrmm(n)-1))=1-a(n-mathrmm(n))$. $blacksquare$
The Prouhet-Thue-Morse constant $0.01101001ldots$ (in binary) (which is based on the Thue-Morse sequence) was shown to be transcendental by Kurt Mahler in 1929. It follows that the constant formed from your sequence is also transcendental.
answered 4 hours ago
Joel Reyes Noche
1,04831531
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $t(i)_0^infty$ be the Thue-Morse sequence. (It starts $0,1,1,0,1,0,0,1,ldots$.).
I claim that your sequence is described by $a(n)=1-t(n-1)$ where $n$ is a positive integer. (It is similar to sequence A010059 in the OEIS, but its index starts at $1$ instead of at $0$.)
The proof is as follows:
For $n=1$, $a(1)=1-t(1-1)=1$. We now consider $nge 2$.
From this OEIS link, let $A_k$ denote the first $2^k$ terms of $t$; then $A_0=0$ and for $kge 0$, $A_k+1=A_k,B_k$, where $B_k$ is obtained from $A_k$ by interchanging $0$'s and $1$'s. That is, $1-t(i)=t(i-2^k)$ where $2^kle ile 2^k+1-1$. Since $mathrmm(i)$ is the largest power of $2$ less than $i$, then $mathrmm(i+1)=2^k$. Thus, $1-t(i)=t(i-mathrmm(i+1))$. Letting $i=n-1$ yields $1-t(n-1)=t(n-1-mathrmm(n))$. Letting $a(n)=1-t(n-1)$ yields $a(n)=t(n-1-mathrmm(n))=1-(1-t(n-mathrmm(n)-1))=1-a(n-mathrmm(n))$. $blacksquare$
The Prouhet-Thue-Morse constant $0.01101001ldots$ (in binary) (which is based on the Thue-Morse sequence) was shown to be transcendental by Kurt Mahler in 1929. It follows that the constant formed from your sequence is also transcendental.
answered 4 hours ago
Joel Reyes Noche
1,04831531
Let $t(i)_0^infty$ be the Thue-Morse sequence. (It starts $0,1,1,0,1,0,0,1,ldots$.).
I claim that your sequence is described by $a(n)=1-t(n-1)$ where $n$ is a positive integer. (It is similar to sequence A010059 in the OEIS, but its index starts at $1$ instead of at $0$.)
The proof is as follows:
For $n=1$, $a(1)=1-t(1-1)=1$. We now consider $nge 2$.
From this OEIS link, let $A_k$ denote the first $2^k$ terms of $t$; then $A_0=0$ and for $kge 0$, $A_k+1=A_k,B_k$, where $B_k$ is obtained from $A_k$ by interchanging $0$'s and $1$'s. That is, $1-t(i)=t(i-2^k)$ where $2^kle ile 2^k+1-1$. Since $mathrmm(i)$ is the largest power of $2$ less than $i$, then $mathrmm(i+1)=2^k$. Thus, $1-t(i)=t(i-mathrmm(i+1))$. Letting $i=n-1$ yields $1-t(n-1)=t(n-1-mathrmm(n))$. Letting $a(n)=1-t(n-1)$ yields $a(n)=t(n-1-mathrmm(n))=1-(1-t(n-mathrmm(n)-1))=1-a(n-mathrmm(n))$. $blacksquare$
The Prouhet-Thue-Morse constant $0.01101001ldots$ (in binary) (which is based on the Thue-Morse sequence) was shown to be transcendental by Kurt Mahler in 1929. It follows that the constant formed from your sequence is also transcendental.
answered 4 hours ago
Joel Reyes Noche
1,04831531
edited 1 hour ago
answered 4 hours ago
Joel Reyes Noche
1,04831531
answered 4 hours ago
Joel Reyes Noche
1,04831531
answered 4 hours ago
Joel Reyes Noche
1,04831531
1,04831531
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