Tricky question related to sequences and series
$begingroup$
I have the following question with me:
"In the sequence $$1,9,7,7,4,7,5,3,9,4,1,....$$ every term from the fifth one is the sum of the previous 4 modulo 10. Do the numbers $1977$(apart from beginning) and $0197$ ever occur in the sequence? If yes, does it occur a finite number of times or not?"
How do I solve this question? The only advancement I have made is that when the sequence is written modulo 2, I get $$1,1,1,1,0,1,1,1,1,0....$$
How do I proceed from here?
sequences-and-series number-theory
asked Dec 15 '18 at 8:08
saisanjeevsaisanjeev
987212
$endgroup$
add a comment |
$begingroup$
I have the following question with me:
"In the sequence $$1,9,7,7,4,7,5,3,9,4,1,....$$ every term from the fifth one is the sum of the previous 4 modulo 10. Do the numbers $1977$(apart from beginning) and $0197$ ever occur in the sequence? If yes, does it occur a finite number of times or not?"
How do I solve this question? The only advancement I have made is that when the sequence is written modulo 2, I get $$1,1,1,1,0,1,1,1,1,0....$$
How do I proceed from here?
sequences-and-series number-theory
asked Dec 15 '18 at 8:08
saisanjeevsaisanjeev
987212
$endgroup$
$begingroup$
Did you look for a period of this sequence ?
$endgroup$
– Peter
Dec 15 '18 at 8:15
$begingroup$
You can work backwards as well as forwards to see if that tells you anything. $2$ is an interesting modulus to choose because it is a factor of $10$.
$endgroup$
– Mark Bennet
Dec 15 '18 at 10:38
add a comment |
$begingroup$
I have the following question with me:
"In the sequence $$1,9,7,7,4,7,5,3,9,4,1,....$$ every term from the fifth one is the sum of the previous 4 modulo 10. Do the numbers $1977$(apart from beginning) and $0197$ ever occur in the sequence? If yes, does it occur a finite number of times or not?"
How do I solve this question? The only advancement I have made is that when the sequence is written modulo 2, I get $$1,1,1,1,0,1,1,1,1,0....$$
How do I proceed from here?
sequences-and-series number-theory
asked Dec 15 '18 at 8:08
saisanjeevsaisanjeev
987212
$endgroup$
I have the following question with me:
"In the sequence $$1,9,7,7,4,7,5,3,9,4,1,....$$ every term from the fifth one is the sum of the previous 4 modulo 10. Do the numbers $1977$(apart from beginning) and $0197$ ever occur in the sequence? If yes, does it occur a finite number of times or not?"
How do I solve this question? The only advancement I have made is that when the sequence is written modulo 2, I get $$1,1,1,1,0,1,1,1,1,0....$$
How do I proceed from here?
sequences-and-series number-theory
sequences-and-series number-theory
asked Dec 15 '18 at 8:08
saisanjeevsaisanjeev
987212
asked Dec 15 '18 at 8:08
saisanjeevsaisanjeev
987212
asked Dec 15 '18 at 8:08
saisanjeevsaisanjeev
987212
asked Dec 15 '18 at 8:08
saisanjeevsaisanjeev
987212
asked Dec 15 '18 at 8:08
saisanjeevsaisanjeev
987212
987212
$begingroup$
Did you look for a period of this sequence ?
$endgroup$
– Peter
Dec 15 '18 at 8:15
$begingroup$
You can work backwards as well as forwards to see if that tells you anything. $2$ is an interesting modulus to choose because it is a factor of $10$.
$endgroup$
– Mark Bennet
Dec 15 '18 at 10:38
add a comment |
$begingroup$
Did you look for a period of this sequence ?
$endgroup$
– Peter
Dec 15 '18 at 8:15
$begingroup$
You can work backwards as well as forwards to see if that tells you anything. $2$ is an interesting modulus to choose because it is a factor of $10$.
$endgroup$
– Mark Bennet
Dec 15 '18 at 10:38
$begingroup$
Did you look for a period of this sequence ?
$endgroup$
– Peter
Dec 15 '18 at 8:15
$begingroup$
Did you look for a period of this sequence ?
$endgroup$
– Peter
Dec 15 '18 at 8:15
$begingroup$
You can work backwards as well as forwards to see if that tells you anything. $2$ is an interesting modulus to choose because it is a factor of $10$.
$endgroup$
– Mark Bennet
Dec 15 '18 at 10:38
$begingroup$
You can work backwards as well as forwards to see if that tells you anything. $2$ is an interesting modulus to choose because it is a factor of $10$.
$endgroup$
– Mark Bennet
Dec 15 '18 at 10:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: you can prove results about sequences like this without explicitly computing anything using two facts:
The number of possible sequences of four digits is finite, while the sequence you are given is infinite.
Any four consecutive digits in the sequence (uniquely) determine the whole sequence both forwards and backwards
Obviously you have to show that these facts are true for the sequence in hand, and then work out how to use them to get the result you need.
The things about explicitly computing residues can help in determining the period of a recurring sequence.
answered Dec 15 '18 at 11:10
Mark BennetMark Bennet
81.1k982179
$endgroup$
$begingroup$
Yes that works and helps me to prove that a certain number will occur infinitely many times. However, I can tell that if I know for a fact that that 4 tuple occurs at least once. I can do that for the first one coz I know 1977 occurs once in the sequence. But how do I solve the second part?
$endgroup$
– saisanjeev
Dec 15 '18 at 11:21
$begingroup$
@saisanjeev Think about when the first repetition happens, and try working backwards.
$endgroup$
– Mark Bennet
Dec 15 '18 at 11:38
$begingroup$
I still didnt understand how to proceed? Can you please elaborate
$endgroup$
– saisanjeev
Dec 15 '18 at 14:32
$begingroup$
@saisanjeev Suppose you get a first repeat at positions $m, m+1, m+2, m+3$ and $n, n+1, n+2, n+3$ - what can you say about positions $m-1$ and $n-1$?
$endgroup$
– Mark Bennet
Dec 15 '18 at 16:05
$begingroup$
m-1 and n-1 would be the same. But to prove that 0197 will occur infinite number of times, I need to prove that it will occur at least once. How do I prove that?
$endgroup$
– saisanjeev
Dec 18 '18 at 15:12
|
show 1 more comment
$begingroup$
First, you can consider the sequence with index in whole integers. For example, we can define $a_{0} = 0$ since $0 + 1 + 9 + 7equiv 7$ mod 10 and $a_{-1} = 7$ since $7 + 0 + 1 + 9 equiv 7$ mod 10. Now try to explain that there exists $mneq n$ such that $(a_{m}, a_{m+1}, a_{m+2}, a_{m+3}) = (a_{n}, a_{n+1}, a_{n+2}, a_{n+3})$ by using the pigeonhole principle.
answered Dec 15 '18 at 8:20
Seewoo LeeSeewoo Lee
6,870927
$endgroup$
$begingroup$
I went ahead like this, please tell whether the method is right. The number of 4-tuples of numbers are only 10^4, however, since the sequence is infinite, one of the sequences will eventually repeat, since the next term of the sequence is determined by the previous four, once one of them repeats, the sequence will repeat itself, is the proof correct?
$endgroup$
– saisanjeev
Dec 15 '18 at 9:36
$begingroup$
@saisanjeev Exactly what I mean.
$endgroup$
– Seewoo Lee
Dec 16 '18 at 2:52
$begingroup$
The pigeonhole principle says there are repeats, but you also have to prove that particular sequences repeat, and this argument does not even show that if a sequence appears once it necessarily repeats.
$endgroup$
– Mark Bennet
Dec 18 '18 at 17:16
$begingroup$
@Mark Bennet I didn’t try to give a full solution. Since there is quadruple that repeats, by the recurrence relation it should be periodic so that every quadruple repeats. Now it is fairly obvious that 1,9,7,7 and 0,1,9,7 are in the sequence (actually I already mentioned in the hint). I don’t think it is always good to give a full solution.
$endgroup$
– Seewoo Lee
Dec 19 '18 at 0:30
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: you can prove results about sequences like this without explicitly computing anything using two facts:
The number of possible sequences of four digits is finite, while the sequence you are given is infinite.
Any four consecutive digits in the sequence (uniquely) determine the whole sequence both forwards and backwards
Obviously you have to show that these facts are true for the sequence in hand, and then work out how to use them to get the result you need.
The things about explicitly computing residues can help in determining the period of a recurring sequence.
answered Dec 15 '18 at 11:10
Mark BennetMark Bennet
81.1k982179
$endgroup$
$begingroup$
Yes that works and helps me to prove that a certain number will occur infinitely many times. However, I can tell that if I know for a fact that that 4 tuple occurs at least once. I can do that for the first one coz I know 1977 occurs once in the sequence. But how do I solve the second part?
$endgroup$
– saisanjeev
Dec 15 '18 at 11:21
$begingroup$
@saisanjeev Think about when the first repetition happens, and try working backwards.
$endgroup$
– Mark Bennet
Dec 15 '18 at 11:38
$begingroup$
I still didnt understand how to proceed? Can you please elaborate
$endgroup$
– saisanjeev
Dec 15 '18 at 14:32
$begingroup$
@saisanjeev Suppose you get a first repeat at positions $m, m+1, m+2, m+3$ and $n, n+1, n+2, n+3$ - what can you say about positions $m-1$ and $n-1$?
$endgroup$
– Mark Bennet
Dec 15 '18 at 16:05
$begingroup$
m-1 and n-1 would be the same. But to prove that 0197 will occur infinite number of times, I need to prove that it will occur at least once. How do I prove that?
$endgroup$
– saisanjeev
Dec 18 '18 at 15:12
|
show 1 more comment
$begingroup$
Hint: you can prove results about sequences like this without explicitly computing anything using two facts:
The number of possible sequences of four digits is finite, while the sequence you are given is infinite.
Any four consecutive digits in the sequence (uniquely) determine the whole sequence both forwards and backwards
Obviously you have to show that these facts are true for the sequence in hand, and then work out how to use them to get the result you need.
The things about explicitly computing residues can help in determining the period of a recurring sequence.
answered Dec 15 '18 at 11:10
Mark BennetMark Bennet
81.1k982179
$endgroup$
$begingroup$
Yes that works and helps me to prove that a certain number will occur infinitely many times. However, I can tell that if I know for a fact that that 4 tuple occurs at least once. I can do that for the first one coz I know 1977 occurs once in the sequence. But how do I solve the second part?
$endgroup$
– saisanjeev
Dec 15 '18 at 11:21
$begingroup$
@saisanjeev Think about when the first repetition happens, and try working backwards.
$endgroup$
– Mark Bennet
Dec 15 '18 at 11:38
$begingroup$
I still didnt understand how to proceed? Can you please elaborate
$endgroup$
– saisanjeev
Dec 15 '18 at 14:32
$begingroup$
@saisanjeev Suppose you get a first repeat at positions $m, m+1, m+2, m+3$ and $n, n+1, n+2, n+3$ - what can you say about positions $m-1$ and $n-1$?
$endgroup$
– Mark Bennet
Dec 15 '18 at 16:05
$begingroup$
m-1 and n-1 would be the same. But to prove that 0197 will occur infinite number of times, I need to prove that it will occur at least once. How do I prove that?
$endgroup$
– saisanjeev
Dec 18 '18 at 15:12
|
show 1 more comment
$begingroup$
Hint: you can prove results about sequences like this without explicitly computing anything using two facts:
The number of possible sequences of four digits is finite, while the sequence you are given is infinite.
Any four consecutive digits in the sequence (uniquely) determine the whole sequence both forwards and backwards
Obviously you have to show that these facts are true for the sequence in hand, and then work out how to use them to get the result you need.
The things about explicitly computing residues can help in determining the period of a recurring sequence.
answered Dec 15 '18 at 11:10
Mark BennetMark Bennet
81.1k982179
$endgroup$
Hint: you can prove results about sequences like this without explicitly computing anything using two facts:
The number of possible sequences of four digits is finite, while the sequence you are given is infinite.
Any four consecutive digits in the sequence (uniquely) determine the whole sequence both forwards and backwards
Obviously you have to show that these facts are true for the sequence in hand, and then work out how to use them to get the result you need.
The things about explicitly computing residues can help in determining the period of a recurring sequence.
answered Dec 15 '18 at 11:10
Mark BennetMark Bennet
81.1k982179
answered Dec 15 '18 at 11:10
Mark BennetMark Bennet
81.1k982179
answered Dec 15 '18 at 11:10
Mark BennetMark Bennet
81.1k982179
answered Dec 15 '18 at 11:10
Mark BennetMark Bennet
81.1k982179
81.1k982179
$begingroup$
Yes that works and helps me to prove that a certain number will occur infinitely many times. However, I can tell that if I know for a fact that that 4 tuple occurs at least once. I can do that for the first one coz I know 1977 occurs once in the sequence. But how do I solve the second part?
$endgroup$
– saisanjeev
Dec 15 '18 at 11:21
$begingroup$
@saisanjeev Think about when the first repetition happens, and try working backwards.
$endgroup$
– Mark Bennet
Dec 15 '18 at 11:38
$begingroup$
I still didnt understand how to proceed? Can you please elaborate
$endgroup$
– saisanjeev
Dec 15 '18 at 14:32
$begingroup$
@saisanjeev Suppose you get a first repeat at positions $m, m+1, m+2, m+3$ and $n, n+1, n+2, n+3$ - what can you say about positions $m-1$ and $n-1$?
$endgroup$
– Mark Bennet
Dec 15 '18 at 16:05
$begingroup$
m-1 and n-1 would be the same. But to prove that 0197 will occur infinite number of times, I need to prove that it will occur at least once. How do I prove that?
$endgroup$
– saisanjeev
Dec 18 '18 at 15:12
|
show 1 more comment
$begingroup$
Yes that works and helps me to prove that a certain number will occur infinitely many times. However, I can tell that if I know for a fact that that 4 tuple occurs at least once. I can do that for the first one coz I know 1977 occurs once in the sequence. But how do I solve the second part?
$endgroup$
– saisanjeev
Dec 15 '18 at 11:21
$begingroup$
@saisanjeev Think about when the first repetition happens, and try working backwards.
$endgroup$
– Mark Bennet
Dec 15 '18 at 11:38
$begingroup$
I still didnt understand how to proceed? Can you please elaborate
$endgroup$
– saisanjeev
Dec 15 '18 at 14:32
$begingroup$
@saisanjeev Suppose you get a first repeat at positions $m, m+1, m+2, m+3$ and $n, n+1, n+2, n+3$ - what can you say about positions $m-1$ and $n-1$?
$endgroup$
– Mark Bennet
Dec 15 '18 at 16:05
$begingroup$
m-1 and n-1 would be the same. But to prove that 0197 will occur infinite number of times, I need to prove that it will occur at least once. How do I prove that?
$endgroup$
– saisanjeev
Dec 18 '18 at 15:12
$begingroup$
Yes that works and helps me to prove that a certain number will occur infinitely many times. However, I can tell that if I know for a fact that that 4 tuple occurs at least once. I can do that for the first one coz I know 1977 occurs once in the sequence. But how do I solve the second part?
$endgroup$
– saisanjeev
Dec 15 '18 at 11:21
$begingroup$
Yes that works and helps me to prove that a certain number will occur infinitely many times. However, I can tell that if I know for a fact that that 4 tuple occurs at least once. I can do that for the first one coz I know 1977 occurs once in the sequence. But how do I solve the second part?
$endgroup$
– saisanjeev
Dec 15 '18 at 11:21
$begingroup$
@saisanjeev Think about when the first repetition happens, and try working backwards.
$endgroup$
– Mark Bennet
Dec 15 '18 at 11:38
$begingroup$
@saisanjeev Think about when the first repetition happens, and try working backwards.
$endgroup$
– Mark Bennet
Dec 15 '18 at 11:38
$begingroup$
I still didnt understand how to proceed? Can you please elaborate
$endgroup$
– saisanjeev
Dec 15 '18 at 14:32
$begingroup$
I still didnt understand how to proceed? Can you please elaborate
$endgroup$
– saisanjeev
Dec 15 '18 at 14:32
$begingroup$
@saisanjeev Suppose you get a first repeat at positions $m, m+1, m+2, m+3$ and $n, n+1, n+2, n+3$ - what can you say about positions $m-1$ and $n-1$?
$endgroup$
– Mark Bennet
Dec 15 '18 at 16:05
$begingroup$
@saisanjeev Suppose you get a first repeat at positions $m, m+1, m+2, m+3$ and $n, n+1, n+2, n+3$ - what can you say about positions $m-1$ and $n-1$?
$endgroup$
– Mark Bennet
Dec 15 '18 at 16:05
$begingroup$
m-1 and n-1 would be the same. But to prove that 0197 will occur infinite number of times, I need to prove that it will occur at least once. How do I prove that?
$endgroup$
– saisanjeev
Dec 18 '18 at 15:12
$begingroup$
m-1 and n-1 would be the same. But to prove that 0197 will occur infinite number of times, I need to prove that it will occur at least once. How do I prove that?
$endgroup$
– saisanjeev
Dec 18 '18 at 15:12
|
show 1 more comment
$begingroup$
First, you can consider the sequence with index in whole integers. For example, we can define $a_{0} = 0$ since $0 + 1 + 9 + 7equiv 7$ mod 10 and $a_{-1} = 7$ since $7 + 0 + 1 + 9 equiv 7$ mod 10. Now try to explain that there exists $mneq n$ such that $(a_{m}, a_{m+1}, a_{m+2}, a_{m+3}) = (a_{n}, a_{n+1}, a_{n+2}, a_{n+3})$ by using the pigeonhole principle.
answered Dec 15 '18 at 8:20
Seewoo LeeSeewoo Lee
6,870927
$endgroup$
$begingroup$
I went ahead like this, please tell whether the method is right. The number of 4-tuples of numbers are only 10^4, however, since the sequence is infinite, one of the sequences will eventually repeat, since the next term of the sequence is determined by the previous four, once one of them repeats, the sequence will repeat itself, is the proof correct?
$endgroup$
– saisanjeev
Dec 15 '18 at 9:36
$begingroup$
@saisanjeev Exactly what I mean.
$endgroup$
– Seewoo Lee
Dec 16 '18 at 2:52
$begingroup$
The pigeonhole principle says there are repeats, but you also have to prove that particular sequences repeat, and this argument does not even show that if a sequence appears once it necessarily repeats.
$endgroup$
– Mark Bennet
Dec 18 '18 at 17:16
$begingroup$
@Mark Bennet I didn’t try to give a full solution. Since there is quadruple that repeats, by the recurrence relation it should be periodic so that every quadruple repeats. Now it is fairly obvious that 1,9,7,7 and 0,1,9,7 are in the sequence (actually I already mentioned in the hint). I don’t think it is always good to give a full solution.
$endgroup$
– Seewoo Lee
Dec 19 '18 at 0:30
add a comment |
$begingroup$
First, you can consider the sequence with index in whole integers. For example, we can define $a_{0} = 0$ since $0 + 1 + 9 + 7equiv 7$ mod 10 and $a_{-1} = 7$ since $7 + 0 + 1 + 9 equiv 7$ mod 10. Now try to explain that there exists $mneq n$ such that $(a_{m}, a_{m+1}, a_{m+2}, a_{m+3}) = (a_{n}, a_{n+1}, a_{n+2}, a_{n+3})$ by using the pigeonhole principle.
answered Dec 15 '18 at 8:20
Seewoo LeeSeewoo Lee
6,870927
$endgroup$
$begingroup$
I went ahead like this, please tell whether the method is right. The number of 4-tuples of numbers are only 10^4, however, since the sequence is infinite, one of the sequences will eventually repeat, since the next term of the sequence is determined by the previous four, once one of them repeats, the sequence will repeat itself, is the proof correct?
$endgroup$
– saisanjeev
Dec 15 '18 at 9:36
$begingroup$
@saisanjeev Exactly what I mean.
$endgroup$
– Seewoo Lee
Dec 16 '18 at 2:52
$begingroup$
The pigeonhole principle says there are repeats, but you also have to prove that particular sequences repeat, and this argument does not even show that if a sequence appears once it necessarily repeats.
$endgroup$
– Mark Bennet
Dec 18 '18 at 17:16
$begingroup$
@Mark Bennet I didn’t try to give a full solution. Since there is quadruple that repeats, by the recurrence relation it should be periodic so that every quadruple repeats. Now it is fairly obvious that 1,9,7,7 and 0,1,9,7 are in the sequence (actually I already mentioned in the hint). I don’t think it is always good to give a full solution.
$endgroup$
– Seewoo Lee
Dec 19 '18 at 0:30
add a comment |
$begingroup$
First, you can consider the sequence with index in whole integers. For example, we can define $a_{0} = 0$ since $0 + 1 + 9 + 7equiv 7$ mod 10 and $a_{-1} = 7$ since $7 + 0 + 1 + 9 equiv 7$ mod 10. Now try to explain that there exists $mneq n$ such that $(a_{m}, a_{m+1}, a_{m+2}, a_{m+3}) = (a_{n}, a_{n+1}, a_{n+2}, a_{n+3})$ by using the pigeonhole principle.
answered Dec 15 '18 at 8:20
Seewoo LeeSeewoo Lee
6,870927
$endgroup$
First, you can consider the sequence with index in whole integers. For example, we can define $a_{0} = 0$ since $0 + 1 + 9 + 7equiv 7$ mod 10 and $a_{-1} = 7$ since $7 + 0 + 1 + 9 equiv 7$ mod 10. Now try to explain that there exists $mneq n$ such that $(a_{m}, a_{m+1}, a_{m+2}, a_{m+3}) = (a_{n}, a_{n+1}, a_{n+2}, a_{n+3})$ by using the pigeonhole principle.
answered Dec 15 '18 at 8:20
Seewoo LeeSeewoo Lee
6,870927
answered Dec 15 '18 at 8:20
Seewoo LeeSeewoo Lee
6,870927
answered Dec 15 '18 at 8:20
Seewoo LeeSeewoo Lee
6,870927
answered Dec 15 '18 at 8:20
Seewoo LeeSeewoo Lee
6,870927
6,870927
$begingroup$
I went ahead like this, please tell whether the method is right. The number of 4-tuples of numbers are only 10^4, however, since the sequence is infinite, one of the sequences will eventually repeat, since the next term of the sequence is determined by the previous four, once one of them repeats, the sequence will repeat itself, is the proof correct?
$endgroup$
– saisanjeev
Dec 15 '18 at 9:36
$begingroup$
@saisanjeev Exactly what I mean.
$endgroup$
– Seewoo Lee
Dec 16 '18 at 2:52
$begingroup$
The pigeonhole principle says there are repeats, but you also have to prove that particular sequences repeat, and this argument does not even show that if a sequence appears once it necessarily repeats.
$endgroup$
– Mark Bennet
Dec 18 '18 at 17:16
$begingroup$
@Mark Bennet I didn’t try to give a full solution. Since there is quadruple that repeats, by the recurrence relation it should be periodic so that every quadruple repeats. Now it is fairly obvious that 1,9,7,7 and 0,1,9,7 are in the sequence (actually I already mentioned in the hint). I don’t think it is always good to give a full solution.
$endgroup$
– Seewoo Lee
Dec 19 '18 at 0:30
add a comment |
$begingroup$
I went ahead like this, please tell whether the method is right. The number of 4-tuples of numbers are only 10^4, however, since the sequence is infinite, one of the sequences will eventually repeat, since the next term of the sequence is determined by the previous four, once one of them repeats, the sequence will repeat itself, is the proof correct?
$endgroup$
– saisanjeev
Dec 15 '18 at 9:36
$begingroup$
@saisanjeev Exactly what I mean.
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– Seewoo Lee
Dec 16 '18 at 2:52
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The pigeonhole principle says there are repeats, but you also have to prove that particular sequences repeat, and this argument does not even show that if a sequence appears once it necessarily repeats.
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– Mark Bennet
Dec 18 '18 at 17:16
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@Mark Bennet I didn’t try to give a full solution. Since there is quadruple that repeats, by the recurrence relation it should be periodic so that every quadruple repeats. Now it is fairly obvious that 1,9,7,7 and 0,1,9,7 are in the sequence (actually I already mentioned in the hint). I don’t think it is always good to give a full solution.
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– Seewoo Lee
Dec 19 '18 at 0:30
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I went ahead like this, please tell whether the method is right. The number of 4-tuples of numbers are only 10^4, however, since the sequence is infinite, one of the sequences will eventually repeat, since the next term of the sequence is determined by the previous four, once one of them repeats, the sequence will repeat itself, is the proof correct?
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– saisanjeev
Dec 15 '18 at 9:36
$begingroup$
I went ahead like this, please tell whether the method is right. The number of 4-tuples of numbers are only 10^4, however, since the sequence is infinite, one of the sequences will eventually repeat, since the next term of the sequence is determined by the previous four, once one of them repeats, the sequence will repeat itself, is the proof correct?
$endgroup$
– saisanjeev
Dec 15 '18 at 9:36
$begingroup$
@saisanjeev Exactly what I mean.
$endgroup$
– Seewoo Lee
Dec 16 '18 at 2:52
$begingroup$
@saisanjeev Exactly what I mean.
$endgroup$
– Seewoo Lee
Dec 16 '18 at 2:52
$begingroup$
The pigeonhole principle says there are repeats, but you also have to prove that particular sequences repeat, and this argument does not even show that if a sequence appears once it necessarily repeats.
$endgroup$
– Mark Bennet
Dec 18 '18 at 17:16
$begingroup$
The pigeonhole principle says there are repeats, but you also have to prove that particular sequences repeat, and this argument does not even show that if a sequence appears once it necessarily repeats.
$endgroup$
– Mark Bennet
Dec 18 '18 at 17:16
$begingroup$
@Mark Bennet I didn’t try to give a full solution. Since there is quadruple that repeats, by the recurrence relation it should be periodic so that every quadruple repeats. Now it is fairly obvious that 1,9,7,7 and 0,1,9,7 are in the sequence (actually I already mentioned in the hint). I don’t think it is always good to give a full solution.
$endgroup$
– Seewoo Lee
Dec 19 '18 at 0:30
$begingroup$
@Mark Bennet I didn’t try to give a full solution. Since there is quadruple that repeats, by the recurrence relation it should be periodic so that every quadruple repeats. Now it is fairly obvious that 1,9,7,7 and 0,1,9,7 are in the sequence (actually I already mentioned in the hint). I don’t think it is always good to give a full solution.
$endgroup$
– Seewoo Lee
Dec 19 '18 at 0:30
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Did you look for a period of this sequence ?
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– Peter
Dec 15 '18 at 8:15
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You can work backwards as well as forwards to see if that tells you anything. $2$ is an interesting modulus to choose because it is a factor of $10$.
$endgroup$
– Mark Bennet
Dec 15 '18 at 10:38