How can I determine if a number is a part of arithmetic sequence
$begingroup$
Sorry if this is a simple question, I am not so good in mathematics and trying to start understanding it better. So thanks is advance.
I have this sequence - 0,1,5,6,10,11,15,16,20,21....
Which is n numbers and then a gap of n+1.
What I need is a way to get a random number and check if it is part of the sequence.
I see that I can check the last number and see if it is (0,1,5,6) but this will be true only for this sequence. So I need more generic way.
Thanks
Shani
edit
The sequence starts from 0
n = 2
Then we start with n numbers (0,1) skip n+1 numbers (2,3,4) then include n numbers (5,6) , skip n+1 (7,8,9)...
sequences-and-series
asked Mar 19 '16 at 9:40
shannogashannoga
10113
$endgroup$
add a comment |
$begingroup$
Sorry if this is a simple question, I am not so good in mathematics and trying to start understanding it better. So thanks is advance.
I have this sequence - 0,1,5,6,10,11,15,16,20,21....
Which is n numbers and then a gap of n+1.
What I need is a way to get a random number and check if it is part of the sequence.
I see that I can check the last number and see if it is (0,1,5,6) but this will be true only for this sequence. So I need more generic way.
Thanks
Shani
edit
The sequence starts from 0
n = 2
Then we start with n numbers (0,1) skip n+1 numbers (2,3,4) then include n numbers (5,6) , skip n+1 (7,8,9)...
sequences-and-series
asked Mar 19 '16 at 9:40
shannogashannoga
10113
$endgroup$
$begingroup$
I know the sequence it is in this case n and then a gap of n+1. Thanks
$endgroup$
– shannoga
Mar 19 '16 at 9:58
$begingroup$
i don't understand the definition of the sequence. You say "$n$ numbers and then a gap of $n+1$" but your sequence looks like $2$ numbers and then a gap of $4$. What is $n$? Can you write out the first few terms if $n$ were, say, $3$?
$endgroup$
– lulu
Mar 19 '16 at 10:29
$begingroup$
Ok. Edited my question. Thanks
$endgroup$
– shannoga
Mar 19 '16 at 10:35
$begingroup$
Got it. So, in this case, the "consecutive strings" start with elements of the progression ${0,5,10,15,dots}$. Given a random $m$ find the nearest multiple of $5$ less than or equal to $m$....then we need $m$ to either be this multiple or $1$ greater. Similar for general $n$.
$endgroup$
– lulu
Mar 19 '16 at 10:37
$begingroup$
The terms of an arithmetic sequence have a common difference, so this is not an arithmetic sequence.
$endgroup$
– N. F. Taussig
Sep 7 '17 at 8:53
add a comment |
$begingroup$
Sorry if this is a simple question, I am not so good in mathematics and trying to start understanding it better. So thanks is advance.
I have this sequence - 0,1,5,6,10,11,15,16,20,21....
Which is n numbers and then a gap of n+1.
What I need is a way to get a random number and check if it is part of the sequence.
I see that I can check the last number and see if it is (0,1,5,6) but this will be true only for this sequence. So I need more generic way.
Thanks
Shani
edit
The sequence starts from 0
n = 2
Then we start with n numbers (0,1) skip n+1 numbers (2,3,4) then include n numbers (5,6) , skip n+1 (7,8,9)...
sequences-and-series
asked Mar 19 '16 at 9:40
shannogashannoga
10113
$endgroup$
Sorry if this is a simple question, I am not so good in mathematics and trying to start understanding it better. So thanks is advance.
I have this sequence - 0,1,5,6,10,11,15,16,20,21....
Which is n numbers and then a gap of n+1.
What I need is a way to get a random number and check if it is part of the sequence.
I see that I can check the last number and see if it is (0,1,5,6) but this will be true only for this sequence. So I need more generic way.
Thanks
Shani
edit
The sequence starts from 0
n = 2
Then we start with n numbers (0,1) skip n+1 numbers (2,3,4) then include n numbers (5,6) , skip n+1 (7,8,9)...
sequences-and-series
sequences-and-series
asked Mar 19 '16 at 9:40
shannogashannoga
10113
asked Mar 19 '16 at 9:40
shannogashannoga
10113
edited Mar 19 '16 at 10:34
shannoga
asked Mar 19 '16 at 9:40
shannogashannoga
10113
asked Mar 19 '16 at 9:40
shannogashannoga
10113
asked Mar 19 '16 at 9:40
shannogashannoga
10113
10113
$begingroup$
I know the sequence it is in this case n and then a gap of n+1. Thanks
$endgroup$
– shannoga
Mar 19 '16 at 9:58
$begingroup$
i don't understand the definition of the sequence. You say "$n$ numbers and then a gap of $n+1$" but your sequence looks like $2$ numbers and then a gap of $4$. What is $n$? Can you write out the first few terms if $n$ were, say, $3$?
$endgroup$
– lulu
Mar 19 '16 at 10:29
$begingroup$
Ok. Edited my question. Thanks
$endgroup$
– shannoga
Mar 19 '16 at 10:35
$begingroup$
Got it. So, in this case, the "consecutive strings" start with elements of the progression ${0,5,10,15,dots}$. Given a random $m$ find the nearest multiple of $5$ less than or equal to $m$....then we need $m$ to either be this multiple or $1$ greater. Similar for general $n$.
$endgroup$
– lulu
Mar 19 '16 at 10:37
$begingroup$
The terms of an arithmetic sequence have a common difference, so this is not an arithmetic sequence.
$endgroup$
– N. F. Taussig
Sep 7 '17 at 8:53
add a comment |
$begingroup$
I know the sequence it is in this case n and then a gap of n+1. Thanks
$endgroup$
– shannoga
Mar 19 '16 at 9:58
$begingroup$
i don't understand the definition of the sequence. You say "$n$ numbers and then a gap of $n+1$" but your sequence looks like $2$ numbers and then a gap of $4$. What is $n$? Can you write out the first few terms if $n$ were, say, $3$?
$endgroup$
– lulu
Mar 19 '16 at 10:29
$begingroup$
Ok. Edited my question. Thanks
$endgroup$
– shannoga
Mar 19 '16 at 10:35
$begingroup$
Got it. So, in this case, the "consecutive strings" start with elements of the progression ${0,5,10,15,dots}$. Given a random $m$ find the nearest multiple of $5$ less than or equal to $m$....then we need $m$ to either be this multiple or $1$ greater. Similar for general $n$.
$endgroup$
– lulu
Mar 19 '16 at 10:37
$begingroup$
The terms of an arithmetic sequence have a common difference, so this is not an arithmetic sequence.
$endgroup$
– N. F. Taussig
Sep 7 '17 at 8:53
$begingroup$
I know the sequence it is in this case n and then a gap of n+1. Thanks
$endgroup$
– shannoga
Mar 19 '16 at 9:58
$begingroup$
I know the sequence it is in this case n and then a gap of n+1. Thanks
$endgroup$
– shannoga
Mar 19 '16 at 9:58
$begingroup$
i don't understand the definition of the sequence. You say "$n$ numbers and then a gap of $n+1$" but your sequence looks like $2$ numbers and then a gap of $4$. What is $n$? Can you write out the first few terms if $n$ were, say, $3$?
$endgroup$
– lulu
Mar 19 '16 at 10:29
$begingroup$
i don't understand the definition of the sequence. You say "$n$ numbers and then a gap of $n+1$" but your sequence looks like $2$ numbers and then a gap of $4$. What is $n$? Can you write out the first few terms if $n$ were, say, $3$?
$endgroup$
– lulu
Mar 19 '16 at 10:29
$begingroup$
Ok. Edited my question. Thanks
$endgroup$
– shannoga
Mar 19 '16 at 10:35
$begingroup$
Ok. Edited my question. Thanks
$endgroup$
– shannoga
Mar 19 '16 at 10:35
$begingroup$
Got it. So, in this case, the "consecutive strings" start with elements of the progression ${0,5,10,15,dots}$. Given a random $m$ find the nearest multiple of $5$ less than or equal to $m$....then we need $m$ to either be this multiple or $1$ greater. Similar for general $n$.
$endgroup$
– lulu
Mar 19 '16 at 10:37
$begingroup$
Got it. So, in this case, the "consecutive strings" start with elements of the progression ${0,5,10,15,dots}$. Given a random $m$ find the nearest multiple of $5$ less than or equal to $m$....then we need $m$ to either be this multiple or $1$ greater. Similar for general $n$.
$endgroup$
– lulu
Mar 19 '16 at 10:37
$begingroup$
The terms of an arithmetic sequence have a common difference, so this is not an arithmetic sequence.
$endgroup$
– N. F. Taussig
Sep 7 '17 at 8:53
$begingroup$
The terms of an arithmetic sequence have a common difference, so this is not an arithmetic sequence.
$endgroup$
– N. F. Taussig
Sep 7 '17 at 8:53
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Okay generally speaking you need to have a formula for the terms of the sequence. In your case the formula seems to be
$$a_{2n} = 5n; , ; a_{2n+1} = 5n+1$$
However it might as well be a completely different formula, this is completely possible!
The basic notion is to try and find the relations between the terms and express them properly.
answered Mar 19 '16 at 10:44
christina_gchristina_g
125213
$endgroup$
add a comment |
$begingroup$
Your description of your sequence is perhaps not very practical when it comes to calculations. But the procedure for producing elements of the sequence is a periodic one. You take an input sequence of natural numbers and pick two and then skips $n$ and repeat. This means that you consume $n+2$ integers from the input sequence for each round through the algorithm.
You can reformulate the produce by working in chunks. You take a chunk of $n+2$ integers and emit the first two.
Now it shouldn't be too hard to see that the $(j+1)$th chunk consist of the numbers $(n+2)j, (n+2)j+1, dots, (n+2)j+n+1$ of which we pick the first two numbers. This results in that we see that the numbers we produce is $(n+2)j$ and $(n+2)j+1$.
So the question is transformed to the question whether a random number $r$ can be written as $r = (n+2)j$ or $r-1=(n+2)j$ for some integer $j$. That is if $r/(n+2)$ or $(r-1)/(n+2)$ are integers (we call that $r$ or $r-1$ is divisible by $n+2$).
answered Aug 8 '17 at 7:43
skykingskyking
14.3k1929
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
Okay generally speaking you need to have a formula for the terms of the sequence. In your case the formula seems to be
$$a_{2n} = 5n; , ; a_{2n+1} = 5n+1$$
However it might as well be a completely different formula, this is completely possible!
The basic notion is to try and find the relations between the terms and express them properly.
answered Mar 19 '16 at 10:44
christina_gchristina_g
125213
$endgroup$
add a comment |
$begingroup$
Okay generally speaking you need to have a formula for the terms of the sequence. In your case the formula seems to be
$$a_{2n} = 5n; , ; a_{2n+1} = 5n+1$$
However it might as well be a completely different formula, this is completely possible!
The basic notion is to try and find the relations between the terms and express them properly.
answered Mar 19 '16 at 10:44
christina_gchristina_g
125213
$endgroup$
add a comment |
$begingroup$
Okay generally speaking you need to have a formula for the terms of the sequence. In your case the formula seems to be
$$a_{2n} = 5n; , ; a_{2n+1} = 5n+1$$
However it might as well be a completely different formula, this is completely possible!
The basic notion is to try and find the relations between the terms and express them properly.
answered Mar 19 '16 at 10:44
christina_gchristina_g
125213
$endgroup$
Okay generally speaking you need to have a formula for the terms of the sequence. In your case the formula seems to be
$$a_{2n} = 5n; , ; a_{2n+1} = 5n+1$$
However it might as well be a completely different formula, this is completely possible!
The basic notion is to try and find the relations between the terms and express them properly.
answered Mar 19 '16 at 10:44
christina_gchristina_g
125213
answered Mar 19 '16 at 10:44
christina_gchristina_g
125213
answered Mar 19 '16 at 10:44
christina_gchristina_g
125213
answered Mar 19 '16 at 10:44
christina_gchristina_g
125213
125213
add a comment |
add a comment |
$begingroup$
Your description of your sequence is perhaps not very practical when it comes to calculations. But the procedure for producing elements of the sequence is a periodic one. You take an input sequence of natural numbers and pick two and then skips $n$ and repeat. This means that you consume $n+2$ integers from the input sequence for each round through the algorithm.
You can reformulate the produce by working in chunks. You take a chunk of $n+2$ integers and emit the first two.
Now it shouldn't be too hard to see that the $(j+1)$th chunk consist of the numbers $(n+2)j, (n+2)j+1, dots, (n+2)j+n+1$ of which we pick the first two numbers. This results in that we see that the numbers we produce is $(n+2)j$ and $(n+2)j+1$.
So the question is transformed to the question whether a random number $r$ can be written as $r = (n+2)j$ or $r-1=(n+2)j$ for some integer $j$. That is if $r/(n+2)$ or $(r-1)/(n+2)$ are integers (we call that $r$ or $r-1$ is divisible by $n+2$).
answered Aug 8 '17 at 7:43
skykingskyking
14.3k1929
$endgroup$
add a comment |
$begingroup$
Your description of your sequence is perhaps not very practical when it comes to calculations. But the procedure for producing elements of the sequence is a periodic one. You take an input sequence of natural numbers and pick two and then skips $n$ and repeat. This means that you consume $n+2$ integers from the input sequence for each round through the algorithm.
You can reformulate the produce by working in chunks. You take a chunk of $n+2$ integers and emit the first two.
Now it shouldn't be too hard to see that the $(j+1)$th chunk consist of the numbers $(n+2)j, (n+2)j+1, dots, (n+2)j+n+1$ of which we pick the first two numbers. This results in that we see that the numbers we produce is $(n+2)j$ and $(n+2)j+1$.
So the question is transformed to the question whether a random number $r$ can be written as $r = (n+2)j$ or $r-1=(n+2)j$ for some integer $j$. That is if $r/(n+2)$ or $(r-1)/(n+2)$ are integers (we call that $r$ or $r-1$ is divisible by $n+2$).
answered Aug 8 '17 at 7:43
skykingskyking
14.3k1929
$endgroup$
add a comment |
$begingroup$
Your description of your sequence is perhaps not very practical when it comes to calculations. But the procedure for producing elements of the sequence is a periodic one. You take an input sequence of natural numbers and pick two and then skips $n$ and repeat. This means that you consume $n+2$ integers from the input sequence for each round through the algorithm.
You can reformulate the produce by working in chunks. You take a chunk of $n+2$ integers and emit the first two.
Now it shouldn't be too hard to see that the $(j+1)$th chunk consist of the numbers $(n+2)j, (n+2)j+1, dots, (n+2)j+n+1$ of which we pick the first two numbers. This results in that we see that the numbers we produce is $(n+2)j$ and $(n+2)j+1$.
So the question is transformed to the question whether a random number $r$ can be written as $r = (n+2)j$ or $r-1=(n+2)j$ for some integer $j$. That is if $r/(n+2)$ or $(r-1)/(n+2)$ are integers (we call that $r$ or $r-1$ is divisible by $n+2$).
answered Aug 8 '17 at 7:43
skykingskyking
14.3k1929
$endgroup$
Your description of your sequence is perhaps not very practical when it comes to calculations. But the procedure for producing elements of the sequence is a periodic one. You take an input sequence of natural numbers and pick two and then skips $n$ and repeat. This means that you consume $n+2$ integers from the input sequence for each round through the algorithm.
You can reformulate the produce by working in chunks. You take a chunk of $n+2$ integers and emit the first two.
Now it shouldn't be too hard to see that the $(j+1)$th chunk consist of the numbers $(n+2)j, (n+2)j+1, dots, (n+2)j+n+1$ of which we pick the first two numbers. This results in that we see that the numbers we produce is $(n+2)j$ and $(n+2)j+1$.
So the question is transformed to the question whether a random number $r$ can be written as $r = (n+2)j$ or $r-1=(n+2)j$ for some integer $j$. That is if $r/(n+2)$ or $(r-1)/(n+2)$ are integers (we call that $r$ or $r-1$ is divisible by $n+2$).
answered Aug 8 '17 at 7:43
skykingskyking
14.3k1929
answered Aug 8 '17 at 7:43
skykingskyking
14.3k1929
answered Aug 8 '17 at 7:43
skykingskyking
14.3k1929
answered Aug 8 '17 at 7:43
skykingskyking
14.3k1929
14.3k1929
add a comment |
add a comment |
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$begingroup$
I know the sequence it is in this case n and then a gap of n+1. Thanks
$endgroup$
– shannoga
Mar 19 '16 at 9:58
$begingroup$
i don't understand the definition of the sequence. You say "$n$ numbers and then a gap of $n+1$" but your sequence looks like $2$ numbers and then a gap of $4$. What is $n$? Can you write out the first few terms if $n$ were, say, $3$?
$endgroup$
– lulu
Mar 19 '16 at 10:29
$begingroup$
Ok. Edited my question. Thanks
$endgroup$
– shannoga
Mar 19 '16 at 10:35
$begingroup$
Got it. So, in this case, the "consecutive strings" start with elements of the progression ${0,5,10,15,dots}$. Given a random $m$ find the nearest multiple of $5$ less than or equal to $m$....then we need $m$ to either be this multiple or $1$ greater. Similar for general $n$.
$endgroup$
– lulu
Mar 19 '16 at 10:37
$begingroup$
The terms of an arithmetic sequence have a common difference, so this is not an arithmetic sequence.
$endgroup$
– N. F. Taussig
Sep 7 '17 at 8:53