How can I determine if a number is a part of arithmetic sequence












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 19 '16 at 10:34







shannoga

















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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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 $n+2$).






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      active

      oldest

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      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 19 '16 at 10:44









          christina_gchristina_g

          125213




          125213























              0












              $begingroup$

              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 $n+2$).






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                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 $n+2$).






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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 $n+2$).






                  share|cite|improve this answer









                  $endgroup$



                  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 $n+2$).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 8 '17 at 7:43









                  skykingskyking

                  14.3k1929




                  14.3k1929






























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