Check if infinite series divisible individually by a number or not?












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We are given first three numbers of a infinite series and the next elements of series would be decided by following formula:



Tn= Tn-1 + Tn-2 + Tn-3



If we are given the first three numbers as 1, 1, 1, then the series would look like this:



1, 1, 1, 3, 5, 9, 17, 31, . . . .



Now I wanted to ask the logic on how to check if all the elements of the series are indivisible by a number individually. For example, we know that the above series is not divisible by 27. Thanks in advance!










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  • This is a bit vague. For small numbers, like $n=27$, you can just write out the entire sequence $pmod n$. After all, it has to be periodic (as there are only finitely many values available). Of course, even for small $n$, the period can be quite large. For $n=27$ you can notice that $(a_{40},a_{41},a_{42})=(10,10,10)$ so from there on the sequence is just $10$ times what it was up to $a_{39}$. As it is not $0$ in that range, it is never $0$.
    – lulu
    2 days ago
















0














We are given first three numbers of a infinite series and the next elements of series would be decided by following formula:



Tn= Tn-1 + Tn-2 + Tn-3



If we are given the first three numbers as 1, 1, 1, then the series would look like this:



1, 1, 1, 3, 5, 9, 17, 31, . . . .



Now I wanted to ask the logic on how to check if all the elements of the series are indivisible by a number individually. For example, we know that the above series is not divisible by 27. Thanks in advance!










share|cite|improve this question







New contributor




Sarques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • This is a bit vague. For small numbers, like $n=27$, you can just write out the entire sequence $pmod n$. After all, it has to be periodic (as there are only finitely many values available). Of course, even for small $n$, the period can be quite large. For $n=27$ you can notice that $(a_{40},a_{41},a_{42})=(10,10,10)$ so from there on the sequence is just $10$ times what it was up to $a_{39}$. As it is not $0$ in that range, it is never $0$.
    – lulu
    2 days ago














0












0








0







We are given first three numbers of a infinite series and the next elements of series would be decided by following formula:



Tn= Tn-1 + Tn-2 + Tn-3



If we are given the first three numbers as 1, 1, 1, then the series would look like this:



1, 1, 1, 3, 5, 9, 17, 31, . . . .



Now I wanted to ask the logic on how to check if all the elements of the series are indivisible by a number individually. For example, we know that the above series is not divisible by 27. Thanks in advance!










share|cite|improve this question







New contributor




Sarques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











We are given first three numbers of a infinite series and the next elements of series would be decided by following formula:



Tn= Tn-1 + Tn-2 + Tn-3



If we are given the first three numbers as 1, 1, 1, then the series would look like this:



1, 1, 1, 3, 5, 9, 17, 31, . . . .



Now I wanted to ask the logic on how to check if all the elements of the series are indivisible by a number individually. For example, we know that the above series is not divisible by 27. Thanks in advance!







sequences-and-series arithmetic-progressions






share|cite|improve this question







New contributor




Sarques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Sarques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Sarques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









SarquesSarques

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New contributor




Sarques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Sarques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Sarques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • This is a bit vague. For small numbers, like $n=27$, you can just write out the entire sequence $pmod n$. After all, it has to be periodic (as there are only finitely many values available). Of course, even for small $n$, the period can be quite large. For $n=27$ you can notice that $(a_{40},a_{41},a_{42})=(10,10,10)$ so from there on the sequence is just $10$ times what it was up to $a_{39}$. As it is not $0$ in that range, it is never $0$.
    – lulu
    2 days ago


















  • This is a bit vague. For small numbers, like $n=27$, you can just write out the entire sequence $pmod n$. After all, it has to be periodic (as there are only finitely many values available). Of course, even for small $n$, the period can be quite large. For $n=27$ you can notice that $(a_{40},a_{41},a_{42})=(10,10,10)$ so from there on the sequence is just $10$ times what it was up to $a_{39}$. As it is not $0$ in that range, it is never $0$.
    – lulu
    2 days ago
















This is a bit vague. For small numbers, like $n=27$, you can just write out the entire sequence $pmod n$. After all, it has to be periodic (as there are only finitely many values available). Of course, even for small $n$, the period can be quite large. For $n=27$ you can notice that $(a_{40},a_{41},a_{42})=(10,10,10)$ so from there on the sequence is just $10$ times what it was up to $a_{39}$. As it is not $0$ in that range, it is never $0$.
– lulu
2 days ago




This is a bit vague. For small numbers, like $n=27$, you can just write out the entire sequence $pmod n$. After all, it has to be periodic (as there are only finitely many values available). Of course, even for small $n$, the period can be quite large. For $n=27$ you can notice that $(a_{40},a_{41},a_{42})=(10,10,10)$ so from there on the sequence is just $10$ times what it was up to $a_{39}$. As it is not $0$ in that range, it is never $0$.
– lulu
2 days ago










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