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!
sequences-and-series arithmetic-progressions
asked Jan 5 at 13:38
SarquesSarques
1
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add a comment |
$begingroup$
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
asked Jan 5 at 13:38
SarquesSarques
1
$endgroup$
$begingroup$
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 $0$.
$endgroup$
– lulu
Jan 5 at 13:49
add a comment |
$begingroup$
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
asked Jan 5 at 13:38
SarquesSarques
1
$endgroup$
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
sequences-and-series arithmetic-progressions
asked Jan 5 at 13:38
SarquesSarques
1
asked Jan 5 at 13:38
SarquesSarques
1
asked Jan 5 at 13:38
SarquesSarques
1
asked Jan 5 at 13:38
SarquesSarques
1
asked Jan 5 at 13:38
SarquesSarques
1
1
$begingroup$
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 $0$.
$endgroup$
– lulu
Jan 5 at 13:49
add a comment |
$begingroup$
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 $0$.
$endgroup$
– lulu
Jan 5 at 13:49
$begingroup$
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 $0$.
$endgroup$
– lulu
Jan 5 at 13:49
$begingroup$
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 $0$.
$endgroup$
– lulu
Jan 5 at 13:49
add a comment |
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$begingroup$
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 $0$.
$endgroup$
– lulu
Jan 5 at 13:49