Check if infinite series divisible individually by a number or not?
$begingroup$
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
$endgroup$
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 |
1 Answer
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One way, is to apply it to remainders. Then the values that are added are never bigger than the number you want to find the remainder on division by. Since odd+odd+odd=odd , we get that the sequence will never be even for example. You could Also expand the reccurence for n greater than the number of terms you have:$$T_n=T_{n-2}+2*T_{n-3}+3*T_{n-4}+2T_{n-5}+T_{n-6}$$ is another version that works for all n>6. For any composite that's not a power of a prime though, it's easier to test its coprime divisors (ironically prime powers or single primes are easiest) if 1 of these fails so will all numbers it's a divisor of. So 2 failing destroys all even numbers divisibility. All we need is to transform it into the correct form of reccurence to test these.
answered Feb 18 at 13:20
Roddy MacPheeRoddy MacPhee
20714
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1 Answer
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$begingroup$
One way, is to apply it to remainders. Then the values that are added are never bigger than the number you want to find the remainder on division by. Since odd+odd+odd=odd , we get that the sequence will never be even for example. You could Also expand the reccurence for n greater than the number of terms you have:$$T_n=T_{n-2}+2*T_{n-3}+3*T_{n-4}+2T_{n-5}+T_{n-6}$$ is another version that works for all n>6. For any composite that's not a power of a prime though, it's easier to test its coprime divisors (ironically prime powers or single primes are easiest) if 1 of these fails so will all numbers it's a divisor of. So 2 failing destroys all even numbers divisibility. All we need is to transform it into the correct form of reccurence to test these.
answered Feb 18 at 13:20
Roddy MacPheeRoddy MacPhee
20714
$endgroup$
add a comment |
$begingroup$
One way, is to apply it to remainders. Then the values that are added are never bigger than the number you want to find the remainder on division by. Since odd+odd+odd=odd , we get that the sequence will never be even for example. You could Also expand the reccurence for n greater than the number of terms you have:$$T_n=T_{n-2}+2*T_{n-3}+3*T_{n-4}+2T_{n-5}+T_{n-6}$$ is another version that works for all n>6. For any composite that's not a power of a prime though, it's easier to test its coprime divisors (ironically prime powers or single primes are easiest) if 1 of these fails so will all numbers it's a divisor of. So 2 failing destroys all even numbers divisibility. All we need is to transform it into the correct form of reccurence to test these.
answered Feb 18 at 13:20
Roddy MacPheeRoddy MacPhee
20714
$endgroup$
add a comment |
$begingroup$
One way, is to apply it to remainders. Then the values that are added are never bigger than the number you want to find the remainder on division by. Since odd+odd+odd=odd , we get that the sequence will never be even for example. You could Also expand the reccurence for n greater than the number of terms you have:$$T_n=T_{n-2}+2*T_{n-3}+3*T_{n-4}+2T_{n-5}+T_{n-6}$$ is another version that works for all n>6. For any composite that's not a power of a prime though, it's easier to test its coprime divisors (ironically prime powers or single primes are easiest) if 1 of these fails so will all numbers it's a divisor of. So 2 failing destroys all even numbers divisibility. All we need is to transform it into the correct form of reccurence to test these.
answered Feb 18 at 13:20
Roddy MacPheeRoddy MacPhee
20714
$endgroup$
One way, is to apply it to remainders. Then the values that are added are never bigger than the number you want to find the remainder on division by. Since odd+odd+odd=odd , we get that the sequence will never be even for example. You could Also expand the reccurence for n greater than the number of terms you have:$$T_n=T_{n-2}+2*T_{n-3}+3*T_{n-4}+2T_{n-5}+T_{n-6}$$ is another version that works for all n>6. For any composite that's not a power of a prime though, it's easier to test its coprime divisors (ironically prime powers or single primes are easiest) if 1 of these fails so will all numbers it's a divisor of. So 2 failing destroys all even numbers divisibility. All we need is to transform it into the correct form of reccurence to test these.
answered Feb 18 at 13:20
Roddy MacPheeRoddy MacPhee
20714
edited Feb 18 at 13:43
answered Feb 18 at 13:20
Roddy MacPheeRoddy MacPhee
20714
answered Feb 18 at 13:20
Roddy MacPheeRoddy MacPhee
20714
answered Feb 18 at 13:20
Roddy MacPheeRoddy MacPhee
20714
20714
add a comment |
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