Mixing
To find a Sequence
$begingroup$
I need to find a sequence of N distinct positive numbers such that each pair of adjacent numbers are not co-prime but each continous triplet should be co-prime (considering array to be in cyclic manner).
I particularly solved the problem by taking the product of Consecutive prime numbers. But the problem is, there is another constraint that each number in the sequence must be less than 10^9 and N can go up to 5*10^4 so my numbers will go beyond 10^9. Can anyone help how to overcome this.
for example N=6; My answer 6 15 35 77 143 26
first six prime Numbers 2 3 5 7 11 13
prime-numbers
asked Jan 5 at 8:52
Vijju12345Vijju12345
6
$endgroup$
add a comment |
$begingroup$
I need to find a sequence of N distinct positive numbers such that each pair of adjacent numbers are not co-prime but each continous triplet should be co-prime (considering array to be in cyclic manner).
I particularly solved the problem by taking the product of Consecutive prime numbers. But the problem is, there is another constraint that each number in the sequence must be less than 10^9 and N can go up to 5*10^4 so my numbers will go beyond 10^9. Can anyone help how to overcome this.
for example N=6; My answer 6 15 35 77 143 26
first six prime Numbers 2 3 5 7 11 13
prime-numbers
asked Jan 5 at 8:52
Vijju12345Vijju12345
6
$endgroup$
$begingroup$
Maybe, you are successful, if you use all products of two primes upto the given limit, for example also the numbers $10$ and $14$. No idea at the moment how to arrange the numbers however to guarantee the given coniditions.
$endgroup$
– Peter
Jan 5 at 9:05
$begingroup$
well I used that idea but doesn't come up with the way to arrange the numbers.
$endgroup$
– Vijju12345
Jan 5 at 9:06
add a comment |
$begingroup$
I need to find a sequence of N distinct positive numbers such that each pair of adjacent numbers are not co-prime but each continous triplet should be co-prime (considering array to be in cyclic manner).
I particularly solved the problem by taking the product of Consecutive prime numbers. But the problem is, there is another constraint that each number in the sequence must be less than 10^9 and N can go up to 5*10^4 so my numbers will go beyond 10^9. Can anyone help how to overcome this.
for example N=6; My answer 6 15 35 77 143 26
first six prime Numbers 2 3 5 7 11 13
prime-numbers
asked Jan 5 at 8:52
Vijju12345Vijju12345
6
$endgroup$
I need to find a sequence of N distinct positive numbers such that each pair of adjacent numbers are not co-prime but each continous triplet should be co-prime (considering array to be in cyclic manner).
I particularly solved the problem by taking the product of Consecutive prime numbers. But the problem is, there is another constraint that each number in the sequence must be less than 10^9 and N can go up to 5*10^4 so my numbers will go beyond 10^9. Can anyone help how to overcome this.
for example N=6; My answer 6 15 35 77 143 26
first six prime Numbers 2 3 5 7 11 13
prime-numbers
prime-numbers
asked Jan 5 at 8:52
Vijju12345Vijju12345
6
asked Jan 5 at 8:52
Vijju12345Vijju12345
6
asked Jan 5 at 8:52
Vijju12345Vijju12345
6
asked Jan 5 at 8:52
Vijju12345Vijju12345
6
asked Jan 5 at 8:52
Vijju12345Vijju12345
6
6
$begingroup$
Maybe, you are successful, if you use all products of two primes upto the given limit, for example also the numbers $10$ and $14$. No idea at the moment how to arrange the numbers however to guarantee the given coniditions.
$endgroup$
– Peter
Jan 5 at 9:05
$begingroup$
well I used that idea but doesn't come up with the way to arrange the numbers.
$endgroup$
– Vijju12345
Jan 5 at 9:06
add a comment |
$begingroup$
Maybe, you are successful, if you use all products of two primes upto the given limit, for example also the numbers $10$ and $14$. No idea at the moment how to arrange the numbers however to guarantee the given coniditions.
$endgroup$
– Peter
Jan 5 at 9:05
$begingroup$
well I used that idea but doesn't come up with the way to arrange the numbers.
$endgroup$
– Vijju12345
Jan 5 at 9:06
$begingroup$
Maybe, you are successful, if you use all products of two primes upto the given limit, for example also the numbers $10$ and $14$. No idea at the moment how to arrange the numbers however to guarantee the given coniditions.
$endgroup$
– Peter
Jan 5 at 9:05
$begingroup$
Maybe, you are successful, if you use all products of two primes upto the given limit, for example also the numbers $10$ and $14$. No idea at the moment how to arrange the numbers however to guarantee the given coniditions.
$endgroup$
– Peter
Jan 5 at 9:05
$begingroup$
well I used that idea but doesn't come up with the way to arrange the numbers.
$endgroup$
– Vijju12345
Jan 5 at 9:06
$begingroup$
well I used that idea but doesn't come up with the way to arrange the numbers.
$endgroup$
– Vijju12345
Jan 5 at 9:06
add a comment |
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$begingroup$
Maybe, you are successful, if you use all products of two primes upto the given limit, for example also the numbers $10$ and $14$. No idea at the moment how to arrange the numbers however to guarantee the given coniditions.
$endgroup$
– Peter
Jan 5 at 9:05
$begingroup$
well I used that idea but doesn't come up with the way to arrange the numbers.
$endgroup$
– Vijju12345
Jan 5 at 9:06