Mixing
Deriving a formula for an arbitrary term in $1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, ldots$
$begingroup$
A sequence of numbers is given as:
$$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, text{and so on}$$
(Each integer $n$ is repeated $n$ times.) What will be the 50th term of that sequence?
Let's say $x=50$. Then solving $x=n(n+1)/2$ gives the value of $n$ (rounded off to nearest integer).
My question is: If I make a small change in the above pattern, can we have a direct formula to calculate $n$?
The new pattern is :
$$1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, ldots$$
i.e., $1$ and $2$ are repeated once, and $n$ is repeated $(n-1)$ times.
How can we derive a formula for this pattern?
pattern-recognition
edited Jan 30 at 19:53
Blue
49.4k870157
asked Jan 30 at 19:19
Maxy DaenMaxy Daen
62
$endgroup$
add a comment |
$begingroup$
A sequence of numbers is given as:
$$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, text{and so on}$$
(Each integer $n$ is repeated $n$ times.) What will be the 50th term of that sequence?
Let's say $x=50$. Then solving $x=n(n+1)/2$ gives the value of $n$ (rounded off to nearest integer).
My question is: If I make a small change in the above pattern, can we have a direct formula to calculate $n$?
The new pattern is :
$$1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, ldots$$
i.e., $1$ and $2$ are repeated once, and $n$ is repeated $(n-1)$ times.
How can we derive a formula for this pattern?
pattern-recognition
edited Jan 30 at 19:53
Blue
49.4k870157
asked Jan 30 at 19:19
Maxy DaenMaxy Daen
62
$endgroup$
1
$begingroup$
You can ignore the existence of the $1$, and then notice that the sequence $2,3,3,4,4,4,5,5,5,5,dots$ is simply the sequence where each entry is one more than the corresponding entry in the earlier mentioned sequence.
$endgroup$
– JMoravitz
Jan 30 at 19:25
$begingroup$
Yeah i get that, but let's say i'm only provided with 'x' and i need to find the corresponding 'n' for it according to the new sequence
$endgroup$
– Maxy Daen
Jan 30 at 19:31
add a comment |
$begingroup$
A sequence of numbers is given as:
$$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, text{and so on}$$
(Each integer $n$ is repeated $n$ times.) What will be the 50th term of that sequence?
Let's say $x=50$. Then solving $x=n(n+1)/2$ gives the value of $n$ (rounded off to nearest integer).
My question is: If I make a small change in the above pattern, can we have a direct formula to calculate $n$?
The new pattern is :
$$1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, ldots$$
i.e., $1$ and $2$ are repeated once, and $n$ is repeated $(n-1)$ times.
How can we derive a formula for this pattern?
pattern-recognition
edited Jan 30 at 19:53
Blue
49.4k870157
asked Jan 30 at 19:19
Maxy DaenMaxy Daen
62
$endgroup$
A sequence of numbers is given as:
$$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, text{and so on}$$
(Each integer $n$ is repeated $n$ times.) What will be the 50th term of that sequence?
Let's say $x=50$. Then solving $x=n(n+1)/2$ gives the value of $n$ (rounded off to nearest integer).
My question is: If I make a small change in the above pattern, can we have a direct formula to calculate $n$?
The new pattern is :
$$1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, ldots$$
i.e., $1$ and $2$ are repeated once, and $n$ is repeated $(n-1)$ times.
How can we derive a formula for this pattern?
pattern-recognition
pattern-recognition
edited Jan 30 at 19:53
Blue
49.4k870157
asked Jan 30 at 19:19
Maxy DaenMaxy Daen
62
edited Jan 30 at 19:53
Blue
49.4k870157
asked Jan 30 at 19:19
Maxy DaenMaxy Daen
62
edited Jan 30 at 19:53
Blue
49.4k870157
edited Jan 30 at 19:53
Blue
49.4k870157
edited Jan 30 at 19:53
Blue
49.4k870157
49.4k870157
asked Jan 30 at 19:19
Maxy DaenMaxy Daen
62
asked Jan 30 at 19:19
Maxy DaenMaxy Daen
62
asked Jan 30 at 19:19
Maxy DaenMaxy Daen
62
62
1
$begingroup$
You can ignore the existence of the $1$, and then notice that the sequence $2,3,3,4,4,4,5,5,5,5,dots$ is simply the sequence where each entry is one more than the corresponding entry in the earlier mentioned sequence.
$endgroup$
– JMoravitz
Jan 30 at 19:25
$begingroup$
Yeah i get that, but let's say i'm only provided with 'x' and i need to find the corresponding 'n' for it according to the new sequence
$endgroup$
– Maxy Daen
Jan 30 at 19:31
add a comment |
1
$begingroup$
You can ignore the existence of the $1$, and then notice that the sequence $2,3,3,4,4,4,5,5,5,5,dots$ is simply the sequence where each entry is one more than the corresponding entry in the earlier mentioned sequence.
$endgroup$
– JMoravitz
Jan 30 at 19:25
$begingroup$
Yeah i get that, but let's say i'm only provided with 'x' and i need to find the corresponding 'n' for it according to the new sequence
$endgroup$
– Maxy Daen
Jan 30 at 19:31
1
1
$begingroup$
You can ignore the existence of the $1$, and then notice that the sequence $2,3,3,4,4,4,5,5,5,5,dots$ is simply the sequence where each entry is one more than the corresponding entry in the earlier mentioned sequence.
$endgroup$
– JMoravitz
Jan 30 at 19:25
$begingroup$
You can ignore the existence of the $1$, and then notice that the sequence $2,3,3,4,4,4,5,5,5,5,dots$ is simply the sequence where each entry is one more than the corresponding entry in the earlier mentioned sequence.
$endgroup$
– JMoravitz
Jan 30 at 19:25
$begingroup$
Yeah i get that, but let's say i'm only provided with 'x' and i need to find the corresponding 'n' for it according to the new sequence
$endgroup$
– Maxy Daen
Jan 30 at 19:31
$begingroup$
Yeah i get that, but let's say i'm only provided with 'x' and i need to find the corresponding 'n' for it according to the new sequence
$endgroup$
– Maxy Daen
Jan 30 at 19:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Using JMoravitz hint above, you can replace $x$ with $x-1$ and $n$ with $n-1$ and now your formula is $x-1=(n-1)(n)/2$. Whatever you get for n should be rounded up to the nearest integer if it isn't an integer already.
For example for $x=5$, we have $4 = (n-1)(n)/2 to 8 = (n-1)(n) to n^2-n-8=0 to n=3.37$ which rounds up to 4.
answered Jan 30 at 19:45
Akash PatelAkash Patel
168110
$endgroup$
1
$begingroup$
Ultimately this amounts to the formula $$ n = left lceil frac 12 (sqrt{8x - 7} + 1) right rceil $$
$endgroup$
– Omnomnomnom
Jan 30 at 19:59
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Using JMoravitz hint above, you can replace $x$ with $x-1$ and $n$ with $n-1$ and now your formula is $x-1=(n-1)(n)/2$. Whatever you get for n should be rounded up to the nearest integer if it isn't an integer already.
For example for $x=5$, we have $4 = (n-1)(n)/2 to 8 = (n-1)(n) to n^2-n-8=0 to n=3.37$ which rounds up to 4.
answered Jan 30 at 19:45
Akash PatelAkash Patel
168110
$endgroup$
1
$begingroup$
Ultimately this amounts to the formula $$ n = left lceil frac 12 (sqrt{8x - 7} + 1) right rceil $$
$endgroup$
– Omnomnomnom
Jan 30 at 19:59
add a comment |
$begingroup$
Using JMoravitz hint above, you can replace $x$ with $x-1$ and $n$ with $n-1$ and now your formula is $x-1=(n-1)(n)/2$. Whatever you get for n should be rounded up to the nearest integer if it isn't an integer already.
For example for $x=5$, we have $4 = (n-1)(n)/2 to 8 = (n-1)(n) to n^2-n-8=0 to n=3.37$ which rounds up to 4.
answered Jan 30 at 19:45
Akash PatelAkash Patel
168110
$endgroup$
1
$begingroup$
Ultimately this amounts to the formula $$ n = left lceil frac 12 (sqrt{8x - 7} + 1) right rceil $$
$endgroup$
– Omnomnomnom
Jan 30 at 19:59
add a comment |
$begingroup$
Using JMoravitz hint above, you can replace $x$ with $x-1$ and $n$ with $n-1$ and now your formula is $x-1=(n-1)(n)/2$. Whatever you get for n should be rounded up to the nearest integer if it isn't an integer already.
For example for $x=5$, we have $4 = (n-1)(n)/2 to 8 = (n-1)(n) to n^2-n-8=0 to n=3.37$ which rounds up to 4.
answered Jan 30 at 19:45
Akash PatelAkash Patel
168110
$endgroup$
Using JMoravitz hint above, you can replace $x$ with $x-1$ and $n$ with $n-1$ and now your formula is $x-1=(n-1)(n)/2$. Whatever you get for n should be rounded up to the nearest integer if it isn't an integer already.
For example for $x=5$, we have $4 = (n-1)(n)/2 to 8 = (n-1)(n) to n^2-n-8=0 to n=3.37$ which rounds up to 4.
answered Jan 30 at 19:45
Akash PatelAkash Patel
168110
answered Jan 30 at 19:45
Akash PatelAkash Patel
168110
answered Jan 30 at 19:45
Akash PatelAkash Patel
168110
answered Jan 30 at 19:45
Akash PatelAkash Patel
168110
168110
1
$begingroup$
Ultimately this amounts to the formula $$ n = left lceil frac 12 (sqrt{8x - 7} + 1) right rceil $$
$endgroup$
– Omnomnomnom
Jan 30 at 19:59
add a comment |
1
$begingroup$
Ultimately this amounts to the formula $$ n = left lceil frac 12 (sqrt{8x - 7} + 1) right rceil $$
$endgroup$
– Omnomnomnom
Jan 30 at 19:59
1
1
$begingroup$
Ultimately this amounts to the formula $$ n = left lceil frac 12 (sqrt{8x - 7} + 1) right rceil $$
$endgroup$
– Omnomnomnom
Jan 30 at 19:59
$begingroup$
Ultimately this amounts to the formula $$ n = left lceil frac 12 (sqrt{8x - 7} + 1) right rceil $$
$endgroup$
– Omnomnomnom
Jan 30 at 19:59
add a comment |
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$begingroup$
You can ignore the existence of the $1$, and then notice that the sequence $2,3,3,4,4,4,5,5,5,5,dots$ is simply the sequence where each entry is one more than the corresponding entry in the earlier mentioned sequence.
$endgroup$
– JMoravitz
Jan 30 at 19:25
$begingroup$
Yeah i get that, but let's say i'm only provided with 'x' and i need to find the corresponding 'n' for it according to the new sequence
$endgroup$
– Maxy Daen
Jan 30 at 19:31