Number of increasing sequences alternating between odd and even involving only integers from the set $[1, 2,...
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I am new to this forum so do correct me if I am doing anything wrong.
(Problem 6 of British Mathematics Olympiad1 - 1994)
An increasing sequence of integers is said to be alternating
if it starts with an odd term, the second term is even, the
third term is odd, the fourth is even, and so on. The empty
sequence (with no term at all!) is considered to be alternating.
Let
$A(n)$ denote the number of alternating sequences which
only involve integers from the set $[1, 2,... ,n]$. Find the value of
$A(20)$ and prove
that your value is correct.
I believe a method is to set up a recursion.
What I have done is as follows:
(I've slightly changed the definition of A(n) to be the set of all possible sequences)
Let $O(n)$ be the number of increasing alternating sequences which only involve integers from the set $[1, 2, ..., n]$ that end in an odd number and $E(n)$ be the number of increasing alternating sequences which only involve integers from the set $[1, 2, ..., n]$ that end in an even number.
For every even $n$, $O(n+2) = O(n)+E(n)+1$.
Proof: For every sequence that ends in an odd number in $A(n)$, it will also be a sequence of $A(n+2)$, hence add $O(n)$. For every sequence that ends in an even number in $A(n)$, $n+1$, which is odd, can be added to the end of that sequence, hence add $E(n)$. The sequence with only $n+1$ also satisfies the conditions hence add $1$.
For every even $n$, $E(n+2)=2E(n)+O(n)+1$.
Proof: For every sequence that ends in an odd number in $A(n)$, $n+2$, which is even, can be added to the end of that sequence, hence add $O(n)$. For every sequence that ends in an even number in $A(n)$, nothing or both $n+1$ and $n+2$ can be added to the end of that sequence, hence add $2E(n)$. The sequence with only $n+2$ also satisfies the conditions hence add $1$.
$A(n) = E(n)+O(n)+1$
Proof: Every sequence either ends in an even number or an odd number or is an empty sequence.
Starting with $E(0)=0$ and $O(0)=0$, we get $E(20)=10945$ and $O(20)=6765$, meaning $A(20)=17711$.
Is there anything wrong with my method? If not is there a more efficient way to get to the answer? Thanks.
combinatorics contest-math
asked Nov 21 at 13:18
3684
1277
add a comment |
up vote
0
down vote
favorite
I am new to this forum so do correct me if I am doing anything wrong.
(Problem 6 of British Mathematics Olympiad1 - 1994)
An increasing sequence of integers is said to be alternating
if it starts with an odd term, the second term is even, the
third term is odd, the fourth is even, and so on. The empty
sequence (with no term at all!) is considered to be alternating.
Let
$A(n)$ denote the number of alternating sequences which
only involve integers from the set $[1, 2,... ,n]$. Find the value of
$A(20)$ and prove
that your value is correct.
I believe a method is to set up a recursion.
What I have done is as follows:
(I've slightly changed the definition of A(n) to be the set of all possible sequences)
Let $O(n)$ be the number of increasing alternating sequences which only involve integers from the set $[1, 2, ..., n]$ that end in an odd number and $E(n)$ be the number of increasing alternating sequences which only involve integers from the set $[1, 2, ..., n]$ that end in an even number.
For every even $n$, $O(n+2) = O(n)+E(n)+1$.
Proof: For every sequence that ends in an odd number in $A(n)$, it will also be a sequence of $A(n+2)$, hence add $O(n)$. For every sequence that ends in an even number in $A(n)$, $n+1$, which is odd, can be added to the end of that sequence, hence add $E(n)$. The sequence with only $n+1$ also satisfies the conditions hence add $1$.
For every even $n$, $E(n+2)=2E(n)+O(n)+1$.
Proof: For every sequence that ends in an odd number in $A(n)$, $n+2$, which is even, can be added to the end of that sequence, hence add $O(n)$. For every sequence that ends in an even number in $A(n)$, nothing or both $n+1$ and $n+2$ can be added to the end of that sequence, hence add $2E(n)$. The sequence with only $n+2$ also satisfies the conditions hence add $1$.
$A(n) = E(n)+O(n)+1$
Proof: Every sequence either ends in an even number or an odd number or is an empty sequence.
Starting with $E(0)=0$ and $O(0)=0$, we get $E(20)=10945$ and $O(20)=6765$, meaning $A(20)=17711$.
Is there anything wrong with my method? If not is there a more efficient way to get to the answer? Thanks.
combinatorics contest-math
asked Nov 21 at 13:18
3684
1277
I think there's a small mistake. You say the sequence consisting only of $n+2$ is alternating, but this is not true unless $n$ is odd, because one of the conditions is that an alternating sequence start with an odd term. Since you eventually compute $A(20),$ you need $n$ to be even. Your general approach looks fine, but you should review the details.
– saulspatz
Nov 21 at 13:51
I believe I have avoided this by stating for every even n, or is this not what you mean.
– 3684
Nov 21 at 14:12
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am new to this forum so do correct me if I am doing anything wrong.
(Problem 6 of British Mathematics Olympiad1 - 1994)
An increasing sequence of integers is said to be alternating
if it starts with an odd term, the second term is even, the
third term is odd, the fourth is even, and so on. The empty
sequence (with no term at all!) is considered to be alternating.
Let
$A(n)$ denote the number of alternating sequences which
only involve integers from the set $[1, 2,... ,n]$. Find the value of
$A(20)$ and prove
that your value is correct.
I believe a method is to set up a recursion.
What I have done is as follows:
(I've slightly changed the definition of A(n) to be the set of all possible sequences)
Let $O(n)$ be the number of increasing alternating sequences which only involve integers from the set $[1, 2, ..., n]$ that end in an odd number and $E(n)$ be the number of increasing alternating sequences which only involve integers from the set $[1, 2, ..., n]$ that end in an even number.
For every even $n$, $O(n+2) = O(n)+E(n)+1$.
Proof: For every sequence that ends in an odd number in $A(n)$, it will also be a sequence of $A(n+2)$, hence add $O(n)$. For every sequence that ends in an even number in $A(n)$, $n+1$, which is odd, can be added to the end of that sequence, hence add $E(n)$. The sequence with only $n+1$ also satisfies the conditions hence add $1$.
For every even $n$, $E(n+2)=2E(n)+O(n)+1$.
Proof: For every sequence that ends in an odd number in $A(n)$, $n+2$, which is even, can be added to the end of that sequence, hence add $O(n)$. For every sequence that ends in an even number in $A(n)$, nothing or both $n+1$ and $n+2$ can be added to the end of that sequence, hence add $2E(n)$. The sequence with only $n+2$ also satisfies the conditions hence add $1$.
$A(n) = E(n)+O(n)+1$
Proof: Every sequence either ends in an even number or an odd number or is an empty sequence.
Starting with $E(0)=0$ and $O(0)=0$, we get $E(20)=10945$ and $O(20)=6765$, meaning $A(20)=17711$.
Is there anything wrong with my method? If not is there a more efficient way to get to the answer? Thanks.
combinatorics contest-math
asked Nov 21 at 13:18
3684
1277
I am new to this forum so do correct me if I am doing anything wrong.
(Problem 6 of British Mathematics Olympiad1 - 1994)
An increasing sequence of integers is said to be alternating
if it starts with an odd term, the second term is even, the
third term is odd, the fourth is even, and so on. The empty
sequence (with no term at all!) is considered to be alternating.
Let
$A(n)$ denote the number of alternating sequences which
only involve integers from the set $[1, 2,... ,n]$. Find the value of
$A(20)$ and prove
that your value is correct.
I believe a method is to set up a recursion.
What I have done is as follows:
(I've slightly changed the definition of A(n) to be the set of all possible sequences)
Let $O(n)$ be the number of increasing alternating sequences which only involve integers from the set $[1, 2, ..., n]$ that end in an odd number and $E(n)$ be the number of increasing alternating sequences which only involve integers from the set $[1, 2, ..., n]$ that end in an even number.
For every even $n$, $O(n+2) = O(n)+E(n)+1$.
Proof: For every sequence that ends in an odd number in $A(n)$, it will also be a sequence of $A(n+2)$, hence add $O(n)$. For every sequence that ends in an even number in $A(n)$, $n+1$, which is odd, can be added to the end of that sequence, hence add $E(n)$. The sequence with only $n+1$ also satisfies the conditions hence add $1$.
For every even $n$, $E(n+2)=2E(n)+O(n)+1$.
Proof: For every sequence that ends in an odd number in $A(n)$, $n+2$, which is even, can be added to the end of that sequence, hence add $O(n)$. For every sequence that ends in an even number in $A(n)$, nothing or both $n+1$ and $n+2$ can be added to the end of that sequence, hence add $2E(n)$. The sequence with only $n+2$ also satisfies the conditions hence add $1$.
$A(n) = E(n)+O(n)+1$
Proof: Every sequence either ends in an even number or an odd number or is an empty sequence.
Starting with $E(0)=0$ and $O(0)=0$, we get $E(20)=10945$ and $O(20)=6765$, meaning $A(20)=17711$.
Is there anything wrong with my method? If not is there a more efficient way to get to the answer? Thanks.
combinatorics contest-math
combinatorics contest-math
asked Nov 21 at 13:18
3684
1277
asked Nov 21 at 13:18
3684
1277
asked Nov 21 at 13:18
3684
1277
asked Nov 21 at 13:18
3684
1277
asked Nov 21 at 13:18
3684
1277
1277
I think there's a small mistake. You say the sequence consisting only of $n+2$ is alternating, but this is not true unless $n$ is odd, because one of the conditions is that an alternating sequence start with an odd term. Since you eventually compute $A(20),$ you need $n$ to be even. Your general approach looks fine, but you should review the details.
– saulspatz
Nov 21 at 13:51
I believe I have avoided this by stating for every even n, or is this not what you mean.
– 3684
Nov 21 at 14:12
add a comment |
I think there's a small mistake. You say the sequence consisting only of $n+2$ is alternating, but this is not true unless $n$ is odd, because one of the conditions is that an alternating sequence start with an odd term. Since you eventually compute $A(20),$ you need $n$ to be even. Your general approach looks fine, but you should review the details.
– saulspatz
Nov 21 at 13:51
I believe I have avoided this by stating for every even n, or is this not what you mean.
– 3684
Nov 21 at 14:12
I think there's a small mistake. You say the sequence consisting only of $n+2$ is alternating, but this is not true unless $n$ is odd, because one of the conditions is that an alternating sequence start with an odd term. Since you eventually compute $A(20),$ you need $n$ to be even. Your general approach looks fine, but you should review the details.
– saulspatz
Nov 21 at 13:51
I think there's a small mistake. You say the sequence consisting only of $n+2$ is alternating, but this is not true unless $n$ is odd, because one of the conditions is that an alternating sequence start with an odd term. Since you eventually compute $A(20),$ you need $n$ to be even. Your general approach looks fine, but you should review the details.
– saulspatz
Nov 21 at 13:51
I believe I have avoided this by stating for every even n, or is this not what you mean.
– 3684
Nov 21 at 14:12
I believe I have avoided this by stating for every even n, or is this not what you mean.
– 3684
Nov 21 at 14:12
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
If a sequence starts with 1, the rest of the sequence can be found by adding 1 every term of a sequence from $A(n-1)$.
If it doesn't begin with 1, you get it by adding 2 to every term of a sequence from $A(n-2)$.
So $A(n)=A(n-1)+A(n-2)$.
answered Nov 21 at 13:48
Empy2
33.3k12261
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 at 14:04
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 at 14:15
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 at 14:17
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 at 14:18
I believe this is not quite right as it does not deal with the empty sequence, it may need to be slightly adjusted but thanks for the comment.
– 3684
Nov 21 at 14:21
|
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
If a sequence starts with 1, the rest of the sequence can be found by adding 1 every term of a sequence from $A(n-1)$.
If it doesn't begin with 1, you get it by adding 2 to every term of a sequence from $A(n-2)$.
So $A(n)=A(n-1)+A(n-2)$.
answered Nov 21 at 13:48
Empy2
33.3k12261
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 at 14:04
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 at 14:15
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 at 14:17
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 at 14:18
I believe this is not quite right as it does not deal with the empty sequence, it may need to be slightly adjusted but thanks for the comment.
– 3684
Nov 21 at 14:21
|
show 4 more comments
up vote
3
down vote
accepted
If a sequence starts with 1, the rest of the sequence can be found by adding 1 every term of a sequence from $A(n-1)$.
If it doesn't begin with 1, you get it by adding 2 to every term of a sequence from $A(n-2)$.
So $A(n)=A(n-1)+A(n-2)$.
answered Nov 21 at 13:48
Empy2
33.3k12261
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 at 14:04
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 at 14:15
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 at 14:17
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 at 14:18
I believe this is not quite right as it does not deal with the empty sequence, it may need to be slightly adjusted but thanks for the comment.
– 3684
Nov 21 at 14:21
|
show 4 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If a sequence starts with 1, the rest of the sequence can be found by adding 1 every term of a sequence from $A(n-1)$.
If it doesn't begin with 1, you get it by adding 2 to every term of a sequence from $A(n-2)$.
So $A(n)=A(n-1)+A(n-2)$.
answered Nov 21 at 13:48
Empy2
33.3k12261
If a sequence starts with 1, the rest of the sequence can be found by adding 1 every term of a sequence from $A(n-1)$.
If it doesn't begin with 1, you get it by adding 2 to every term of a sequence from $A(n-2)$.
So $A(n)=A(n-1)+A(n-2)$.
answered Nov 21 at 13:48
Empy2
33.3k12261
answered Nov 21 at 13:48
Empy2
33.3k12261
answered Nov 21 at 13:48
Empy2
33.3k12261
answered Nov 21 at 13:48
Empy2
33.3k12261
33.3k12261
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 at 14:04
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 at 14:15
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 at 14:17
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 at 14:18
I believe this is not quite right as it does not deal with the empty sequence, it may need to be slightly adjusted but thanks for the comment.
– 3684
Nov 21 at 14:21
|
show 4 more comments
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 at 14:04
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 at 14:15
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 at 14:17
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 at 14:18
I believe this is not quite right as it does not deal with the empty sequence, it may need to be slightly adjusted but thanks for the comment.
– 3684
Nov 21 at 14:21
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 at 14:04
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 at 14:04
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 at 14:15
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 at 14:15
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 at 14:17
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 at 14:17
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 at 14:18
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 at 14:18
I believe this is not quite right as it does not deal with the empty sequence, it may need to be slightly adjusted but thanks for the comment.
– 3684
Nov 21 at 14:21
I believe this is not quite right as it does not deal with the empty sequence, it may need to be slightly adjusted but thanks for the comment.
– 3684
Nov 21 at 14:21
|
show 4 more comments
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I think there's a small mistake. You say the sequence consisting only of $n+2$ is alternating, but this is not true unless $n$ is odd, because one of the conditions is that an alternating sequence start with an odd term. Since you eventually compute $A(20),$ you need $n$ to be even. Your general approach looks fine, but you should review the details.
– saulspatz
Nov 21 at 13:51
I believe I have avoided this by stating for every even n, or is this not what you mean.
– 3684
Nov 21 at 14:12