Number of increasing sequences alternating between odd and even involving only integers from the set $[1, 2,...
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 '18 at 13:18
3684
1277
add a comment |
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 '18 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 '18 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 '18 at 14:12
add a comment |
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 '18 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 '18 at 13:18
3684
1277
asked Nov 21 '18 at 13:18
3684
1277
asked Nov 21 '18 at 13:18
3684
1277
asked Nov 21 '18 at 13:18
3684
1277
asked Nov 21 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 14:12
add a comment |
1 Answer
1
active
oldest
votes
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 '18 at 13:48
Empy2
33.5k12261
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 '18 at 14:04
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 '18 at 14:15
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 '18 at 14:17
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 '18 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 '18 at 14:21
|
show 4 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007719%2fnumber-of-increasing-sequences-alternating-between-odd-and-even-involving-only-i%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 '18 at 13:48
Empy2
33.5k12261
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 '18 at 14:04
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 '18 at 14:15
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 '18 at 14:17
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 '18 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 '18 at 14:21
|
show 4 more comments
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 '18 at 13:48
Empy2
33.5k12261
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 '18 at 14:04
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 '18 at 14:15
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 '18 at 14:17
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 '18 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 '18 at 14:21
|
show 4 more comments
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 '18 at 13:48
Empy2
33.5k12261
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 '18 at 13:48
Empy2
33.5k12261
answered Nov 21 '18 at 13:48
Empy2
33.5k12261
answered Nov 21 '18 at 13:48
Empy2
33.5k12261
answered Nov 21 '18 at 13:48
Empy2
33.5k12261
33.5k12261
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 '18 at 14:04
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 '18 at 14:15
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 '18 at 14:17
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 '18 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 '18 at 14:21
|
show 4 more comments
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 '18 at 14:04
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 '18 at 14:15
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 '18 at 14:17
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 '18 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 '18 at 14:21
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 '18 at 14:04
And you need to start things off with $A(0)=1, A(1)=2$.
– TonyK
Nov 21 '18 at 14:04
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 '18 at 14:15
@Empy2 How were you able to come up with that recursion, have you done similar questions before?
– 3684
Nov 21 '18 at 14:15
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 '18 at 14:17
I recognized 6765 and 17711, and looked for that recursion.
– Empy2
Nov 21 '18 at 14:17
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 '18 at 14:18
If I had not provided those values do you think you could have got there?
– 3684
Nov 21 '18 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 '18 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 '18 at 14:21
|
show 4 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007719%2fnumber-of-increasing-sequences-alternating-between-odd-and-even-involving-only-i%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 '18 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 '18 at 14:12