How many ways can a number be written as a sum of two non negative integers?
$begingroup$
How many ways can a number be written as a sum of two non negative integers?
For example there is $4$ way for $7$. $ 7=0+7=1+6=2+5=3+4$
I think there is $[ frac{N}{2}]+1$ way for number$N$. Is it true?
combinatorics combinations
asked Dec 11 '18 at 5:42
ilenilen
226
$endgroup$
add a comment |
$begingroup$
How many ways can a number be written as a sum of two non negative integers?
For example there is $4$ way for $7$. $ 7=0+7=1+6=2+5=3+4$
I think there is $[ frac{N}{2}]+1$ way for number$N$. Is it true?
combinatorics combinations
asked Dec 11 '18 at 5:42
ilenilen
226
$endgroup$
1
$begingroup$
Your example shows your rule does not work. Also $0$ is not positive so $0+7$ should not count.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:46
$begingroup$
@Ross Millikan, yes. I forgot to divide by 2. And I use nonnegative numbers. Thanks for your note.
$endgroup$
– ilen
Dec 11 '18 at 5:58
add a comment |
$begingroup$
How many ways can a number be written as a sum of two non negative integers?
For example there is $4$ way for $7$. $ 7=0+7=1+6=2+5=3+4$
I think there is $[ frac{N}{2}]+1$ way for number$N$. Is it true?
combinatorics combinations
asked Dec 11 '18 at 5:42
ilenilen
226
$endgroup$
How many ways can a number be written as a sum of two non negative integers?
For example there is $4$ way for $7$. $ 7=0+7=1+6=2+5=3+4$
I think there is $[ frac{N}{2}]+1$ way for number$N$. Is it true?
combinatorics combinations
combinatorics combinations
asked Dec 11 '18 at 5:42
ilenilen
226
asked Dec 11 '18 at 5:42
ilenilen
226
edited Dec 11 '18 at 6:06
ilen
asked Dec 11 '18 at 5:42
ilenilen
226
asked Dec 11 '18 at 5:42
ilenilen
226
asked Dec 11 '18 at 5:42
ilenilen
226
226
1
$begingroup$
Your example shows your rule does not work. Also $0$ is not positive so $0+7$ should not count.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:46
$begingroup$
@Ross Millikan, yes. I forgot to divide by 2. And I use nonnegative numbers. Thanks for your note.
$endgroup$
– ilen
Dec 11 '18 at 5:58
add a comment |
1
$begingroup$
Your example shows your rule does not work. Also $0$ is not positive so $0+7$ should not count.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:46
$begingroup$
@Ross Millikan, yes. I forgot to divide by 2. And I use nonnegative numbers. Thanks for your note.
$endgroup$
– ilen
Dec 11 '18 at 5:58
1
1
$begingroup$
Your example shows your rule does not work. Also $0$ is not positive so $0+7$ should not count.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:46
$begingroup$
Your example shows your rule does not work. Also $0$ is not positive so $0+7$ should not count.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:46
$begingroup$
@Ross Millikan, yes. I forgot to divide by 2. And I use nonnegative numbers. Thanks for your note.
$endgroup$
– ilen
Dec 11 '18 at 5:58
$begingroup$
@Ross Millikan, yes. I forgot to divide by 2. And I use nonnegative numbers. Thanks for your note.
$endgroup$
– ilen
Dec 11 '18 at 5:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you consider 3+4 and 4+3 as two different ways then yes it will be N+1. Think of it as placing a partition in a row of N objects you can place it right in the beginning, right at the end and any of the N-1 locations. However if 3+4 and 4+3 are considered same then we have only 4 ways of writing 7 as sum of two numbers. In this case the answer will be ceil((N+1)/2) where ceil(x) is smallest integer greater is Han or equal to x.
answered Dec 11 '18 at 5:52
CuriousCurious
889516
$endgroup$
1
$begingroup$
No, because we are talking of positive integers. $0$ is not allowed. You have a typo with Han at the end.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:55
1
$begingroup$
Yes I have requested an edit in question changing positive to negative. As seen in example 0+7 is seen as a valid breakup
$endgroup$
– Curious
Dec 11 '18 at 5:57
$begingroup$
Thanks for correcting that was a typo I will edit
$endgroup$
– Curious
Dec 11 '18 at 6:09
add a comment |
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%2f3034937%2fhow-many-ways-can-a-number-be-written-as-a-sum-of-two-non-negative-integers%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
$begingroup$
If you consider 3+4 and 4+3 as two different ways then yes it will be N+1. Think of it as placing a partition in a row of N objects you can place it right in the beginning, right at the end and any of the N-1 locations. However if 3+4 and 4+3 are considered same then we have only 4 ways of writing 7 as sum of two numbers. In this case the answer will be ceil((N+1)/2) where ceil(x) is smallest integer greater is Han or equal to x.
answered Dec 11 '18 at 5:52
CuriousCurious
889516
$endgroup$
1
$begingroup$
No, because we are talking of positive integers. $0$ is not allowed. You have a typo with Han at the end.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:55
1
$begingroup$
Yes I have requested an edit in question changing positive to negative. As seen in example 0+7 is seen as a valid breakup
$endgroup$
– Curious
Dec 11 '18 at 5:57
$begingroup$
Thanks for correcting that was a typo I will edit
$endgroup$
– Curious
Dec 11 '18 at 6:09
add a comment |
$begingroup$
If you consider 3+4 and 4+3 as two different ways then yes it will be N+1. Think of it as placing a partition in a row of N objects you can place it right in the beginning, right at the end and any of the N-1 locations. However if 3+4 and 4+3 are considered same then we have only 4 ways of writing 7 as sum of two numbers. In this case the answer will be ceil((N+1)/2) where ceil(x) is smallest integer greater is Han or equal to x.
answered Dec 11 '18 at 5:52
CuriousCurious
889516
$endgroup$
1
$begingroup$
No, because we are talking of positive integers. $0$ is not allowed. You have a typo with Han at the end.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:55
1
$begingroup$
Yes I have requested an edit in question changing positive to negative. As seen in example 0+7 is seen as a valid breakup
$endgroup$
– Curious
Dec 11 '18 at 5:57
$begingroup$
Thanks for correcting that was a typo I will edit
$endgroup$
– Curious
Dec 11 '18 at 6:09
add a comment |
$begingroup$
If you consider 3+4 and 4+3 as two different ways then yes it will be N+1. Think of it as placing a partition in a row of N objects you can place it right in the beginning, right at the end and any of the N-1 locations. However if 3+4 and 4+3 are considered same then we have only 4 ways of writing 7 as sum of two numbers. In this case the answer will be ceil((N+1)/2) where ceil(x) is smallest integer greater is Han or equal to x.
answered Dec 11 '18 at 5:52
CuriousCurious
889516
$endgroup$
If you consider 3+4 and 4+3 as two different ways then yes it will be N+1. Think of it as placing a partition in a row of N objects you can place it right in the beginning, right at the end and any of the N-1 locations. However if 3+4 and 4+3 are considered same then we have only 4 ways of writing 7 as sum of two numbers. In this case the answer will be ceil((N+1)/2) where ceil(x) is smallest integer greater is Han or equal to x.
answered Dec 11 '18 at 5:52
CuriousCurious
889516
edited Dec 11 '18 at 6:09
answered Dec 11 '18 at 5:52
CuriousCurious
889516
answered Dec 11 '18 at 5:52
CuriousCurious
889516
answered Dec 11 '18 at 5:52
CuriousCurious
889516
889516
1
$begingroup$
No, because we are talking of positive integers. $0$ is not allowed. You have a typo with Han at the end.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:55
1
$begingroup$
Yes I have requested an edit in question changing positive to negative. As seen in example 0+7 is seen as a valid breakup
$endgroup$
– Curious
Dec 11 '18 at 5:57
$begingroup$
Thanks for correcting that was a typo I will edit
$endgroup$
– Curious
Dec 11 '18 at 6:09
add a comment |
1
$begingroup$
No, because we are talking of positive integers. $0$ is not allowed. You have a typo with Han at the end.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:55
1
$begingroup$
Yes I have requested an edit in question changing positive to negative. As seen in example 0+7 is seen as a valid breakup
$endgroup$
– Curious
Dec 11 '18 at 5:57
$begingroup$
Thanks for correcting that was a typo I will edit
$endgroup$
– Curious
Dec 11 '18 at 6:09
1
1
$begingroup$
No, because we are talking of positive integers. $0$ is not allowed. You have a typo with Han at the end.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:55
$begingroup$
No, because we are talking of positive integers. $0$ is not allowed. You have a typo with Han at the end.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:55
1
1
$begingroup$
Yes I have requested an edit in question changing positive to negative. As seen in example 0+7 is seen as a valid breakup
$endgroup$
– Curious
Dec 11 '18 at 5:57
$begingroup$
Yes I have requested an edit in question changing positive to negative. As seen in example 0+7 is seen as a valid breakup
$endgroup$
– Curious
Dec 11 '18 at 5:57
$begingroup$
Thanks for correcting that was a typo I will edit
$endgroup$
– Curious
Dec 11 '18 at 6:09
$begingroup$
Thanks for correcting that was a typo I will edit
$endgroup$
– Curious
Dec 11 '18 at 6:09
add a comment |
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.
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%2f3034937%2fhow-many-ways-can-a-number-be-written-as-a-sum-of-two-non-negative-integers%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
1
$begingroup$
Your example shows your rule does not work. Also $0$ is not positive so $0+7$ should not count.
$endgroup$
– Ross Millikan
Dec 11 '18 at 5:46
$begingroup$
@Ross Millikan, yes. I forgot to divide by 2. And I use nonnegative numbers. Thanks for your note.
$endgroup$
– ilen
Dec 11 '18 at 5:58