An IMO inspired problem
$begingroup$
This problem from IMO 1988 is said to be one of the most elegant ones in functional equations.
Problem : The function $f$ is defined on the set of all positive integers as follows:
begin{align}
f(1) = 1, f(3) &= 3, f(2n) = f(n), \
f(4n+1) &= 2f(2n+1) - f(n), \
f(4n+3) &= 3 f(2n+1) - 2f(n)
end{align}
Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$.
The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$.
My question is the following: How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a general base $k$?
Thanks.
contest-math functional-equations
asked Mar 16 '14 at 20:15
Sandeep ThilakanSandeep Thilakan
1,721715
$endgroup$
add a comment |
$begingroup$
This problem from IMO 1988 is said to be one of the most elegant ones in functional equations.
Problem : The function $f$ is defined on the set of all positive integers as follows:
begin{align}
f(1) = 1, f(3) &= 3, f(2n) = f(n), \
f(4n+1) &= 2f(2n+1) - f(n), \
f(4n+3) &= 3 f(2n+1) - 2f(n)
end{align}
Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$.
The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$.
My question is the following: How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a general base $k$?
Thanks.
contest-math functional-equations
asked Mar 16 '14 at 20:15
Sandeep ThilakanSandeep Thilakan
1,721715
$endgroup$
add a comment |
$begingroup$
This problem from IMO 1988 is said to be one of the most elegant ones in functional equations.
Problem : The function $f$ is defined on the set of all positive integers as follows:
begin{align}
f(1) = 1, f(3) &= 3, f(2n) = f(n), \
f(4n+1) &= 2f(2n+1) - f(n), \
f(4n+3) &= 3 f(2n+1) - 2f(n)
end{align}
Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$.
The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$.
My question is the following: How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a general base $k$?
Thanks.
contest-math functional-equations
asked Mar 16 '14 at 20:15
Sandeep ThilakanSandeep Thilakan
1,721715
$endgroup$
This problem from IMO 1988 is said to be one of the most elegant ones in functional equations.
Problem : The function $f$ is defined on the set of all positive integers as follows:
begin{align}
f(1) = 1, f(3) &= 3, f(2n) = f(n), \
f(4n+1) &= 2f(2n+1) - f(n), \
f(4n+3) &= 3 f(2n+1) - 2f(n)
end{align}
Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$.
The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$.
My question is the following: How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a general base $k$?
Thanks.
contest-math functional-equations
contest-math functional-equations
asked Mar 16 '14 at 20:15
Sandeep ThilakanSandeep Thilakan
1,721715
asked Mar 16 '14 at 20:15
Sandeep ThilakanSandeep Thilakan
1,721715
asked Mar 16 '14 at 20:15
Sandeep ThilakanSandeep Thilakan
1,721715
asked Mar 16 '14 at 20:15
Sandeep ThilakanSandeep Thilakan
1,721715
asked Mar 16 '14 at 20:15
Sandeep ThilakanSandeep Thilakan
1,721715
1,721715
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In general for base $k$:
$$ eqalign{f(kn ) &= f(n)cr
f(k^2 n + a k + b) &= k f(kn+b) + a f(kn+1) - (k - a -1) f(n)cr}$$
for $a = 0, ldots, k-1$, $b = 1, ldots, k-1$.
answered Mar 17 '14 at 1:26
Robert IsraelRobert Israel
330k23219473
$endgroup$
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%2f714652%2fan-imo-inspired-problem%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
$begingroup$
In general for base $k$:
$$ eqalign{f(kn ) &= f(n)cr
f(k^2 n + a k + b) &= k f(kn+b) + a f(kn+1) - (k - a -1) f(n)cr}$$
for $a = 0, ldots, k-1$, $b = 1, ldots, k-1$.
answered Mar 17 '14 at 1:26
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
In general for base $k$:
$$ eqalign{f(kn ) &= f(n)cr
f(k^2 n + a k + b) &= k f(kn+b) + a f(kn+1) - (k - a -1) f(n)cr}$$
for $a = 0, ldots, k-1$, $b = 1, ldots, k-1$.
answered Mar 17 '14 at 1:26
Robert IsraelRobert Israel
330k23219473
$endgroup$
add a comment |
$begingroup$
In general for base $k$:
$$ eqalign{f(kn ) &= f(n)cr
f(k^2 n + a k + b) &= k f(kn+b) + a f(kn+1) - (k - a -1) f(n)cr}$$
for $a = 0, ldots, k-1$, $b = 1, ldots, k-1$.
answered Mar 17 '14 at 1:26
Robert IsraelRobert Israel
330k23219473
$endgroup$
In general for base $k$:
$$ eqalign{f(kn ) &= f(n)cr
f(k^2 n + a k + b) &= k f(kn+b) + a f(kn+1) - (k - a -1) f(n)cr}$$
for $a = 0, ldots, k-1$, $b = 1, ldots, k-1$.
answered Mar 17 '14 at 1:26
Robert IsraelRobert Israel
330k23219473
edited Mar 17 '14 at 2:16
answered Mar 17 '14 at 1:26
Robert IsraelRobert Israel
330k23219473
answered Mar 17 '14 at 1:26
Robert IsraelRobert Israel
330k23219473
answered Mar 17 '14 at 1:26
Robert IsraelRobert Israel
330k23219473
330k23219473
add a comment |
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%2f714652%2fan-imo-inspired-problem%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
