An IMO inspired problem
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
1,706615
add a comment |
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
1,706615
add a comment |
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
1,706615
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
1,706615
asked Mar 16 '14 at 20:15
Sandeep ThilakanSandeep Thilakan
1,706615
asked Mar 16 '14 at 20:15
Sandeep ThilakanSandeep Thilakan
1,706615
asked Mar 16 '14 at 20:15
Sandeep ThilakanSandeep Thilakan
1,706615
asked Mar 16 '14 at 20:15
Sandeep ThilakanSandeep Thilakan
1,706615
1,706615
add a comment |
add a comment |
1 Answer
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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
319k23208458
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
319k23208458
add a comment |
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
319k23208458
add a comment |
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
319k23208458
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
319k23208458
edited Mar 17 '14 at 2:16
answered Mar 17 '14 at 1:26
Robert IsraelRobert Israel
319k23208458
answered Mar 17 '14 at 1:26
Robert IsraelRobert Israel
319k23208458
answered Mar 17 '14 at 1:26
Robert IsraelRobert Israel
319k23208458
319k23208458
add a comment |
add a comment |
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