Prove that function on naturals defined recursively is idempotent on odd numbers
$begingroup$
Consider the function $f$ on natural numbers defined by the following recursion:
- $f(1)=1$
- $f(3)=3$
- $f(2n)=f(n)$
- $f(4n+1)=2f(2n+1)-f(n)$
- $f(4n+3)=3f(2n+1)-2f(n)$
Numerical evidence shows that for odd $k$ we have $f(f(k))=k$, but I have no clue on how to prove it. Any ideas?
recurrence-relations recursion
asked Dec 19 '18 at 18:37
A. BellmuntA. Bellmunt
895515
$endgroup$
add a comment |
$begingroup$
Consider the function $f$ on natural numbers defined by the following recursion:
- $f(1)=1$
- $f(3)=3$
- $f(2n)=f(n)$
- $f(4n+1)=2f(2n+1)-f(n)$
- $f(4n+3)=3f(2n+1)-2f(n)$
Numerical evidence shows that for odd $k$ we have $f(f(k))=k$, but I have no clue on how to prove it. Any ideas?
recurrence-relations recursion
asked Dec 19 '18 at 18:37
A. BellmuntA. Bellmunt
895515
$endgroup$
add a comment |
$begingroup$
Consider the function $f$ on natural numbers defined by the following recursion:
- $f(1)=1$
- $f(3)=3$
- $f(2n)=f(n)$
- $f(4n+1)=2f(2n+1)-f(n)$
- $f(4n+3)=3f(2n+1)-2f(n)$
Numerical evidence shows that for odd $k$ we have $f(f(k))=k$, but I have no clue on how to prove it. Any ideas?
recurrence-relations recursion
asked Dec 19 '18 at 18:37
A. BellmuntA. Bellmunt
895515
$endgroup$
Consider the function $f$ on natural numbers defined by the following recursion:
- $f(1)=1$
- $f(3)=3$
- $f(2n)=f(n)$
- $f(4n+1)=2f(2n+1)-f(n)$
- $f(4n+3)=3f(2n+1)-2f(n)$
Numerical evidence shows that for odd $k$ we have $f(f(k))=k$, but I have no clue on how to prove it. Any ideas?
recurrence-relations recursion
recurrence-relations recursion
asked Dec 19 '18 at 18:37
A. BellmuntA. Bellmunt
895515
asked Dec 19 '18 at 18:37
A. BellmuntA. Bellmunt
895515
asked Dec 19 '18 at 18:37
A. BellmuntA. Bellmunt
895515
asked Dec 19 '18 at 18:37
A. BellmuntA. Bellmunt
895515
asked Dec 19 '18 at 18:37
A. BellmuntA. Bellmunt
895515
895515
add a comment |
add a comment |
1 Answer
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Hint: Consider a binary expansion of natural numbers. Take a detailed solution here https://artofproblemsolving.com/community/c6h60400p365112
answered Dec 19 '18 at 18:41
Maria MazurMaria Mazur
49.6k1361124
$endgroup$
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– A. Bellmunt
Dec 19 '18 at 18:54
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Consider a binary expansion of natural numbers. Take a detailed solution here https://artofproblemsolving.com/community/c6h60400p365112
answered Dec 19 '18 at 18:41
Maria MazurMaria Mazur
49.6k1361124
$endgroup$
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– A. Bellmunt
Dec 19 '18 at 18:54
add a comment |
$begingroup$
Hint: Consider a binary expansion of natural numbers. Take a detailed solution here https://artofproblemsolving.com/community/c6h60400p365112
answered Dec 19 '18 at 18:41
Maria MazurMaria Mazur
49.6k1361124
$endgroup$
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– A. Bellmunt
Dec 19 '18 at 18:54
add a comment |
$begingroup$
Hint: Consider a binary expansion of natural numbers. Take a detailed solution here https://artofproblemsolving.com/community/c6h60400p365112
answered Dec 19 '18 at 18:41
Maria MazurMaria Mazur
49.6k1361124
$endgroup$
Hint: Consider a binary expansion of natural numbers. Take a detailed solution here https://artofproblemsolving.com/community/c6h60400p365112
answered Dec 19 '18 at 18:41
Maria MazurMaria Mazur
49.6k1361124
edited Dec 19 '18 at 18:51
answered Dec 19 '18 at 18:41
Maria MazurMaria Mazur
49.6k1361124
answered Dec 19 '18 at 18:41
Maria MazurMaria Mazur
49.6k1361124
answered Dec 19 '18 at 18:41
Maria MazurMaria Mazur
49.6k1361124
49.6k1361124
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– A. Bellmunt
Dec 19 '18 at 18:54
add a comment |
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– A. Bellmunt
Dec 19 '18 at 18:54
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– A. Bellmunt
Dec 19 '18 at 18:54
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– A. Bellmunt
Dec 19 '18 at 18:54
add a comment |
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