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Recursion: computing $f(3^100)$


Recursion function question againcomputer theory recursionMaths/Programming recursion questionA Problem about RecursionHow to solve recursion?Lin Alg 100-Level Recursion ProblemSolve this recursionA positive integer is equal to the sum of digits of a multiple of itself.Recursion RelationsKilling Flies with a Checkerboard Flyswatter













0












$begingroup$


Given the function $f:N^+ rightarrow N$ such that: $f(2^n)=n^2$, $f(3n)=f(2n)+5$, Find the value of $f(3^100)$:



I tried calculating some smaller $f(3^k)$ for integer $k$, to look for a pattern, If I haven't done silly mistakes I got:$$f(3)=6,f(3^2)=9,f(3^3)=24,f(3^4)=36$$
I noticed that $f(3)cdot4=f(3^3)$ and $f(3^2)cdot4=f(3^4)$, nonetheless, I don't think this kind of pattern holds, I wrote in general that: $$f(3cdot2^n)=n^2+2n+6$$
Hints or partial help is much appreciated!






contest-math recursion







share|cite|improve this question









$endgroup$











  • $begingroup$
    Let me know if I made some silly mistake!
    $endgroup$
    – Spasoje Durovic
    Mar 20 at 19:55















0












$begingroup$


Given the function $f:N^+ rightarrow N$ such that: $f(2^n)=n^2$, $f(3n)=f(2n)+5$, Find the value of $f(3^100)$:



I tried calculating some smaller $f(3^k)$ for integer $k$, to look for a pattern, If I haven't done silly mistakes I got:$$f(3)=6,f(3^2)=9,f(3^3)=24,f(3^4)=36$$
I noticed that $f(3)cdot4=f(3^3)$ and $f(3^2)cdot4=f(3^4)$, nonetheless, I don't think this kind of pattern holds, I wrote in general that: $$f(3cdot2^n)=n^2+2n+6$$
Hints or partial help is much appreciated!






contest-math recursion







share|cite|improve this question









$endgroup$











  • $begingroup$
    Let me know if I made some silly mistake!
    $endgroup$
    – Spasoje Durovic
    Mar 20 at 19:55













0












0








0





$begingroup$


Given the function $f:N^+ rightarrow N$ such that: $f(2^n)=n^2$, $f(3n)=f(2n)+5$, Find the value of $f(3^100)$:



I tried calculating some smaller $f(3^k)$ for integer $k$, to look for a pattern, If I haven't done silly mistakes I got:$$f(3)=6,f(3^2)=9,f(3^3)=24,f(3^4)=36$$
I noticed that $f(3)cdot4=f(3^3)$ and $f(3^2)cdot4=f(3^4)$, nonetheless, I don't think this kind of pattern holds, I wrote in general that: $$f(3cdot2^n)=n^2+2n+6$$
Hints or partial help is much appreciated!






contest-math recursion







share|cite|improve this question









$endgroup$




Given the function $f:N^+ rightarrow N$ such that: $f(2^n)=n^2$, $f(3n)=f(2n)+5$, Find the value of $f(3^100)$:



I tried calculating some smaller $f(3^k)$ for integer $k$, to look for a pattern, If I haven't done silly mistakes I got:$$f(3)=6,f(3^2)=9,f(3^3)=24,f(3^4)=36$$
I noticed that $f(3)cdot4=f(3^3)$ and $f(3^2)cdot4=f(3^4)$, nonetheless, I don't think this kind of pattern holds, I wrote in general that: $$f(3cdot2^n)=n^2+2n+6$$
Hints or partial help is much appreciated!






contest-math recursion




contest-math recursion






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 20 at 19:54









Spasoje DurovicSpasoje Durovic

40811




40811











  • $begingroup$
    Let me know if I made some silly mistake!
    $endgroup$
    – Spasoje Durovic
    Mar 20 at 19:55
















  • $begingroup$
    Let me know if I made some silly mistake!
    $endgroup$
    – Spasoje Durovic
    Mar 20 at 19:55















$begingroup$
Let me know if I made some silly mistake!
$endgroup$
– Spasoje Durovic
Mar 20 at 19:55




$begingroup$
Let me know if I made some silly mistake!
$endgroup$
– Spasoje Durovic
Mar 20 at 19:55










1 Answer
1






active

oldest

votes


















2












$begingroup$

You can easily show by induction that for all $n in lbrace 0, ..., 100 rbrace$,
$$f(3^100)=f(2^n times 3^100-n) + 5n$$



Using that with $n=100$, you get
$$f(3^100) = f(2^100) + 500 = 100^2+ 500 = 10500$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    $f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
    $endgroup$
    – TheSilverDoe
    Mar 20 at 20:10










  • $begingroup$
    Right! I had deleted my comment because I was trying exactly that! great, thanks
    $endgroup$
    – Spasoje Durovic
    Mar 20 at 20:11











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

You can easily show by induction that for all $n in lbrace 0, ..., 100 rbrace$,
$$f(3^100)=f(2^n times 3^100-n) + 5n$$



Using that with $n=100$, you get
$$f(3^100) = f(2^100) + 500 = 100^2+ 500 = 10500$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    $f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
    $endgroup$
    – TheSilverDoe
    Mar 20 at 20:10










  • $begingroup$
    Right! I had deleted my comment because I was trying exactly that! great, thanks
    $endgroup$
    – Spasoje Durovic
    Mar 20 at 20:11















2












$begingroup$

You can easily show by induction that for all $n in lbrace 0, ..., 100 rbrace$,
$$f(3^100)=f(2^n times 3^100-n) + 5n$$



Using that with $n=100$, you get
$$f(3^100) = f(2^100) + 500 = 100^2+ 500 = 10500$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    $f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
    $endgroup$
    – TheSilverDoe
    Mar 20 at 20:10










  • $begingroup$
    Right! I had deleted my comment because I was trying exactly that! great, thanks
    $endgroup$
    – Spasoje Durovic
    Mar 20 at 20:11













2












2








2





$begingroup$

You can easily show by induction that for all $n in lbrace 0, ..., 100 rbrace$,
$$f(3^100)=f(2^n times 3^100-n) + 5n$$



Using that with $n=100$, you get
$$f(3^100) = f(2^100) + 500 = 100^2+ 500 = 10500$$






share|cite|improve this answer









$endgroup$



You can easily show by induction that for all $n in lbrace 0, ..., 100 rbrace$,
$$f(3^100)=f(2^n times 3^100-n) + 5n$$



Using that with $n=100$, you get
$$f(3^100) = f(2^100) + 500 = 100^2+ 500 = 10500$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 20 at 20:02









TheSilverDoeTheSilverDoe

5,066215




5,066215











  • $begingroup$
    $f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
    $endgroup$
    – TheSilverDoe
    Mar 20 at 20:10










  • $begingroup$
    Right! I had deleted my comment because I was trying exactly that! great, thanks
    $endgroup$
    – Spasoje Durovic
    Mar 20 at 20:11
















  • $begingroup$
    $f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
    $endgroup$
    – TheSilverDoe
    Mar 20 at 20:10










  • $begingroup$
    Right! I had deleted my comment because I was trying exactly that! great, thanks
    $endgroup$
    – Spasoje Durovic
    Mar 20 at 20:11















$begingroup$
$f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
$endgroup$
– TheSilverDoe
Mar 20 at 20:10




$begingroup$
$f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
$endgroup$
– TheSilverDoe
Mar 20 at 20:10












$begingroup$
Right! I had deleted my comment because I was trying exactly that! great, thanks
$endgroup$
– Spasoje Durovic
Mar 20 at 20:11




$begingroup$
Right! I had deleted my comment because I was trying exactly that! great, thanks
$endgroup$
– Spasoje Durovic
Mar 20 at 20:11

















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