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Find all functions, $f:\mathbb{N} \to \mathbb{N}$, for which $f(1) = 1, f(2n) < 6f(n)$, and

$$3f(n)f(2n+1) = f(2n)(3f(n)+1).$$

My first approach is to try to play around and set values equal to 0, to test for even-ness, that kind of thing, but this gets me absolutely nowhere. I've looked for help on problem solving forums and it hasn't really got anywhere, so I'm thinking that maybe this is a problem of another type, not necessarily hard, but covering topics in functional equations that I have not yet covered.

Was wondering if someone could give the solution and thought-process behind steps, as this is really driving me nuts.

Thanks.

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For $n=1$, $3 f(3) = 4 f(2)$ and $f(2) < 6$. Since $f(2)$ must be divisible by $3$, $f(2) = 3$ and $f(3) = 4$.

For $n=2$, $9 f(5) = 10 f(4)$ and $f(4) < 18$. Since $f(4)$ must be divisible by $9$, $f(4) = 9$ and $f(5) = 10$.

See a possible pattern starting?

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  • $\begingroup$ Letting n = 1, wouldn't we get, $3f(1)2f(3) = f(2)(3f(1) + 1))$? How did you get 3f(3) = 4f(2), and make the claim that f(2) < 6? We get f(2) < 6f(1). $\endgroup$ – user164403 Dec 28 '14 at 4:59
  • $\begingroup$ For $n=1$, it's $3 f(1) f(2\cdot 1 + 1) = f(2\cdot 1)(3 f(1)+1)$ and $f(1) = 1$ so it's $3 f(3) = 4 f(2)$. Yes, $f(2) < 6 f(1)$, but $f(1) = 1$. $\endgroup$ – Robert Israel Dec 28 '14 at 6:36
  • $\begingroup$ Full solution can be seen here: www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=618989 $\endgroup$ – Tintarn Dec 30 '14 at 14:26

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