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This is related to problem in graph theory.

OEIS defines A033485 as

$a(1)=1$ and $a(n)=a(n-1)+a(\lfloor n/2 \rfloor)$.

Q1 what are upper bounds and asymptotics for $a(n)$, can we get $\exp(o(n))$?

For real $A,0 < A < 1$ define:

$b(0)=b(1)=1$ and $b(n)=b(n-1)+b(\lfloor A n\rfloor)$

If $A \le \frac12$ then $b(n) \le a(n)$.

Q2 what are upper bounds and asymptotics for $b(n)$?

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  • $\begingroup$ Numerical evidence appears to suggest $\log{a(n)}=C \log^2{n}$ $\endgroup$
    – joro
    Commented Aug 8, 2021 at 17:31

5 Answers 5

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This recurrence was the subject of one of Donald Knuth's earliest papers! See https://tensen.net/research/static/logos/number-theory/partitions-restricted/knuth.pdf .

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  • $\begingroup$ Many thanks. Did you notice that Emil solved generalization of a(n)? Also there is graph theoretic problem related to the sequence. $\endgroup$
    – joro
    Commented Aug 16, 2021 at 14:11
  • $\begingroup$ Very interesting: I was always looking for a really elementary argument for the intermediate growth of the partition function. This gives an elementary lower bound of intermediate growth! $\endgroup$ Commented Aug 17, 2021 at 21:09
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Sure, for each fixed $k$ both of them are (up to a constant) eventually dominated by $c(n)=c(n-1)+c(n-k)$ and this sequence is $O(t^n)$ for $t=t(k)\to 1$ as $k\to \infty$ (one has to prove a simple estimate for the roots of $t^k=t^{k-1}+1$).

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  • $\begingroup$ Thanks, but I am not sure this is correct. t^n appears to be disproved by experimental evidence. For n=10^3, a(n)<n^4 and this can't be exponential. $\endgroup$
    – joro
    Commented Aug 8, 2021 at 12:43
  • $\begingroup$ @joro did you see the word dominated in my answer? Sure, sequence $n^4$ is dominated by $t^n$ for every $t>1$, I did not claim any lower bounds (though it’s easy to see that your bound $n^4$ or any polynomial bound really can not be true but it’s unrelated to the matter at hand). $\endgroup$ Commented Aug 8, 2021 at 12:57
  • $\begingroup$ @joro if it’s not clear, I proved the requested $exp(o(n))$ upper bound, nothing more, nothing less. $\endgroup$ Commented Aug 8, 2021 at 13:30
  • $\begingroup$ Thanks. It is not clear to me still. $\endgroup$
    – joro
    Commented Aug 8, 2021 at 13:31
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    $\begingroup$ @joro If $a(n) \in O(t^n)$, then it is also $o(s^n)$ for any $s > t$. Therefore, if $a(n) \in O(t^n)$ for all $t > 1$, it is also $o(t^n)$ for all $t > 1$. $\endgroup$
    – user44191
    Commented Aug 8, 2021 at 15:28
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Set $a_k(1)=r_k$ and $a_k(n)=a_k(n-1)+2^k a_k(\lfloor n/2\rfloor)$ for $n\geq 2$ where $r_0=1$ and $r_k=a_{k-1}(2)=(1+2^{k-1})r_{k-1}\leq 2^kr_{k-1}$.

We have $a_0(n)=a(n)$ for $a(n)$ defined by the OP.

An easy induction shows $a_k(2n),a_k(2n+1)\leq a_{k+1}(n)$.

We get thus $$a(n)=a_0(n)\leq a_1(\lfloor n/2\rfloor)\leq a_2(\lfloor n/4\rfloor)\leq \ldots\leq a_{\lfloor \log_2 n\rfloor}(1)=r_{\lfloor \log_2(n)\rfloor}\ .$$

Using $r_k\leq 2^k r_{k-1}\leq 2^{k+(k-1)}r_{k-2}\leq\ldots$, we get finally $$a(n)\leq 2^{1+2+\ldots+\lfloor\log_2(n)\rfloor}\sim \sqrt{2}^{(\log_2 n)^2}$$ as observed by the OP.

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  • $\begingroup$ Many thanks :)) $\endgroup$
    – joro
    Commented Aug 11, 2021 at 15:34
  • $\begingroup$ It is true that $1+2+\ldots+\lfloor\log_2(n)\rfloor$ is asymptotic to $(\log_2 n)^2/2$, but it is not true that $2^{1+2+\ldots+\lfloor\log_2(n)\rfloor}$ is asymptotic to $2^{(\log_2 n)^2/2}$. Exponentiation is sensitive. $\endgroup$
    – GH from MO
    Commented Aug 11, 2021 at 17:05
  • $\begingroup$ I agree with GH, it gives however the correct logarithmic asymptotic for the lower bound. Moreover, my upper bounds $r_k\leq 2^kr_{k-1}$ are rather savage ($r_k\sim 2^{k-1}r_{k-1}$ asymptotically) and overcompensate asymptotically the previous error. $\endgroup$ Commented Aug 11, 2021 at 19:32
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Confirming the numerical evidence, the $b(n)$ sequences have rate of growth $$b(n)=n^{\Theta(\log n)}.$$

On the one hand, since $b(n)$ is nondecreasing, we have $$\begin{align}b(n)&=b(\lfloor An\rfloor)+b(n-1)\\ &\le2b(\lfloor An\rfloor)+b(n-2)\\ &\le\cdots\\ &\le(n-1)b(\lfloor An\rfloor)+b(1)\\ &\le nb(\lfloor An\rfloor). \end{align}$$ Since it takes at most $(\log n)/\log A^{-1}$ iterations of $n\mapsto\lfloor An\rfloor$ to get from $n$ down to $1$, we obtain $$b(n)\le n^{C\log n},\qquad C=(\log A^{-1})^{-1}.$$

On the other hand, we have, say, $$b(n)\ge\frac n2b(\lfloor An/2\rfloor)$$ by the same argument, which easily yields $$b(n)=\Omega(n^{c\log n})$$ for any $c<\frac12\log\frac2A$.

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  • $\begingroup$ Many thanks. Would you please elaborate the transition to b(n)<=n^(C log(n))? Are you using this answer: mathoverflow.net/questions/401203/… $\endgroup$
    – joro
    Commented Aug 11, 2021 at 15:33
  • $\begingroup$ Which part of it is unclear? The final step just uses that it takes $(\log n)/\log A^{-1}$ iterations of $b(n)\le n b(An)$ to get from $n$ down to $1$, thus $b(n)$ is bounded by the product of $(\log n)/\log A^{-1}$ many $n$’s. $\endgroup$ Commented Aug 11, 2021 at 15:39
  • $\begingroup$ FYI: did you see Shallit's answer that this is one Knuth earliest papers? $\endgroup$
    – joro
    Commented Aug 16, 2021 at 14:13
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Philippe Dumas has work deriving very precise asymptotic results for 'divide-and-conquor' sequences such as this. For instance, see 'Asymptotic expansions for linear homogeneous divide-and-conquer recurrences: Algebraic and analytic approaches collated' from 2014 (https://specfun.inria.fr/dumas/Publications/Dumas14.pdf).

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  • $\begingroup$ Many thanks and welcome to mathoverflow. $\endgroup$
    – joro
    Commented Aug 18, 2021 at 8:55

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