Let $p(n)$ be the number of partitions of $n$.  Prove that growth rate of $p(n)$ is superpolynomial, meaning that for every given $k$ there is $p(n)= \omega (n^k)$.
I'm currently a Combinatorics student trying to parse through this solution. I do not understand the proof currently. Any help understanding it is greatly appreciated.    Question Let the number of partitions be given by the function $p(n)$, i.e. the number of ways of writing the integer $n$ as a sum of positive integers. Prove that that if $g$ is any polynomial function in $n$, then there exists an integer $N$ such that $g(n)&lt;p(n)$ for all $n&gt;N$. Do not use the asymptotic Hardy-Ramanujan formula.    Provided Proof.    Enough to show $n^a&lt;p(n)$ for fixed positive integer $a$. (This is just saying that we want to show $p(n)$ grows faster than any degree polynomial of $n$, right?) We have that $(i_1,i_2,\ldots,i_{a+1},1^{n-i_1-\cdots -i_{a+1}})$ is a partition of $n$ as long as $i_{a+1}\leq i_a\leq \cdots i_1 \leq \lfloor\frac{n}{a+1}\rfloor$. (I really don't understand this step. What is this partition? Of the coefficients or terms of polynomial $p(n)$? And how do we know this is $\leq \lfloor\frac{n}{a+1}\rfloor$?) The number of partitions $n$ of this form is multiset binomial coefficient $$\left(\dbinom{\lfloor \frac{n}{a+1}\rfloor}{a+1}\right)=\frac{\lfloor\frac{n}{a+1}\rfloor(\lfloor\frac{n}{a+1}\rfloor+1)\cdots(\lfloor\frac{n}{a+1}\rfloor+a-1)}{(a+1)!}$$ (I'm assuming that if I can understand (2.) then I can understand this line) This means $p(n)&gt;\left(\binom{\lfloor \frac{n}{a+1}\rfloor}{a+1}\right)$ for all $n&gt;0$. Since $a$ is constant, the given fraction grows with order $n^{a+1}$, which means $p(n)&gt;n^a$ for $n$ sufficiently large. (Why does it grow with order $n^{a+1}$?)   Thanks all again in advance!
Does a man contain all the genes needed to make a woman? This question is brought on by a Sci Fi novel I am thinking about writing. The plot device involves a colonist in charge of building a population on a new planet who loses his supply of embryos and so forth in the landing. With the artificial "wombs" intact, he populates the planet with clones of himself.  If he wanted to introduce enough variation (without making up new genes, only rearranging) in the population to allow them to be able to reproduce naturally without inbreeding problems, would he be able to? In particular, could he make a woman?  Alternately, could a woman (without a Y chromosome) do the the inverse?  At the risk of introducing bias into the answers, the plot as it stands assumes he is a male and he CAN'T do any of the above and he is restricted to clones. What file sizes and formats are most appropriate for enlarging and projecting an image onto a large wallspace for painting? Do users get notifications when @ was edited in I always wondered if users get notified about comments were their name was edited in afterwards.  Example:  I write a comment     See this link here   then I edit it to     @someuser: See this link here   Does the user get notified now? Why is "women" pronounced the way it is? As far as I know, it is the only word where wo is pronounced as wee. What is the reason for this? Does it have to do with the origin of the word? I think I've got it down to the most basic case:  int main(int argc, char ** argv) {   int * arr;    foo(arr);   printf("car[3]=%d\n",arr[3]);   free (arr);   return 1; }  void foo(int * arr) {   arr = (int*) malloc( sizeof(int)*25 );   arr[3] = 69; }   The output is this:  &gt; ./a.out   car[3]=-1869558540  a.out(4100) malloc: *** error for object 0x8fe01037: Non-aligned pointer                          being freed  *** set a breakpoint in malloc_error_break to debug &gt;   If anyone can shed light on where my understanding is failing, it'd be greatly appreciated. Can ANOVA be significant when none of the pairwise t-tests is? Is it possible for one-way (with $N&gt;2$ groups, or "levels") ANOVA to report a significant difference when none of the $N(N-1)/2$ pairwise t-tests does?   In  @whuber wrote:      It is well known that a global ANOVA F test can detect a difference of means even in cases where no individual [unadjusted pairwise] t-test of any of the pairs of means will yield a significant result.   so apparently it is possible, but I do not understand how. When does it happen and what the intuition behind such a case would be? Maybe somebody can provide a simple toy example of such a situation?  Some further remarks:   The opposite is clearly possible: overall ANOVA can be non-significant while some of the pairwise t-tests erroneously report significant differences (i.e. those would be false positives). My question is about standard, non-adjusted for multiple comparisons t-tests. If adjusted tests are used (like e.g. Tukey's HSD procedure), then it is possible that none of them turns out to be significant even though the overall ANOVA is. This is covered here in several questions, e.g.  and . Update. My question originally referred to the usual two-sample pairwise t-tests. However, as @whuber pointed out in the comments, in the ANOVA context, t-tests are usually understood as post hoc contrasts using the ANOVA estimate of the within-group variance, pooled across all groups (which is not what happens in a two-sample t-test). So there are actually two different versions of my question, and the answer to both of them turns out to be positive. See below. Update help center article on search to reflect the new position of the search box The page  says:      Enter search terms in the search box that appears in the upper right corner of every page    But the box is not there. Keyboard layout indicator sometimes disappears Ubuntu 18.04.1 after some update Rounded caps on a Bezier Curve pipe This may be a simple question.  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Which is the most recent one What's the purpose of this split packaging? Sometimes it is convenient to refer to a group of equations by a letter and a number.  For example:   2+1   (A1) 2+4   (A2) x^4   (B3) x^e   (B4)   Placing  \tag{A\theequation}\label{eqn:thisone}\stepcounter{equation}} instead of just \label{eqn:thisone} works, but seems hacky (especially manually stepping the counter).   Is there a better way? Given an integer $n$, we are asked to investigate about the existence of integer divisors of $n^2+1$ of the form $4k+3$. Can you provide some insights about it? Can a spellcaster use spell slots above their normal limit if their ability scores allow bonus spells of that level? This question is inspired by discussion , about how Heighten Spell works with Reserve Feats.  The question that I am asking here deals with whether or not players can cast spells using spell slots for spell levels that they don't have access to yet.    I feel like an example would explain this concept best.  Take Xylitol, the first-level wizard with an intelligence of 18.  He has 5 0th-level spells per day and 3 1st level spells per day from his class.  He also has 1 bonus 1st level spell per day from his high intelligence.  The  table says that he also has 1 bonus spell of each 2nd, 3rd, and 4th levels.  The question is this:  Can Xylitol cast 1st level spells using the bonus slots for his 2nd, 3rd, and 4th level spells?  He clearly can't cast spells that are inherently higher level than 1st, since he can't learn them, but can he cast 1st level spells using bonus 2nd, 3rd, and 4th level slots?  This has come up on these boards before, .  What I'm looking for here is a clear rules citation that clarifies this one way or the other.  The wording in the section on ability scores makes it quite clear that characters cannot cast spells that are actually higher level than their class allows them to cast, but it is not clear that these higher level slots are barred until the appropriate spell level is unlocked.  If a character has access to a spell level that they can't use because their ability score isn't high enough, they can freely use those slots for lower level spells, so why can't they use their bonus spell slots for lower level spells?  As KRyan's answer says, the Basics section covers this.  Based on questions and comments both here and elsewhere, that section seems to be worded ambiguously enough that it's caused confusion.  I'd like to find another, more clear wording to this rule if at all possible.  Answers from either D&amp;D 3.5 or Pathfinder are acceptable, but please note which one your answer applies to.  N.B. 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The issue was that console.log takes its parameters like a printf statement, and since the string I was logging ("%A") had a special value, it was trying to find the value of the next parameter.