identity proof for partitions of natural numbers

Definition:

A tuple $\lambda = (\lambda_1, \cdots, \lambda_k)$ of Natural Numbers is called a numerical dividing of n if $1 bkslshleq bkslshlambda_1 bkslshleq bkslshcdots bkslshleq bkslshlambda_k$ and $bkslshlambda_1+bkslshcdots+bkslshlambda_k = n$ and is written as $bkslshlambda bkslshvdash n$. A numeric partition $bkslshlambda = (bkslshlambda_1, bkslshcdots, bkslshlambda_k) bkslshvdash n$ can be as well written as $bkslshlambda = [m_1, bkslshcdots, m_n ] $ with $m_i = bkslsh # bkslsh bkslshlambda_j = ibkslsh $ and $bkslshsum _ k = 1 ^ n m_k *k = n$.


Workout:

$g_k(\lambda) := \# \{i | m_i(\lambda) \geq k\}$.

Instance :

$\lambda = [5,3,3,3,2,2,0,\cdots,0] \Rightarrow m_3(\lambda) = 3, g_2(\lambda) = 6$

Prove for solution $n,k \geq 1$ $$\sum\limits_{\lambda \vdash n} m_k(\lambda) = \sum\limits_{\lambda \vdash n} g_k(\lambda)$$


First I attempted to list all numerical dividings of $n = 3$

$$ \{\lambda | \lambda \vdash 3 \} = \{(1,1,1),(1,2),(3)\}$$

$$ \begin{array}{l|cccccc} \lambda & m_1 & m_2 & m_3 & g_1 & g_2 & g_3 \\\hline (1,1,1) & 3 & 0 & 0 & 1 & 1 & 1 \\ (1,2) & 1 & 1 & 0 & 2 & 0 & 0 \\ (3) & 0 & 0 & 1 & 1 & 0 & 0 \\\hline \sum & 4 & 1 & 1 & 4 & 1 & 1 \end{array} $$

5
2022-06-07 14:30:51
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Answers: 1

Suppose we can show that the variety of dividings with $m_l \geq k$ amounts to the variety of dividings with $m_k \geq l$. It after that adheres to that $$ \sum_{p \vdash n} m_k(p) = \sum_{l \geq 1} \sum_{p \vdash n}[m_k(p) \geq l] = \sum_{l \geq 1} \sum_{p \vdash n}[m_l(p) \geq k] = \sum_{p \vdash n} g_k(p). $$ In order to confirm the case, it suffices to limit ourselves to dividings whose only components are $k,l$. Without a doubt, an approximate dividing of $n$ decays distinctly as a dividing of $n_1+n_2$, where $n_1$ is a dividing making use of $k,l$, and also $n_2$ is a dividing not making use of $k,l$. In addition, we can think that $k,l$ are reasonably prime (or else, terminate the GCD).

We show that the variety of $k,l$ - dividings with $m_l < k$ amounts to the variety of $k,l$ - dividings with $m_k < l$. Notification that given that $k,l$ are reasonably prime, there goes to the majority of one $k,l$ - dividing with $m_l < k$ and also at the majority of one $k,l$ - dividing with $m_k < l$. We show that if there is a $k,l$ - dividing with $m_l < k$ after that there is one with $m_k < l$.

Intend there is a $k,l$ - dividing with $m_k < l$. If for that dividing additionally $m_l < k$, we are done. Or else, repetitively change $k$ components of dimension $l$ with $l$ components of dimension $k$ till you get to a dividing with $m_l < k$.


Below is a different evidence for the case concerning $k,l$ - dividings, for coprime $k,l$.

Allow $$\alpha = (k^{-1}n) \bmod{l}, \quad \beta = (l^{-1}n) \bmod{k}.$$ Here $k^{-1}$ is taken relative to $l$, and also the other way around.

Specify the adhering to procedure ("conjugation") on sets $(a,b)$ that address $ak+bl=n$: $$(a,b)' = \left(\frac{l}{k}(b - \beta) + \alpha, \frac{k}{l}(a - \alpha) + \beta\right).$$ One can examine the adhering to buildings:

  1. Conjugation takes a remedy of $ak+bl=n$ to an additional remedy
  2. Conjugation is an involution (it is its very own inverse)
  3. If there is an indispensable remedy $(a,b)$ with $0 \leq a < l$ after that its conjugate $(a',b')$ has $0 \leq b < k$, and also the other way around
  4. If $kl|n$ after that $(a,b)' = (b,a)$

Here are some instances, with $k = 5$ and also $l = 3$. One can examine that $k^{-1} \bmod{l} = 2$ and also $l^{-1} \bmod{k} = 2$.

If $n = 30$ after that $(a,b)'=(b,a)$. We have $(0,10)' = (6,0)$ and also $(3,5)' = (3,5)$.

If $n = 40$ after that $\alpha = 80 \bmod{3} = 2$ and also $\beta = 80 \bmod{2} = 0$. Hence $(a,b)' = (3/5b + 2, 5/3(a - 2))$. We have $(8,0)' = (2,10)$ and also $(5,5)' = (5,5)$.

If $n = 7$ after that $\alpha = 14 \bmod{3} = 2$ and also $\beta = 14 \bmod{2} = 0$. This moment there are no marginal remedies. Rather, we have both $(2,-1)' = (7/5,0)$.


Below is an evidence making use of creating collection. The benefit of this evidence is that it calls for no idea.

The common creating collection for dividings is $$P = \frac{1}{(1-x)(1-x^2)(1-x^3)\cdots}.$$ The coefficient of $x^n$ is the variety of dividings of $n$.

Intend we intend to sum over $m_k$. So we require to change the variable $1/(1-x^k)$ with $$ x^k + 2x^{2k} + 3x^{3k} + \cdots = \frac{x^k}{(1-x^k)^2}. $$ So the creating collection for $\sum m_k$ is $$ M = \frac{x^k}{1-x^k} P. $$

Now allow is amount over $g_k$. Just how much do $1$ - components add to the amount? We just count them if there go to the very least $k$ of them. So we require to change $1/(1-x)$ with $$ x^k + x^{k+1} + \cdots = \frac{x^k}{1-x}. $$ So $1$ - components add $x^k P$. In a similar way, $2$ - components add $x^{2k} P$, and more. In total amount, the creating collection for $\sum g_k$ is $$ G = (x^k + x^{2k} + x^{3k} + \cdots) P = \frac{x^k}{1-x^k} P = M. $$

6
2022-06-07 14:45:17
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