Product of Multiple Products?

Generalisation:-

Let's tackle each product stepwise. Recall: $\color{#D61F06}{\small{\sum r=\dfrac{n(n+1)}{2},\sum r^2=\dfrac{n(n+1)(2n+1)}6,\sum r^3=\left(\dfrac{n(n+1)}{2}\right)^2}}$

$\large\star~\displaystyle\prod_{i=1}^j n=n^j$

$\large\star ~\displaystyle\prod_{j=1}^k n^j= n^{\small{\color{#D61F06}{\displaystyle\sum_{j=1}^k j}}}= n^{\small{\color{#D61F06}{\frac{k(k+1)}{2}}}}$

$\large\star ~\displaystyle\prod_{k=1}^l n^{\small{\frac{k(k+1)}{2}}}=n^{\small{\displaystyle\sum_{k=1}^l\frac{k(k+1)}{2}}}$

$\small{ \displaystyle\sum_{k=1}^l\dfrac{k(k+1)}{2}=(1/2)\displaystyle\sum_{k=1}^l (\color{#D61F06}{k^2+k})~~~~\\=(1/2)\left(\color{#D61F06}{ \dfrac{l(l+1)(2l+1)}{6}}+\color{#D61F06}{\dfrac{l(l+1)}{2}}\right) ~~~~\\=\dfrac{l(l+1)(l+2)}{6}}~~~~$ $\small{\text{Hence}}~\large\displaystyle\prod_{k=1}^l n^{\small{\frac{k(k+1)}{2}}}=n^{\small{\frac{l(l+1)(l+2)}{6}}}$

Moving towards last product :

$\large\star ~\displaystyle\prod_{l=1}^r n^{\frac{l(l+1)(l+2)}{6}}=n^{\small{\displaystyle\sum_{l=1}^r\frac{l(l+1)(l+2)}{6}}}$

$\small{\displaystyle\sum_{l=1}^r n^{\frac{l(l+1)(l+2)}{6}}=(1/6)\displaystyle\sum_{l=1}^r (\color{#D61F06}{l^3}+3\color{#D61F06}{l^2}+2\color{#D61F06}{l})~~~~~~~~~\\=(1/6)\left(\color{#D61F06}{\left(\dfrac{r(r+1)}{2}\right)^2}+3\color{#D61F06}{ \dfrac{r(r+1)(r+2)}{6}}+2\color{#D61F06}{\dfrac{r(r+1)}{2}}\right) \\=\dfrac{r(r+1)(r+2)(r+3)}{24}}~~~~~~~~~~$

$\small{\text{Hence}}~\displaystyle\prod_{l=1}^r n^{\frac{l(l+1)(l+2)}{6}}=n^{\small{\frac{r(r+1)(r+2)(r+3)}{24}}}$

Hence, $\color{#0C6AC7}{\displaystyle\prod_{l=1}^{r}\prod_{k=1}^{l}\prod_{j=1}^{k}\prod_{i=1}^{j} \left(n\right)=n^{\small{\dfrac{r(r+1)(r+2)(r+3)}{24}}}}$

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Put $n=64$ and $r=7$ to get $\large\mathfrak{A}=2^{1260}$ .

Hence,

$\large 1260+2=\huge\boxed{1262}$

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Indeed you could put as many $\prod$ as you want and can be easily evaluated by using $\displaystyle\sum_{m=1}^p\dfrac{m(m+1)(m+2)\cdots(m+q)}{(q-1)!}=\dfrac{p(p+1)(p+2)\cdots(p+q+1)}{q!}$ .

It's a product of many 64s, so obviously the answer will be a power of 2. So we can conclude $\alpha = 2$ straight away, and take the logarithm to base 2 of the expression to get $\gamma = \sum_{l=1}^7 \sum_{k=1}^l \sum_{j=1}^k \sum_{i=1}^j 6 \\ = 6 \sum_{l=1}^7 \sum_{k=1}^l \sum_{j=1}^k j$ And then, by using the standard formulae $\sum_{r=1}^n r = \frac{n(n+1)}{2} \\ \sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6} \\ \sum_{r=1}^n r^3 = \frac{n^2(n+1)^2}{4}$ we can continue the summation: $\gamma = 3 \sum_{l=1}^7 \sum_{k=1}^l (k^2 + k) \\ = \sum_{l=1}^7 (l^3 + 3l^2 + 2l) \\ = \frac{7^4 + 6 \cdot 7^3 + 11 \cdot 7^2 + 6 \cdot 7}{4} = 1260$

Hence $\mathfrak{A} = \boxed{2^{1260}}$ .

When l = 1, we get 1 possibility for l,k,j and i. All indices should be 1. Let us note this as 1-1-1-1.

When l = 2, we get the possibilities :

2-1-1-1

2-2-1-1

2-2-2-1

2-2-2-2

and hence 4 possibilities.

When l = 3, we get :

3-1-1-1

3-2-1-1

3-2-2-1

3-2-2-2

3-3-1-1

3-3-2-1

3-3-2-2

3-3-3-1

3-3-3-2

3-3-3-3

and hence 10 possibilities.

Now you can write it out l =4 to note the following, or you can note it from the previous examples :

the first 4 sequences when l = 3 resemble the sequences when l = 2 a lot. In stead of a 2 in front, we get a 3, but the rest is the same. The other 6 sequences come from k = 3. We can reason where they come from. j can be 1, so i must be 1. j can be 2, so i has 2 options. j can be 3, so i has 3 options. This gives us 6, which is also the sum of 1,2 and 3, which is logical : the number j determines the number of options for i. Since j goes from 1 to l, this number is the sum of all natural numbers from 1 to l.

We thus come to the conclusion that when l gets l+1, the total amount of possibilities gets determined by the sum of

the number of possibilities for l + (when k=l) the sum of all natural numbers from 1 to l+1.

So for l = 1 we get 1 option.

For l = 2 we get 1 + 3 = 4 options.

For l = 3 we get 4 + 6 = 10 options.

For l = 4 we get 10 + 10 = 20 options.

For l = 5 we get 20 + 15 = 35 options.

For l = 6 we get 35 + 21 = 56 options.

For l = 7 we get 56 + 28 = 74 options.

Now 1+4+10+20+35+56+74 = 210 total options.

This means we get $(2^6)^{210} = 2^{1260},$ so that $\alpha + \gamma = 1262$ .

(I admit this is more combinatorial than algebraïc, but it gives you a different approach to these things).

Exactly one 64 is generated for each ordered pair (i,j,k,l) for which $1 \leq i \leq j \leq k \leq l \leq 7$

Then we use stars and bars to generate the number of ordered quintuplets with $i-1, j-i, k-j, l-k, 7-l \geq 0$ with sum 6.

By stars and bars this is $\dbinom{6+5 -1}{5-1} =210$ .

So we have $64^{210} = 2^{1260}$ and so the final answer is 1262.

This is easily generalized if we have $p^k$ instead of 64, n variables instead of 4 (i,j,k,l), and the cap on the largest variable is x instead of 7 as $k* \dbinom{x+n -1}{n} +p$