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What is the recursive relation showing the possible ways of putting n = $2^k -1$ elements in a stack where always the ith element is in ith place(the element number one is always in first place in the stack). we define the amount of element i, $X(i)$. Always $X(2i)$ or $X(2i+1)$ should not be less than $X(i)$.($X(2i) orX(2i + 1) >= X(i)$) and the minimum amount is 1 unit and maximum amount is n = $2^k -1$ unit.

*In the original problem, the question asked of possible ways to make a syrup with fixed order of elements.

MMM
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  • if the $i$-th element is always in the $i$-th place, then that completely specifies the arrangement. All the other conditions are pointless, because this condition alone tells you exactly where everything is. – Paul Sinclair Jun 07 '20 at 15:38
  • @PaulSinclair But the amount of each is not constant. – MMM Jun 09 '20 at 17:58
  • Your question is so poorly explained, I can't make any sense of it or your reply. What "amount" are you referring to? You say the $i$-th element is in the $i$-th place. For that to be true, there can be only one $i$-th element. So what is $X(i)$ measuring? If there is more than one $i$-th element, where do the rest of them go without violating the $i$-th element in $i$-th place rule? What the heck is a "syrup"? Sounds sticky to me. – Paul Sinclair Jun 09 '20 at 18:12

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