Sigma Ceilings

Let the sum be $S$ and $a_n = \left \lceil \dfrac{11}{n} \right \rceil$ , $b_n = \left \lceil \dfrac{7}{a} \right \rceil$ , $c_n = \left \lceil \dfrac{5}{b} \right \rceil$ , $d_n = \left \lceil \dfrac{3}{c} \right \rceil$ and $x_n = \left \lceil \dfrac{2}{d} \right \rceil$ .

Therefore, $S = \displaystyle \sum_{n=1}^{4096} \lceil \frac{2}{\lceil \frac{3}{\lceil \frac{5}{ \lceil \frac{7}{\lceil \frac{11}{n} \rceil} \rceil} \rceil} \rceil} \rceil = \sum_{n=1}^{4096} x_n$

We note that there are only $6$ values of $a_n$ for all $n$ , therefore $x_n$ takes only limited values.

$\begin{array} {cccccc} n & a_n & b_n & c_n & d_n & x_n \\ \hline 1 & 11 & 1 & 5 & 1 & 2 \\ 2 & 6 & 2 & 3 & 1 & 2 \\ 3 & 4 & 2 & 3 & 1 & 2 \\ 4 \text{ to } 5 & 3 & 3 & 2 & 2 & 1 \\ 6 \text{ to } 10 & 2 & 4 & 2 & 2 & 1 \\ 11 \text{ to } 4096 & 1 & 7 & 1 & 3 & 1 \end{array}$

Therefore, $S = \displaystyle \sum_{n=1}^{4096} x_n = \sum_{n=1}^{3} 2 + \sum_{n=4}^{4096} 1 = 6 + 4093 = \boxed{4099}$

$Note\quad that\quad the\quad final\quad term\quad below\quad 2\quad will\quad be\quad a\quad positive\quad integer,\quad since\quad Range(\left\lceil x \right\rceil )=\quad { Z }^{ + }\quad \forall \quad x\quad \in \quad { R }^{ + }\\ If\quad the\quad denominator\quad is\quad =\quad 1,\quad then\quad the\quad final\quad term\quad will\quad be\quad 2.\quad \\ If\quad it\quad is\quad greater \quad\ than \quad 1,\quad then\quad the\quad final\quad term\quad will\quad be\quad 1.\\ Thus\quad we\quad note\quad all\quad terms=\quad { X }_{ i }\quad =\quad Either\quad 1\quad or\quad 2\\ Now\quad let\quad us\quad determine\quad the\quad max(n)\quad for\quad which\quad { X }_{ i }=\quad 2\\ \\ Since\quad { X }_{ i }=2,\quad the\quad final\quad term\quad below\quad 2\quad =1\\ Thus\quad the\quad term\quad below\quad 3\quad must\quad be\quad greter\quad than\quad than\quad or\quad equal\quad to\quad 3\\ \Longrightarrow \quad The\quad term\quad below\quad 5\quad must\quad be\quad less\quad than\quad or\quad equal\quad to\quad 2\\ \Longrightarrow The\quad term\quad below\quad 7\quad must\quad be\quad greater\quad than\quad or\quad equal\quad to\quad 4\\ Thus\quad \left\lceil \frac { 11 }{ n } \right\rceil \quad \ge 4\Longrightarrow \quad n=\left\{ 1,2,3 \right\} \\ Therfore\quad S=\quad 2+2+2+4093\times 1=\boxed { 4099 }$