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Consider the series:
$N-(\left \lceil N(1-\alpha) \right \rceil-1)-(\left \lceil \left \lfloor N \alpha \right \rfloor (1-\alpha) \right \rceil -1)-( \left \lceil \left \lfloor N \alpha ^2\right\rfloor (1-\alpha) \right \rceil-1)-(\left \lceil \left \lfloor N \alpha ^3 \right \rfloor (1-\alpha) \right \rceil-1)-...-( \left \lceil \left \lfloor N \alpha ^n \right \rfloor (1-\alpha)\right\rceil-1)$
Where $\alpha <1$ and $n$ is value of the power of $\alpha$ when the last expression:
$\left \lceil \left \lfloor N \alpha ^n \right \rfloor (1-\alpha)\right\rceil-1$
is zero, whose value comes out to be:
$n>\frac{log 3-log N - log(1-\alpha)}{log \alpha}$
(See Solving A Simple Exponential floor Equation (a similar question)) Now (again) removing the floors and the ceilings directly would affect the answer. We can also remove them by introducing inequalities, but they don't seem to solve to something straight forward.

Any Help Appreciated.
Moon.

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    $\begingroup$ So that would be $N-\sum_{k=1}^\infty(\lceil \lfloor N\alpha^k\rfloor(1-\alpha)\rceil-1)$? And I assume $0<\alpha<1$? -- If $1-\alpha$ is not the reciprocal of an integer, we can "simplify" to $N-\sum_{k=1}^\infty\lfloor \lfloor N\alpha^k\rfloor(1-\alpha)\rfloor$ $\endgroup$ – Hagen von Eitzen Mar 1 '17 at 9:05
  • $\begingroup$ @HagenVonEitzen Eye opening! $\endgroup$ – Mooncrater Mar 1 '17 at 12:26

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