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Given $n\in\mathbb{N}$, I need to get just enough more than half of it. For example (you can think this is : number of games $\rightarrow$ minimum turns to win)

$$ 1 \rightarrow 1 $$ $$ 2 \rightarrow 2 $$ $$ 3 \rightarrow 2 $$ $$ 4 \rightarrow 3 $$ $$ 5 \rightarrow 3 $$ $$ 6 \rightarrow 4 $$ $$ 7 \rightarrow 4 $$ $$ \vdots $$ $$ 2i \rightarrow i+1 $$ $$ 2i+1 \rightarrow i+1 $$ $$ \vdots $$

Is it possible to create a simple formula without piecewise it into odd and even? Sorry for my bad English.

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    $\begingroup$ One approach: Are you familiar with the floor and/or ceiling functions? See here: en.wikipedia.org/wiki/Floor_and_ceiling_functions and consider using division by 2. $\endgroup$
    – Xoque55
    Commented Jan 31, 2016 at 1:20
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    $\begingroup$ By splitting the formula into cases for the input being either even or odd you can describe the sequence quite concisely using only the most elementary language and symbols. Why do you want to avoid splitting it? What is gained? $\endgroup$
    – gnyrinn
    Commented Feb 1, 2016 at 16:29

3 Answers 3

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How about: $$ \frac{3+2n+(-1)^n}{4} $$ or (continuous function of $n \in \mathbb R$ or even $\mathbb C$): $$ \frac{3+2n+\cos(\pi n)}{4} $$

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    $\begingroup$ Cute! Not all that useful in practice, but a fresh way to look at the question. I give it a thumbs up. $\endgroup$ Commented Jan 31, 2016 at 2:03
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    $\begingroup$ I like continuity. $\endgroup$ Commented Jan 31, 2016 at 22:08
  • $\begingroup$ From the question, $n\in\mathbb{N}$. Any function $\mathbb{N} \to \mathbb{N}$ is continuous. The two formulas represent the same continuous function on $\mathbb{N}$ (as required when they are both correct). But the second expression generalizes to $\mathbb{R} \to \mathbb{R}$ or $\mathbb{C} \to \mathbb{C}$ and this extension is still continuous (and even analytic). $\endgroup$ Commented Feb 1, 2016 at 14:30
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    $\begingroup$ @JeppeStigNielsen ... thanks for that clarification $\endgroup$
    – GEdgar
    Commented Feb 1, 2016 at 14:36
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The formula is $$n\to \left\lfloor \frac{n}{2} \right\rfloor +1$$ where the notation $\lfloor x \rfloor$ means the greatest integer not exceeding $x$.

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    $\begingroup$ And in some programming languages the expression $\left\lfloor \frac{n}{2} \right\rfloor$ is written as a single infix operator, for example n/2 (where integer division with discarding the remainder is implied) or n\2. $\endgroup$ Commented Feb 1, 2016 at 14:36
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The formula below uses the floor function. $\lfloor x\rfloor$ denotes the floor function of x. For example, $\lfloor2\rfloor=2$ while $\lfloor2.5\rfloor=2$. For more on the floor and ceiling functions, go checkout Wikipedia's page on them.

$\frac{2i}{2}=i$, so $\lfloor\frac{2i}{2}\rfloor=i$. Meanwhile, $\frac{2i+1}{2}=i+\frac 1 2$, so $\lfloor\frac{2i+1}{2}\rfloor=i$. Therefore, for any $x$ that is either $2i$ or $2i+1$, $\lfloor\frac x 2\rfloor+1=i+1$. Thus, the function you are looking for is $f(x)=\lfloor\frac x 2\rfloor+1$.

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