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This is not a direct answer to your main question, but it does answer something very important. What is the nature of relationship between taking the fractional part of a number and the mod function? Just for fun, let's use the notation you've built above. Take $n=1$ and $m$. The function is $\sqrt{1+m}$. Let's look at the fractional part ...


I don't know if this will help (or even if it is in your text or links): If $n = m^2+k$, where $0 \le k \le 2m$, $\begin{array}\\ \sqrt{n} - \lfloor \sqrt{n} \rfloor &= (\sqrt{n} - \lfloor \sqrt{n} \rfloor)\frac{\sqrt{n} + \lfloor \sqrt{n} \rfloor}{\sqrt{n} + \lfloor \sqrt{n} \rfloor}\\ &= \frac{n - \lfloor \sqrt{n} \rfloor^2}{\sqrt{n} + \lfloor ...


This sequence can be described by formula: $$ u(n)=\left\{ \begin{array}{l} n\cdot 2^{n-1}, \qquad\; n=0,1,2,3,4,5,6;\\ n\cdot 2^{n-1}+4, \;\; n=7,8,9,...;\end{array} \right. $$ $u(0)=0$, $u(1)=1$, $u(2)=4$, $u(3)=12$, $\ldots$, $u(6)=192$; $u(7)=448+4$, $u(8)=1024+4$, $\ldots$, $u(13)=53248+4$.


It is the OEIS sequence A002024 and the formula is $a_n=[\sqrt{2n} + 1/2].$ So $a_{800}=40.$


This is the same as asking for (the index of) the smallest triangular number that is no less than 800 (a triangular number is a positive integer of the form $\frac{n(n+1)}{2}$). Solving $\frac{x(x+1)}{2} = 800$ gives $x=39.50312, x=-40.50312$. Since $\frac{39 \times 40}{2} = 780$ and $\frac{40 \times 41}{2} = 820$ we see that the 800th term of the sequence ...


a(n(n-1)/2 + 1 ) , ... , a(n(n+1)/2) are equal to n put n=40 : a(781) , ... , a(820) are equal to 40 so a(800) = 40


The choice of 1 or 6 is clear based on structure and repetition. Observing the position of the black point makes the selection of 1 somewhat preferable when compared the positions selected. So number 1 is the most appropriate selection.

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