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1

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 ...

1

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 ... 3 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$. 0 It is the OEIS sequence A002024 and the formula is$a_n=[\sqrt{2n} + 1/2].$So$a_{800}=40.$5 This is the same as asking for (the index of) the smallest triangular number that is no less than 800 (a triangular number is a positive integer of the form$\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 ...

0

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

0

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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