Questions and Comments

Comment 1) From my academic brother, Moa Apagodu, Nov. 17, 2018

Correction on Lemma 2:

The number of Schur triples on [1,n] should be

1/2 *(binomial(n,2)-floor(n/2)). # (x,y,z): x+y=z, with (x,y,z)=(y,x,z) and x not equal to y.

and 1/2 *(binomial(n,2)+floor(n/2)). # (x,y,z): x+y=z, with (x,y,z)=(y,x,z) and x can be equal to y.

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Question 2) From Gang Yang, a master student in University of Shanghai for Science and Technology, China, June 3, 2022

Question: I do not know that how to get the curves in Figure 7?

Answer: First, please let me remind you about the notations that are needed.

X = {x,n+1-x}, 1 <=x <= n/2

Y_1 = {y,n/2+1-y}, 1 <= y <= n/4

The interaction of X and Y_1 are under the condition that 2y < x (as in def.17)

Now in figure 7a), we consider the quantity mu_{BB}*mu^(1)_{RB}-mu_{RR}*mu^(1)_{RB} as mentioned in the bottom of page 10.

The quantity on the vertical axis (the pairs gained) represents the quantity: mu_{BB}-mu_{RR} from all x with n/2 >= x > 2y (i.e. all possible elements in set X) for each specific y value i.e. specific set of Y_1.

(We also disregard the color in {y,n/2+1-y} and only regard the colors for the number of pairs in X)

At y =0 we have mu_{BB}-mu_{RR}.

As y increases, mu_{RR} decreases as x>2y (consider only x that is greater than 2y).

Once y=mu_{RR}/2, the quantity mu_{RR}=0 as there are no more (R,R) pairs in X (for x>2y), hence the corresponding number of pairs from set X at this y is mu_{BB}.

Now you can see that the quantity mu_{BB} stays the same for a while until y=n/4-mu_{BB}/2. It will start to decrease down to 0 at y = n/4 (from condition x>2y).

I hope this helps. Sorry if you think that the coloring of the interval is misleading.

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