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What is the rule to get this list of ordered pairs : (0,0),(1,1),(2,1),(3,2),(4,2),(5,3),(6,3),...

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perhaps whenever x is odd add one to y ?
..or you want an equation?
\(\Large \color{purple}{ 1) x + 1,y + 1 }\) \(\Large \color{purple}{ 2) x + 1, y }\) It keeps repeating.

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Other answers:

and expression for the n-th term.
Well I actually got it from a while loop, and I wanted to know if I can translate it to a mathematical expression.
No, you have to use the programming if - else, or while - do
should be like (n,n/2) when n is even and (n,n/2+1) when n is odd how to combine them (you using python?)
@TuringTest lol, Do - while is in every language
I ask because I am learning python and wanted tips if No-Data is an expert ..sort of a separate Q
This is what it does: x = 0, y = 0 y = x - y x = x + 1
Sorry Turing I'm not a expert, I'm actually learning some Java, but I got distracted with this thing hehe.
thank you both.
what you have typed up is already a recurrsive definition; and it might have to be piecewise defined to be explicit as has already been pointed out
There is no way of expressing it without in the piecewise way?. I'm so fool because I can't see it from the code =/
if you are trying to confine yourself to a limited method, your going to be limited in your results .... the piecewise functions are real and useable function ... as is it looks like y is increaseing at half the rate of x, and if you can use a ceiling or floor function to weed it up or down as needed, you might be able to code it up sufficiently with a 0,0 condition as separate
if we ratio it out: \[1,2,1\frac{1}{2},2,1\frac{1}{3},2,1\frac{1}{4},2,1\frac{1}{5},2,1\frac{1}{6},...\]
\[f(x')=1+\frac{1}{x};\ x\in Odd\]\[f(x')=2;\ x\in Even\] prolly bad notations, but the slope of each discrete point would be something like this
1 + 1/n maybe better; for x in Odd
Thank you amistre64 I'll try to understand that.
I just th
I just thought it would be easier to do without piecewise functions, but now I see it is not.
computers are great with piecewises ;) good luck with it
sorry no-Data I had to take your medal away and give it to amistre... his was the best answer :P
No problem TuringTest I actually was asking for an answer and that is what amistre64 gave me.
Although I didn't know someone could change his mind about a medal.

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