What is the formula for this curve?

[ramsey2879]

I have a formula F that takes an ordered pair (a,b) of positive integer variables and outputs an unique positive integer c for each pair. Then I was working around with the formula and noticed that where i is added to a and subtracted from b to get (a',b') and in every case F(a',b') equals c + i . Then I thought that since i is both added and subtracted, the sum of the pair does not change with the value i and you can make i equal a+b-c so that F(a',b') = a' + b'.

This gave the following ordered pairs for each integer:

.
.
.
-4 (6,-10)
-3 (3,-6)
-2 (1,-3)
-1 (0,-1)
0 (0, 0)
1 (1, 0)
2 (3, -1)
3 (6, -3)

Here a and b are the positive and negative triangular numbers. The ordered pairs form a curve. In between points could be filled in by making a and b real instead of integers. Now if i is added to each a and subtracted from each b the shape of the curve does not change.
What is the formula for such a curve that passes through the above points? Please help me.

 

[HallsofIvy]

I have a formula F that takes an ordered pair (a,b) of positive integer variables and outputs an unique positive integer c for each pair. Then I was working around with the formula and noticed that where i is added to a and subtracted from b to get (a',b') and in every case F(a',b') equals c + i . Then I thought that since i is both added and subtracted, the sum of the pair does not change with the value i and you can make i equal a+b-c so that F(a',b') = a' + b'.

This gave the following ordered pairs for each integer:

.
.
.
-4 (6,-10)
-3 (3,-6)
-2 (1,-3)
-1 (0,-1)
0 (0, 0)
1 (1, 0)
2 (3, -1)
3 (6, -3)

Here a and b are the positive and negative triangular numbers. The ordered pairs form a curve. In between points could be filled in by making a and b real instead of integers. Now if i is added to each a and subtracted from each b the shape of the curve does not change.
What is the formula for such a curve that passes through the above points? Please help me.

So you are saying that F(a+i,b-i)= F(a,b)+ i? Clearly F(a,b)= a+ b does not fit that: (a+i)+ (b-i)= a+ b, not a+ b+ i. But you certainly could try F(a, b)= 2a+ b. Then F(a+i, b-i)= 2(a+ i)+ (b- i)= 2a+ b+ i= F(a,b)+ i.

 

[ramsey2879]

So you are saying that F(a+i,b-i)= F(a,b)+ i? Clearly F(a,b)= a+ b does not fit that: (a+i)+ (b-i)= a+ b, not a+ b+ i. But you certainly could try F(a, b)= 2a+ b. Then F(a+i, b-i)= 2(a+ i)+ (b- i)= 2a+ b+ i= F(a,b)+ i.

No I am not saying that F(a,b) = a+b! Only when a = T(a) and b = -T(a-1) for integers does F(a,b) = a+b. I know what the formula F(a,b) is but that is not what I am looking for. I want a formula in "a" that gives -T(x-1) when a = T(x). P.S. I can do it by treating positive and negative a separately by using the square root of 2*|a| and the floor function but I am trying to avoid that since it does not work for non-triangular a.

 

[ramsey2879]

If it helps F(a,b) = (a^2 + 2ab + b^2 +a -b)/2

Is it possible to provide a formula in a for b such that F(a,b) = a+b without using the summation sign?

Edit.

OK I believe I answered my question

a^2 + 2ab + b^2 +a -b = 2a+ 2b
a^2 + 2ab + b^2 -a - 3b = 0

b^2 + b*(2a-3) +(a^2 - a) = 0

solve for b using the quadratic equation

 

[ramsey2879]

If it helps F(a,b) = (a^2 + 2ab + b^2 +a -b)/2

Is it possible to provide a formula in a for b such that F(a,b) = a+b without using the summation sign?

Edit.

OK I believe I answered my question

a^2 + 2ab + b^2 +a -b = 2a+ 2b
a^2 + 2ab + b^2 -a - 3b = 0

b^2 + b*(2a-3) +(a^2 - a) = 0

solve for b using the quadratic equation

b = \frac{3-2a + \sqrt{-8a + 9}}{2}
(I mean plus/minus for plus here)
this gives the following integer points and corresponding values

F(1,0) = 1, F(0,0)=0, F(-2,1)=-1, F(-5,3)=-2, F(-9,6)=-3, ...
F(1,1) = 2, F(0,3)=3, F(-2,6)= 4, F(-5,10)=5, F(-9,15)=6, ...

The reason these points are different from what I gave in the first post is on the order of not using the correct choice of coordinates either when computing the formula or when plotting the points or both. But it can be seen that the stated properties of the formula can be obtained by choosing an appropriate coordinate system.

 

[ramsey2879]

If it helps F(a,b) = (a^2 + 2ab + b^2 +a -b)/2

Is it possible to provide a formula in a for b such that F(a,b) = a+b without using the summation sign?

Edit.

OK I believe I answered my question

a^2 + 2ab + b^2 +a -b = 2a+ 2b
a^2 + 2ab + b^2 -a - 3b = 0

b^2 + b*(2a-3) +(a^2 - a) = 0

solve for b using the quadratic equation

I found that using a spreadsheet it is simple to find two imaginary numbers a' and b' which when plugged into the formula F(a,b) above will give a specific predetermined imaginary number c' and at the same time the imaginary numbers a' and b' add to c'.

Anyone want to give me an imaginary number c' = A + Bi (although A and B need not be integers or any specific value, please select integer A and B between -500 and +500). P.S., the solution is unique for a given c'.

 

[ramsey2879]

solving the equation (a+b)^2 - 3b - a = 0 for b gives the following integer points and corresponding values

F(1,0) = 1, F(0,0)=0, F(-2,1)=-1, F(-5,3)=-2, F(-9,6)=-3, ...
F(1,1) = 2, F(0,3)=3, F(-2,6)= 4, F(-5,10)=5, F(-9,15)=6, ...

I was looking for points where the equation F(a,b) = \frac{(a+b)^{2} + a - b}{2} gives the same value as "Sum = a + b", which as seen above is true. These points can be numbered from -3 to 6 according to their value. A property of these points is that a vector having the angle and length from the point (1,0) to any of the numbered points (N) has the property that when added to the vector from the orgin to any of the other numbered points (e,g, point M) gives the coordinates that have the value in F of N*M

For instance V(2) = <0,1>, when added to any of the above points it gives

F(1,1) = 2*F(1,0), F(0,4) = 2*F(0,3), F(-2,7) = 2*F(-2,6) F(-5,11) = 2*F(-5,10), etc.

Although I mentioned points numbered by integers, the union of function F(a,b) and Sum(a,b) actually gives a surface in the complex coordinate system having both real and imaginary coordinates.

For instance point (4.5 + 4.5i, -4,5 - 1.5i) is on the surface since F(..) = Sum(..) = 3i.

Now then F(4.5 + 4.5i, (1-4.5) -1.5i) should equal 2*3i = 6i.
also V(5) = <-6,10> so F((4.5-6) + 4.5i,(10-4.5) - 1.5i) should equal 15i
and V(3i) = <3.5 +4.5i,-4.5 -1.5i> so F(8+9i,-9-3i) should equal -9
Checking it out

((a+b)^2 +a-b)/2 = ((1+3i)^2 +4.5 + 3.5 +4.5i +1.5i)/2 = (-8+6i +8 +6i)/2 = 6i !

and ((4+3i)^2 -1.5 -5.5 + 6i)/2 = (16-9-7 + 24i +6i)/2 = 15i !

and ((-1+6i)^2 +8 +9 +9i + 3i)/2 = (-35 -12i +17 +12i)/2 = -9 !

It is possible to easily determine V(a + bi) for any real "a" and "b" and there is a one to one mapping of a complex number to a point on the Union surface. The shifting of the Union surface by the vector <a + bi> gives a surface for which the value of any point over the function F is a+bi times the value over F of a point separated by the vector <a+bi>.

Unfortunately I haven't found any usefulness as yet of this result.

 

[ramsey2879]

I found that using a spreadsheet it is simple to find two imaginary numbers a' and b' which when plugged into the formula F(a,b) above will give a specific predetermined imaginary number c' and at the same time the imaginary numbers a' and b' add to c'.

Anyone want to give me an imaginary number c' = A + Bi (although A and B need not be integers or any specific value, please select integer A and B between -500 and +500). P.S., the solution is unique for a given c'.

I turned my spreadsheet routine into a formula for complex numbers a' and b' as a function of A and B.

a&#039; = (B^{2} - A*(A-3))/2 + (1.5 - A)*B*i
b&#039; = (A*(A-1) - B^{2})/2 + (A - 0.5)*B*i

This computes as a&#039; + b&#039; = A + Bi = F(a&#039;,b&#039;) where
F(a,b) = ((a+b)^{2} +a - b)/2

Also, If the above formula is used to find a" and b" corresponding to A' and B' then
F(V&lt;a&#039;,b&#039;&gt; + V&lt;a&#039;&#039;,b&#039;&#039;&gt; - V(1,0)) = (A + Bi)*(A&#039; + B&#039;i)