User talk:Chris Kunferman

a(n)=[n*cos(α)−sin(a(0))sin(a)cos(α)(-sin(n)^2 - cos(n)^2)/(sin(n)^2 + cos(n)^2)^(3/2)

simplified a(n)=[ncos(α)−sin(a(0))sin(a)cos(α)⋅(−1)] attaching +(n+1) a(n)=(n+1)

For n=-1 a(-1) = 0 For n = 0 a(0) = 1 For n=1 a(1) = 2

-1, 0, 1,2,3,4,5,6,7,8...

Concept of Counting in 3D In a 3D space, each point can be represented by coordinates (𝑥,𝑦,𝑧)(x,y,z). Counting in 3D could mean generating a sequence of these coordinates in a systematic way.

Common Methods for 3D Counting: Linear Counting in One Direction:

Increment one coordinate while keeping others constant, then move to the next coordinate. Example: (1,1,1)(1,1,1),(2,1,1)(2,1,1),(3,1,1)(3,1,1), ...

Layer-by-Layer Counting:

Increment within a layer, then move to the next layer. Example: First layer (𝑥,𝑦,0))(x,y,0), then (𝑥,𝑦,1)(x,y,1), ...

Spiral or Radial Counting: Move in a spiral or radial pattern from the origin. Example: Start at (0,0,0)(0,0,0), then (1,0,0)(1,0,0),(1,1,0)(1,1,0),(0,1,0)(0,1,0), ...

Modified Equation: Given the original 2D-like behavior, extending to 3D would involve each term representing a 3D coordinate. If the modified equation is: a(n)=[ncos(α)−sin(a(0))sin(a)cos(α)/(sin(n)^2+cos(n)2)^3/2(−sin(n)^2 −cos(n)^2)]+(n+1)

Counting logic can be derived X-Coordinate:x(𝑛)=𝑛cos⁡(𝛼)+(𝑛+1)

Y-Coordinate: 𝑦(𝑛)=𝑛sin(𝛼)+(𝑛+1)

Z-Coordinate: 𝑧(𝑛)=[−sin(𝑎(0))sin⁡(𝑎(cos⁡(𝛼)⋅(−1)]+(𝑛+1)

For simplicity, assuming α=0 (where

cos(0)=1 and

sin(0)=0), the coordinates would be:

X-Coordinate: x(n)=n+(n+1)=2n+1

Y-Coordinate: y(n)=0+(n+1)=n+1

Z-Coordinate: z(n)=[−sin(a(0))sin(a)]+(n+1)

Assuming sin(a(0))sin(a)=0:

z(n)=(n+1) Resulting Sequence: For n=1,2,3,…, the 3D coordinates might look like:

n=1:

(x(1),y(1),z(1))=(2⋅1+1,1+1,1+1)=(3,2,2) n=2

(x(2),y(2),z(2))=(2⋅2+1,2+1,2+1)=(5,3,3)

n=3

(x(3),y(3),z(3))=(2⋅3+1,3+1,3+1)=(7,4,4)

or

(x(n),y(n),z(n))=(2n+1,n+1,n+1)

And so on...

(x(4),y(4),z(4))=(9,5,5)

(x(5),y(5),z(5))=(11,6,6)

(x(6),y(6),z(6))=(13,7,7)

(x(7),y(7),z(7))=(15,8,8)

And thus we are left with a 3D integer sequence with coordinates:

(1,2,2),(3,3,3),(5,4,4),(7,5,5),(9,6,6),(11,7,7),(13,8,8),(15,9,9)

Constructing the 3D Fibonacci-Like Sequence

X-Coordinate:

X(n)=X(n−1)+2

Initial value:

X(1)=1.

Y-Coordinate:

Y(n)=Y(n−1)+Y(n−2)

Initial values:

Y(0)=0 and

Y(1)=1.

Z-Coordinate:

Z(n)=Z(n−1)+Z(n−2)

Initial values:

Z(0)=0 and

Z(1)=1.

Sequence Rules:

X-Coordinate:

X(n)=X(n−1)+2

Initial value:

Y(n)=Y(n−1)+Y(n−2)

Initial values:

Y(0)=0 and

Y(1)=1.

Z-Coordinate: Z(n)=Z(n−1)+Z(n−2)

Initial values:

Z(0)=0 and

Z(1)=1.

(X(1),Y(1),Z(1))=(1,1,1)

(X(2),Y(2),Z(2))=(3,1,1)

(X(3),Y(3),Z(3))=(5,2,2)

(X(4),Y(4),Z(4))=(7,3,3)

For n=5:

X(5)=X(4)+2=7+2=9

Y(5)=Y(4)+Y(3)=3+2=5

Z(5)=Z(4)+Z(3)=3+2=5

(X(5),Y(5),Z(5))=(9,5,5).

For n=6:

X(6)=X(5)+2=9+2=11

Y(6)=Y(5)+Y(4)=5+3=8

Z(6)=Z(5)+Z(4)=5+3=8

(X(6),Y(6),Z(6))=(11,8,8).

For n=7:

X(7)=X(6)+2=11+2=13

Y(7)=Y(6)+Y(5)=8+5=13

Z(7)=Z(6)+Z(5)=8+5=13

(X(7),Y(7),Z(7))=(13,13,13)

For n=8:

X(8)=X(7)+2=13+2=15

Y(8)=Y(7)+Y(6)=13+8=21

Z(8)=Z(7)+Z(6)=13+8=21

(X(8),Y(8),Z(8))=(15,21,21)