[Guest]

the first circular number is 1, next is 4.

the 3rd is 8

  x x

x x x x

x x x x

  x x

the 4th is 24

    x x

  x x x x

x x x x x x

x x x x x x

  x x x x

    x x

and so on.

1. write a formula that goes through every circular number.

2. write a formula that only goes through numbers that are both square and circular. (assume integer input)

[superprismatic]

Your third circular number looks like 12, not 8. What is your definition of a circular number?

[Guest]

the first circular number is 1, next is 4.

the 3rd is 8

  x x

x x x x

x x x x

  x x

the 4th is 24

    x x

  x x x x

x x x x x x

x x x x x x

  x x x x

    x x

and so on.

1. write a formula that goes through every circular number.

2. write a formula that only goes through numbers that are both square and circular. (assume integer input)

[Guest]

You may want to check the definiton of circular numbers, numbers whose square end in the same digit as the original number, such a 0, 1, 5 or 6. Are you asking for something other than this.

[Guest]

you're right, 3rd number should be 12.

[Guest]

what is a circular number?

[Guest]

Round numbers, indeed!

Except for 1, the formula x(2x-2)works.

Edited February 3, 2010 by seajay

[EventHorizon]

Round numbers, indeed!

Except for 1, the formula x(2x-2)works.

Yup...

Each quadrant of the circle is a triangle, so circular numbers are simply 4*triangular numbers (with a slight change due to the first circular number).

Triangular numbers are x(x+1)/2. Multiplying by 4 is 2x(x+1). Replacing x by (x-1) results in 2x(x-1), which is the equation given by seajay.

[EventHorizon]

Let b = (1+sqrt(2))^4 = 17+12*sqrt(2)

F(0) = 1

F(x) = (round(sqrt(4.246320343559*(b^(x-1)))))^2

I'm pretty sure this works, but here's a recurrence relation that is exact...

s₀ = 1

s₁ = 4

s_x+1 = ( sqrt(s_x) + sqrt(2*s_x+1) + sqrt(2*(sqrt(s_x) + sqrt(2*s_x+1))^2 - 1) )^2

Have fun trying to figure out how I derived it!

Edited February 4, 2010 by EventHorizon

[Guest]

questions 1 and 2 answered

1) F(n) = 2(n^2-n) gives the set of all circular numbers. this can be found by adding the diagonal lengths for a circle.

2) since G(n) = n^2 gives the set of all square numbers, the numbers that are both circular and square can be found by setting G(n) = F(n) giving 2(n^2)-2n = n^2 ==> n^2 - 2n = 0 ==> n(n-2) = 0 ==> n = 0,2

so F(2) and F(0) are the only square and circular numbers. since F(0) is trivial since it was established in the question that F(1) is the first circular number, 4 is the only non-trivial number that is both square and circular

[Guest]

The numbers are 1,4,8,24..... They follow 1 is not multipled by any number, the second number is multiple by 1, third is multiplied by 2 and so no.

i.e. 1, 4x1,4x2,8x3,24x4, 96x5

so the numbers are 1,4,8,24,96,480

Please reply whether my answer is correct or not

the first circular number is 1, next is 4.

the 3rd is 8

  x x

x x x x

x x x x

  x x

the 4th is 24

    x x

  x x x x

x x x x x x

x x x x x x

  x x x x

    x x

and so on.

1. write a formula that goes through every circular number.

2. write a formula that only goes through numbers that are both square and circular. (assume integer input)

[EventHorizon]

The numbers are 1,4,8,24..... They follow 1 is not multipled by any number, the second number is multiple by 1, third is multiplied by 2 and so no.

i.e. 1, 4x1,4x2,8x3,24x4, 96x5

so the numbers are 1,4,8,24,96,480

Please reply whether my answer is correct or not

Read through the posts, and you'll see that the OP made a mistake and the third circular number is 12, not 8.

[EventHorizon]

questions 1 and 2 answered

1) F(n) = 2(n^2-n) gives the set of all circular numbers. this can be found by adding the diagonal lengths for a circle.

2) since G(n) = n^2 gives the set of all square numbers, the numbers that are both circular and square can be found by setting G(n) = F(n) giving 2(n^2)-2n = n^2 ==> n^2 - 2n = 0 ==> n(n-2) = 0 ==> n = 0,2

so F(2) and F(0) are the only square and circular numbers. since F(0) is trivial since it was established in the question that F(1) is the first circular number, 4 is the only non-trivial number that is both square and circular

You assume that the same number will need to be fed into both equations. (e.g., same 'n')

How about F(9) = 2(9)(8) = 2*72 = 144 = 12*12 = G(12)

It would be better to say F(n) = G(m), and see what you can say about that.