Nth term for integers resulting from an algorithm...

That's the best title i could do that's descriptive...

Here is the algorithm..

Step 1: Input a positive integer n
Step 2: Draw a graph consisting of n vertices and no arcs.
Step 3: Create a new vertex and join it directly to every vertex of order 0.
Step 4: Create a new vertex and join it directly to every vertex of odd order.
Step 5: Stop.

I put in the integers 1 to 5 in the algorithm and got the results of the number of arcs as...

n=1, a=3
n=2, a=4
n=3, a=7
n=4, a=8
n=5, a=11

The question asks:

State how many arcs are in the graphs that results when the algorithm is applied to a set of n vertices.

So basically the question is asking what is the nth term of these numbers:

3, 4, 7, 8, 11

Please could someone help me understand the method of how they got the nth term?

Thanks guys!

Quote:

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Originally Posted by anthmoo

That's the best title i could do that's descriptive...

Here is the algorithm..

Step 1: Input a positive integer n
Step 2: Draw a graph consisting of n vertices and no arcs.
Step 3: Create a new vertex and join it directly to every vertex of order 0.
Step 4: Create a new vertex and join it directly to every vertex of odd order.
Step 5: Stop.

I put in the integers 1 to 5 in the algorithm and got the results of the number of arcs as...

n=1, a=3
n=2, a=4
n=3, a=7
n=4, a=8
n=5, a=11

The question asks:

State how many arcs are in the graphs that results when the algorithm is applied to a set of n vertices.

So basically the question is asking what is the nth term of these numbers:

3, 4, 7, 8, 11

Please could someone help me understand the method of how they got the nth term?

Thanks guys!

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Suppose n is odd.

at step 2 there are n verts of order 0
at step 3 there is 1 vert of order n, and n or order 1
at step 4 there are 2 verts of order n+1, and n of order 2.

Total or orders at step 5=2(n+1)+2n=4n+2

Number of arcs is half the sum of the orders so a=2n+1.

Suppose n is even.

at step 2 there are n verts of order 0
at step 3 there is 1 vert of order n, and n or order 1
at step 4 there are 2 verts of order n, and n of order 2.

Total or orders at step 5=2n+2n=4n

Number of arcs is half the sum of the orders so a=2n.

RonL

Thankyou! But...

Is there no possible way for there to be one nth term for this sequence??

Quote:

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Originally Posted by anthmoo

Thankyou! But...

Is there no possible way for there to be one nth term for this sequence??

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if n is odd ceil(n/2)-floor(n/2)=1
and if n even ceil(n/2)-floor(n/2)=0

we may write:

a=2n+ceil(n/2)-floor(n/2)

RonL

(ceiling function ceil(x) is the smallest integer greater than or equal to
x, and the floor function floor(x) is the largest integer less than or equal to
x).

Quote:

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Originally Posted by CaptainBlack

if n is odd ceil(n/2)-floor(n/2)=1
and if n even ceil(n/2)-floor(n/2)=0

we may write:

a=2n+ceil(n/2)-floor(n/2)

RonL

(ceiling function ceil(x) is the smallest integer greater than or equal to
x, and the floor function floor(x) is the largest integer less than or equal to
x).

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Thankyou very much man! I didn't know much about the floor/ceiling function, I figured it was a lot more complicated than that!

Thanks man!