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- Oct 16th 2006, 07:55 AM #1

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- Oct 16th 2006, 09:50 AM #2

- Oct 16th 2006, 05:26 PM #3

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- Oct 17th 2006, 06:42 AM #4

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- Oct 17th 2006, 07:26 AM #5

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- Oct 17th 2006, 08:23 AM #6

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Beautiful job Soroban +rep+

Never seen it done like that.

Your own solution?

I have modified (not really the right word) your recurrence relation into a different form:

---

I always tried to show that the cubic sequence and triangular sequence only have a trivial solution i.e. 1. And it seems to be true for higher orders also but I was unable to prove that, you know how?

- Oct 17th 2006, 01:37 PM #7

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Hello again, TPHacker!

I got a different closed form for the recurrence.

It's not my own solution; I learned the method from a few of my books.

You may have seen something similar while exploring.

Now that LaTeX is back, I'll revise this post.

We have the recurrence: .

Assume that is exponential: .

The equation becomes: .

Divide by and we have: .

. . which has roots: .

We form a linear combination of the two roots:

. .

To solve for and , we use the first two values of the sequence:

. .

Solve the system and get: .

Therefore: .

By the way: .

. . so you can rewrite the formula if you like . . .

- Oct 23rd 2006, 08:36 AM #8

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Besides for Soroban's solution.... I have 3 more (1 of which I am too lazy to post).

Let be the triangular number.

Solution 1)

If t_n is a square then t_{4n(n+1)} is a square. How you are suppopsed to see that, I have no idea.

Solution 2)

Following the start of Soroban;s solution we arrive at the square root of 8k² +1, which is the "Pellian equation". Thus it has infinitely many solutions.