How do I write the general term ?

[gortwell]

Hi,

I have a sequence of numbers but can't figure how to write the general term of the sequence

\(\displaystyle T_n=\dfrac{(a)p+(b)q}{(b)p+(c)q}\ \ \ \ \ \ \ \ \ \ T_{n+1}=\dfrac{(b)p+(c)q}{(c)p+(a+2b)q}\ \ \ \ \ \ \ \ \ \ T_{n+2}=\dfrac{(c)p+(a+2b)q}{(a+2b)p+(b+2c)q}\)

If the equations aren't clear enough, the next term \(\displaystyle T_{n+3}\) is calculated like this...

the denominator of \(\displaystyle T_{n+2}\) becomes the numerator of \(\displaystyle T_{n+3}\)

the \(\displaystyle p\) term of the denominator of \(\displaystyle T_{n+3}\) is the same value as the \(\displaystyle q\) term of the numerator of \(\displaystyle T_{n+3}\)

the \(\displaystyle q\) term of the denominator of \(\displaystyle T_{n+3}\) is (from the numerator of \(\displaystyle T_{n+2}\)) the \(\displaystyle p\) value + 2 x the \(\displaystyle q\) value

Thanks for any help you can give

 

[HallsofIvy]

The problem is you don't have "a sequence of numbers". You have a formula from one term to the next but to have numbers you will need to have a specific value for the first number. And even that may not be enough. Given a fixed number for \(\displaystyle T_1\) there are still an infinite number of ways to divide a number into numerator and denominator and then there may be an infinite number of way to select "p" and "q" from the numerator and denominator.

 

[gortwell]

The problem is you don't has "a sequence of numbers".

Fair enough, badly worded.

I'll try again...

p and q are integers q > p > 0

The first term \(\displaystyle T_1\) is \(\displaystyle \dfrac{p}{q}\)

\(\displaystyle T_2 = \dfrac{q}{p+2q}\)

\(\displaystyle T_3 = \dfrac{p+2q}{2p+5q}\)

\(\displaystyle T_4 = \dfrac{2p+5q}{5p+12q}\)

As can be seen from the above ( and the wordy OP), each term can be calculated in exactly the same way, using the previous term.

I can't figure how to write \(\displaystyle T_n\) in terms of \(\displaystyle T_{n-1}\)

 

[gortwell]

Nearest I can seem to come...

If each term \(\displaystyle T_n\) is split into seperate numerators and denominators \(\displaystyle N_n\) and \(\displaystyle D_n\) respectively then we have (for \(\displaystyle n>2\))

\(\displaystyle T_n=\dfrac{N_n}{D_n}=\dfrac{N_{n-2}+2(N_{n-1})}{D_{n-2}+2(D_{n-1})}\)

Can I make any improvements on this ?

 

[gortwell]

What you're doing (I think!) is obtaining a certain "term"

I'm not doing anything ;-)

The sequence comes from a method of calculating values of 3 squares in arithmetic progression with the same starting figure

i.e

1, 24, 49
1, 841, 1681
1, 28561, 57121

Are you smoking funny stuff?

Not yet, lol

 

[gortwell]

But you said earlier:
"p and q are integers q > p > 0
The first term is p/q " ............................yes

So how can the 1st term be 1 ?...............would take a post longer than you'd like to read ;-)

Btw, there's loads of 3term solutions; like:....................... infinite
81 2025 3969
3 solutions with 49 as 1st term:
49 169 289
49 289 529
49 1225 2401

But I don't think there's any 4term solutions.........correct. Proposed by Fermat in 17th century, proven in late 20th century (if I remember correctly) or possibly early 21st.

.