# New recursive series identity

1. Oct 17, 2007

### ramsey2879

I derived the following identity after considering my thread "Recursive series equality" but the result is so clean and neat that I post it as a new topic.

Let $$S_{0} = 0 \quad S_{1} = 1 \quad S_{n} = b*S_{n-1} - S_{n-2}$$

Then $$S_{n}*(S_{n+b} -S_{b-2}) = (S_{n+1}+1)*(S_{n+b-1} -S_{b-1})$$

Has anyone seen this before?

2. Oct 18, 2007

### dodo

Ramsey, not to rain on your party, but it *might* turn out to be a little of a tautology. Since 0=S0 and 1=S1, it can be rewritten as

$$(S_{n} + S_0)*(S_{n+b} -S_{b-2}) = (S_{n+1}+S_1)*(S_{n+b-1} -S_{b-1})$$

which just means that the LHS formula has the same value when the indexes displace. Now, I don't know if this is common or not, the fact of having this kind of (quadratic) invariants along the sequence.

Edit: Doh, this note is probably rubbish, since two terms are fixed while the other two vary with n. It was a hasty comment.

Last edited: Oct 18, 2007
3. Oct 18, 2007

### ramsey2879

Now I can generalize this identity such that $$S_{0}$$ and $$S_{1}$$ can each be any integer!

$$(S_{n} - S_{0})*(S_{n+b} - S_{b-2}) = (S_{n+1} - S_{0} + 1)*(S_{n+b-1}-S_{b-1})$$

Edit. sorry for misinformation, but this works only for b=3 if $$S_{0} <> 0$$

Last edited: Oct 19, 2007
4. Oct 19, 2007

### dodo

Allow me some more deranging. Note that your last equation comes out by replacing each term of the sequence, say S_i, by a constant displacement, S_i + c. (In your case this constant is minus S0, thus bringing the generic sequence back to a "canonical" sequence where S0=0.). The places in the equation where two terms are subtracting cancel the constant out, so it only shows up where terms are by themselves.

I can but wonder what happens to recurrence sequences when a constant is added to each term. Your last edit suggests that for b=3 (and for -1 as the coefficient of S_n-2) the original equation in Post#1 is invariant to addition of a constant.

Deranging end. : )

5. Oct 19, 2007

### ramsey2879

I don't understand what you are saying

6. Oct 19, 2007

### dodo

Define a new sequence Rn by the rule: Rn = Sn + c, where c is a constant.

Then R0 = S0 + c, R1 = S1 + c, and the new recurrence rule for R is given by
Rn = Sn + c = b . Sn-1 - Sn-2 + c = b(Rn-1 - c) - (Rn-2 - c) + c
Rn = b . Rn-1 - Rn-2 - c(b - 2)​

which, when b=2, becomes the same rule that defines Sn (why did you say b=3?).

So you can displace by any constant a sequence defined by the rule 2 . Xn-1 - Xn-2, and it will be transformed in another sequence with the same rule.

If c = -S0, R0 becomes 0, and you can apply your first identity (on post#1) to R:

Rn (Rn+b - Rb-2) = (Rn+1 + 1)(Rn+b-1 - Rb-1)​

and substituting Rn = Sn + c = Sn - S0, you get the last identity (on post #3).

But you will obtain a similar effect for any constant c, not only for c = -S0 (as long as b=2). (Edit: And as long as R1 is not arbitrary, but displaced by the same constant as the others.)

Last edited: Oct 19, 2007
7. Oct 19, 2007

### ramsey2879

Fact is that my identity in the third post doesn't work for b = 2 so you need to consider this further.

8. Oct 19, 2007

### ramsey2879

My new identity in fact only works when $$S_{1} = (b-1)*S_{0} +1$$ and it works for all b. Dodo noted that you can add or subtract a constant to the series when b = 2 but if $$S_1 <> S_0 + 1$$ it doesn't work for b = 2 which I noted in the last post.

Last edited: Oct 20, 2007
9. Oct 20, 2007

### dodo

It is a neat equation, anyway. How come the b coefficient does not appear multiplying anything, but as an index displacement? Funny.

It seems to stem from the conjectures

\begin{align*} S_{n-1} \ S_{b-1} \ +\ S_{n+b-1} &= S_n \ S_b \\ S_{n-1} \ S_{n+b-1} \ +\ S_{b-1} &= S_n \ S_{n+b-2} \end{align*}

Last edited: Oct 20, 2007
10. Oct 20, 2007

### ramsey2879

Edit
should have been
Then $$S_{n}*(S_{n+a} -S_{a-2}) = (S_{n+1}+1)*(S_{n+a-1} -S_{a-1})$$

11. Oct 20, 2007

### ramsey2879

Actually I derived a proof for the case of S(0) = 0 based upon your proof in the previous thread. Now that I have my new conjecture for S(0) not equal to 0, I will work on a proof for the general case of S(1) = bS(0) +1. Your conjectures seem to be off base but I excuse that in view of the confusion I created with the dual use of the b variable.
If you start with series $$S$$ with for d =0 as in my previous thread and consider what happens when you subtract S(n) from each term of the series you get a new series with $$d = (b-2)*S(n)$$ but the index n becomes the zero index in the equations of my last post. Play with the equation that you proved in the last thread and you get something like S(a) - bS(a-1) wich is the same as S(a-2).

12. Oct 21, 2007

### Gib Z

I have also found a remarkable identity:

$$a+6 = 8$$, but it only works when a=2 !!

:D:D:D

13. Oct 21, 2007

### ramsey2879

But I can prove your identity, can you prove mine?

14. Oct 21, 2007

### ramsey2879

OK this identity works for all S(0), S(1) and b:

$$(S_{n} - S_{0})*(S_{n+a} - S_{a-2}) = (S_{n+1} + S_{1} -bS_{0})*(S_{n+a-1} - S_{a-1})$$

15. Oct 22, 2007

### ramsey2879

Changing the index (other than n and a) by adding 1 and noting that $$S_{2} - bS_{1} = -S_{0}$$ we have the following rewrite of the identity

$$(S_{n} -S_{1}) \cdot (S_{n+a} - S{a-1}) = (S_{n+1} - S_{0}) \cdot (S_{n+a-1} - S_{a})$$

Now since the the bijections of recursive series $$R_{n} = bR_{n-1} + R_{n-2}$$ (one bijection S1 consists of every even term and the other bijection S2 consists of every odd term) each have the recursive formula $$S_{n} = (b^2+2)S_{n-1} - S_{n-1}$$ we have the following two recursive formulas

$$(R_{2n} - R_{2}) \cdot (R_{2n+2a} -R_{2a-2}) = (R_{2n+2} - R_{0}) \cdot (R_{2n+2a -2) - R_{2a})$$

$$(R_{2n+1} - R_{3}) \cdot (R_{2n+1 +2a} - R_{2a-1}) = (R_{2n+3} - R_{1}) \cdot (R_{2n+2a-1} -R_{2a+1})$$

This has some implications for Fibonacci type series and possibly situations where
$$a^{2} + b^{2} = c^{2}$$