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My aim is to find a rigorous proof of the following conjectured identity.Given

$$1+q+q^2-q^4-q^5+q^7+q^8-q^{10}-q^{11}+\ddots=\cfrac{1}{1-q+\cfrac{(q^3)}{1-q^3+\cfrac{q^2(1+q)(1+q^3)}{1-q^5+\cfrac{q^3(1+q^2)(1+q^4)}{1-q^7+\cfrac{q^4(1+q^3)(1+q^5)}{1-q^9+\ddots}}}}}$$

The exponents on the left-hand side $${1,2,4,5,7,8,\ddots}$$ are positive integers not divisible by $3$.

How do we show that, the identity is true?

Independent verification and attempts at proving/disproving the result are very welcome.

Martin Sleziak
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Nicco
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  • You mean your LHS equals $1+\frac{q,(1+q)}{1+q^3} = \frac{1+q^2}{1-q+q^2}$? – ccorn Aug 18 '15 at 17:39
  • @ccorn well since they have the same series ,I believe they are equal,unless I'm flawed. – Nicco Aug 18 '15 at 17:55
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    This is the same as your previous question http://math.stackexchange.com/questions/1400700/a-conjectured-new-generating-function-of-narayanas-sequence Why are you asking a new question without referring to the old one, or commenting on the answer there? – Robert Israel Aug 18 '15 at 18:25
  • @Robert Israel sorry for that,I have commented,it's quite a good answer. – Nicco Aug 18 '15 at 18:56
  • @Robert Israel this question is very different from my previous question,see Tito's answer [here][http://math.stackexchange.com/questions/1400244/conjectured-new-generating-function-of-fibonacci-numbers?noredirect=1#comment2856083_1400244]. – Nicco Aug 19 '15 at 08:32
  • @RobertIsrael: Not exactly the same, but in the same family as the Fibonacci and Narayana. Kindly see answer/comment below. – Tito Piezas III Aug 20 '15 at 07:21

1 Answers1

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Regarding R. Israel's remark, three of your continued fractions, while not exactly the same, are variants of a common form discussed here for $|q|<1$,

$$\frac{1}{1-q} =\cfrac{1}{1+q-\cfrac{\color{brown}{2q(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}\tag0$$

First, the one for the Fibonacci numbers in this post,

$$\chi(q)=\cfrac{1}{1+q-\cfrac{\color{brown}{(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}\tag1$$

Second, the one for the Narayana numbers in this post,

$$\psi\Big(q\Big)=\cfrac{1}{1+q-\cfrac{\color{brown}{q^2}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}\tag2$$

Third, the one for this post with $q \to -q$,

$$M(q) = \cfrac{1}{1-q+\cfrac{\color{brown}{q^3}}{1-q^3+\cfrac{q^2(1+q)(1+q^3)}{1-q^5+\cfrac{q^3(1+q^2)(1+q^4)}{1-q^7+\cfrac{q^4(1+q^3)(1+q^5)}{1-q^9+\ddots}}}}}\tag3$$

Arranged in this way, one can see their common form. Notice that the brown part is the only "level" that changes, as well as letting $q\to-q$. (Based on this observation, I experimented with using $\color{brown}{-1+q^2, -q+q^2, q+q^2, q^4}$ instead of the ones above, and they also yield generating functions of known sequences in the OEIS.)

P.S. Transforming $(0)$ to $(3)$, I get,

$$\begin{aligned}M(q) &=\frac{1+q^2}{1-q+q^2}\\ &= 1+q+q^2-q^4-q^5+q^7+q^8-q^{10}-q^{11}+\dots \end{aligned}$$

and the sequence is A163806 (the Gf is corrected by B. Berselli). It states that,

$$a(3n) = 0$$

thus your observation that the exponents with non-zero coefficients are the integers not divisible by 3 is correct. (Note that this formula is only valid for $|q|<1$.)

Community
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Tito Piezas III
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  • Nicco, let me reiterate my comment that it is very nice you found a q-continued fraction that is a generating function for both the Fibonacci and Narayana numbers. (Perhaps you can find one for the Perrin or Padovan sequence as well.) – Tito Piezas III Aug 20 '15 at 07:36