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In the 14th century ,an Indian mathematician T.V Narayana came up with a sequence now named after him.The sequence satisfies the recurrence $$a_{n}=a_{n-1}+a_{n-3}$$

Starting with $a_{0}=a_{1}=1$, and $a_{2}=2$ hence, $a_n = 1, 1, 2, 3, 4, 6, 9, 13,\dots$

I would like to conjecture a new generating function for narayana's sequence $a_{n}$. Given

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

How can we show that

$$\psi\Big(q\Big)= \sum_{n=0}^\infty (-1)^{n} a_{n}q^n = 1 -1q +2q^2-3q^3+4q^4-6q^5+\dots$$

is true,such that the functional equation holds

$$\psi\Big(q\Big)=\frac{1}{q}\psi\Big(\frac{1}{q}\Big)$$

Tito Piezas III
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Nicco
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2 Answers2

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The ordinary generating function for your recurrence is $$ g(x) = \dfrac{1+x^2}{1-x-x^3}$$ Thus $$\sum_{n=0}^\infty (-1)^n a_n q^n = g(-q) = \frac{1+q^2}{1+q+q^3}\tag1$$

If that is $\phi(q)$, then indeed $$ \phi(1/q) = \dfrac{1/q^2+1}{1/q^3 + 1/q + 1} = \dfrac{q (1 + q^2)}{1+q^2 + q^3} = q \phi(q)$$

Now let's try to get your continued fraction. $$ \phi(q) = \dfrac{1}{1+q - \phi_1(q)}$$ where $$ \phi_1(q) = \dfrac{q^2}{q^2+1}$$ $$\phi_1(q) = \dfrac{q^2}{1 + q^3 + \phi_2(q)}$$ where $$ \phi_2(q) = q^2 - q^3$$ $$ \phi_2(q) = \dfrac{q^2 (1-q)(1-q^3)}{1+q^5 - \phi_3(q)}$$ where $$ \phi_3(q) = q^3 + q^5$$ $$\phi_3(q) = \dfrac{q^2(1+q^2)(1+q^4)}{1+q^7 + \phi_4(q)}$$ where $$ \phi_4(q) = q^4 - q^7$$ Hmm. Looks like a pattern developing, and should be possible to prove it by induction.

Tito Piezas III
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Robert Israel
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  • this is indeed a good answer,very illuminating.The result can be proven by induction. – Nicco Aug 18 '15 at 18:54
  • OK to make a slight edit? $\phi(1/q)$ will have denominator $1+q^2+q^3$, not $1+q+q^3$, as defined in $\phi(q)$, though all this changes is that $\phi_1(q)$ will have numerator $q$ rather than $q^2$. – Antonio DJC Nov 27 '17 at 05:41
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This supplements R. Israel's answer. Given the continued fraction discussed in this post for $|q|<1$,

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

and using a little algebraic manipulation to transform the brown part to the form in the post,

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

one then gets,

$$\psi\Big(q\Big)= \frac{1+q^2}{1+q+q^3}$$

the same answer $(1)$ given by R. Israel.

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Tito Piezas III
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  • Unlike the cfrac in the other post about fibonacci numbers,this one is different in that it satisfies a functional equation. – Nicco Aug 20 '15 at 06:58
  • By letting the brown part be $$n(1+q^2)$$,we obtain generalized fibonacci numbers for every number $n$ ,such that the recurrence relation $$F_{k}=F_{k-1}+nF_{k-2}$$ holds . – Nicco Aug 20 '15 at 07:09
  • @Nicco: Actually, it's very nice that you found a common form for generating functions for the Fibonacci and Narayana numbers. I'm hoping you can find one for the Padovan or Perrin sequence as well. – Tito Piezas III Aug 20 '15 at 07:18
  • @Nicco: Yes, the functional equation is interesting, too. – Tito Piezas III Aug 20 '15 at 07:25