Let, $a_1=1, b_1=3, c_1=3$ and $d_1=1$


$a_2=a_1+c_1,b_2=b_1+d_1, c_2=a_1+b_1+c_1 $ and $d_2=a_1+d_1$

and we defined

$$a_n=a_{n-1}+c_{n-1}$$ $$b_n=b_{n-1}+d_{n-1}$$ $$c_n=a_{n-1}+b_{n-1}+c_{n-1}$$ $$d_n=a_{n-1}+d_{n-1}$$

then we have this approximate relation of involvement of $\phi$

$$a_n\phi^{3/2}+b_n\phi^{1/2}-c_n\phi\approx d_n$$

Where $\phi={\sqrt{5}+1\over2}$, which is known as the golden ration, related to the well-known Fibonacci numbers.

Apparently $d_n$ it is almost an integer for any values of n. As $n\to \infty$ we see $d_n \to$ an integer

My question is: Can anyone please explain why does this part $a_n\phi^{3/2}+b_n\phi^{1/2}-c_n\phi$ always result in almost an integer values for any n?

$\phi$ it is an irrational number and it is interesting to see $\phi$ turn the above equation into an almost integer result.

Examples of above approximate equation, give

$$\phi^{3/2}+3\phi^{1/2}-3\phi \approx 1$$ $$4\phi^{3/2}+4\phi^{1/2}-7\phi \approx 2$$ $$11\phi^{3/2}+6\phi^{1/2}-15\phi \approx 6$$ $$26\phi^{3/2}+12\phi^{1/2}-32\phi \approx 17$$ $$58\phi^{3/2}+29\phi^{1/2}-70\phi \approx 43$$


$$\left( a_n\phi^{3/2}+b_n\phi^{1/2}-c_n\phi - d_n \right) = -(1 - \phi^{1/2})^{n+2} $$

Since $1 - \varphi^{1/2} \approx -0.272$, its powers rapidly converge to zero.

I found this formula from:

  • The knowledge that these sorts of phenomena usually boil down to exponential functions with a small base
  • Recognizing that $1, \phi^{1/2}, \phi, \phi^{3/2}$ form a basis for a number field
  • Running the recursion backwards to find the $-2$ and $-1$ terms are $(0,0,0,1)$ and $(0,1,0,1)$, corresponding to the numbers $-1$ and $\varphi^{1/2} - 1$ respectively.
  • Numerically checking the formula

To rigorously prove the identity, one might simply show that stepping the recursion has the same effect as multiplying by $(1 - \phi^{1/2})$.

There was some good fortune that the formula was particularly simple; I expected to have to do more work to find it.

A more systematic approach to obtain the closed form would be to apply the theory of linear difference equations.

  • $\begingroup$ nice of you to came with the formula @Hurkyl $\endgroup$ – user515439 Dec 23 '17 at 16:45

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