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Conversion between Zeckendorf and Golden Ratio

Zeckendorf and Golden Ratio Base are clearly closely related, but it still seems difficult to move from one to the other. I know that this is the work of Frugna and Sakarovich, but I did not quite understand this. One problem is that the Golden Aspect representations are fairly symmetrical around the base point, which suggests that these representations may be free of context. Sakarovich and Frugny cope with this, using the "folded" base numbers of the "Golden Ratio". With this modified view, they can supposedly perform the transformation using the final state converter, but I did not understand how this should work.

As for the partial symmetry of the base of the golden ratio, this is due to the fact that the roots fall in pairs (there is a longer explanation that I have from George Bergman (pc)).

One thing I know about the relationship between the two representations is that for each base Golden Ratio representation of the form d-1 ... d_i * d_j ... d_n (using "*" as the number) the corresponding equation with Fibonacci numbers:

Example 4 = 101.01 <=> 4f_n = f_{n+2} + f_n + f_{n-2}   (with f_0 = f_1 = 1
                                                          and f_n = f_{n-1} + f_{n-2})
For n=3,  f_n=3:  12 =   10101
for n=4,  f_n=5:  20 =  101010
for n=5   f_n=8:  32 = 1010100

(Etc. There is a whole series of numbers, each of which has the same Zeckendorf bit diagram as the base Golden Ratio representation for 4). It certainly looks like it should be useful, but how?

This pattern is discussed in D. Gerdemann, Combinatorial Evidence of Seckendorf Fibonacci Family Identities Quarterly, 2008/2009.

: , "" , . , , .

+3
1

, 1.75 , , , , , , :

phi phinary .

phi , , , . :

L[n] = F[n-1] + F[n+1]

5 * F(n) = (L[n-1] + L[n+1])

Lucas phi :

L[n] = phi^n + (-1/phi)^n, n'th -n'th phi Lucas.

F[n] phi:

F[n] = ( phi^n - (-1/phi)^n )/sqrt(5) ( )

phinary:

F[n] = ( 10^n - (-0.1)^n )/10.1

sprt(5) phinary 10.1, , 5, 5 - sqrt(5) . , phi 5 , sqrt(5) ( ). sqrt(5) . , phi, Dirichlet Integer - .

-, , .

, . . , Zeckendorf phi- Zeckendorf 0 n Phinary -n n, C-like:

for (int n = 0; n < length(Zeckendorf); n++) {
    if (Zeckendorf[n] == 1) {
        Phinary[n] = 1;
        /* in a real array, the negative n needs to be offset like fixed point */
        Phinary[-n] = -1; /* negative phinary digits
        can be converted to positive ones later
        (see Golden Ratio Base article on wikipedia) */
    }
}
Standardize(Phinary); /* Change -1 to 1 with 0,-1,0 -> -1,0,1
negatives will eventually cancel with their positive 1 neighbors to the left. */
/* Divide by sqrt 5 = 10.1 in phinary */
Sqrt5[-1 .. 1] = {1, 0, 1}
PhinalNumber = PhiDivide(Phinary, Sqrt5);

wikipedia .

Balanced Ternary Tau, " " " 0- ", ( ). " , - " .

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