Display powers of Phi with Fibonacci precision

Question

Write some code that takes a single non-negative integer \$n\$ and outputs the \$n\$th power of phi (\$\phi\$, the Golden Ratio, approximately 1.61803398874989) with the same number of decimal digits as the \$n\$th Fibonacci number.

Your code must produce the correct sequence of digits for all inputs up to at least 10 (55 decimal digits). The output must be human-readable decimal. You may choose whether to round the last digit to the closest value, or truncate the value. Please specify which one your code uses.

\$n\$ and output, up to 10, rounding down:

 0   1
 1   1.6
 2   2.6
 3   4.23
 4   6.854
 5  11.09016
 6  17.94427190
 7  29.0344418537486
 8  46.978713763747791812296
 9  76.0131556174964248389559523684316960
10 122.9918693812442166512522758901100964746170048893169574174

\$n\$ and output, up to 10, rounding to the closest value:

 0   1
 1   1.6
 2   2.6
 3   4.24
 4   6.854
 5  11.09017
 6  17.94427191
 7  29.0344418537486
 8  46.978713763747791812296
 9  76.0131556174964248389559523684316960
10 122.9918693812442166512522758901100964746170048893169574174

The 7th Fibonacci number is 13, so the output for \$n=7\$, \$\phi^7\$, has 13 decimal places. You must not truncate trailing zeros that would display too few digits; see output for 6 in the first table, which ends in a single zero to keep the decimal precision at 8 digits.

Maybe as a bonus, say what the highest number your program can correctly output is.

Answer 1

dc, 26 bytes

99k5v1+2/?^d5v/.5+0k1/k1/p

Due to the initial precision of 99 digits after the comma, this will work up yo input 11. A dynamic (or higher static) precision is possible, but would elevate the byte count.

Test cases

$ for ((i = 0; i < 11; i++)) { dc -e '99k5v1+2/?^d5v/.5+0k1/k1/p' <<< $i; }
1
1.6
2.6
4.23
6.854
11.09016
17.94427190
29.0344418537486
46.978713763747791812296
76.0131556174964248389559523684316960
122.9918693812442166512522758901100964746170048893169574174

How it works

Since the desired output is φⁿ, we can calculate the the Fibonacci number F(n) as ⌊φⁿ ÷ √5 + 0.5⌋ with little additional effort.

99k                         Set the precision to 99.
   5v                       Compute the square root of 5.
     1+                     Add 1.
       2/                   Divide by 2.
                            This pushes the golden ratio.
         ?                  Read the input from STDIN.
          ^                 Elevate the golden ratio to that power.
           d                Push a copy.
            5v/             Divide it by the square root of 5.
               .5+          Add 0.5.
                  0k        Set the precision to 0.
                    1/      Divide by 1, truncating to the desired precision.
                            This pushes F(n).
                      k     Set the precision to F(n).
                       1/   Divide by 1, truncating to the desired precision.
                         p  Print.

Answer 2

Mathematica, 50 bytes

N[GoldenRatio^#,2^#]~NumberForm~{2^#,Fibonacci@#}&

Basic solution. Rounds to the nearest value. Still verifying the highest value that won't make my computer run out of memory. Input 32 works, but it takes 45 minutes and uses 16GiB of RAM. However, given infinite time and memory, this could theoretically run for any value.

Answer 3

Javascript, 65 bytes

a=>Math.pow((1+Math.sqrt(5))/2,a).toFixed(k(a));k=b=>b?b+k(b-1):0

Getting Phi by calculating (1+Math.sqrt(5))/2.

Getting Fib value using the function: k=b=>b?b+k(b-1):0.

Try it out