mathematics - Find the word with the highest normalized product of letter-prime values where A=2, B=3, C=5, ..., Z=101

Which word has the highest normalized score according to the following:

• Assign each individual letter A..Z of an (English) word a value corresponding to the n'th prime (see table below).


• Take the product of its letter values.


• Normalization: Take the log10 of that product, and divide by the wordlength.


• (See Python example code below)


• Please state the wordlength along with its score

Examples:

• SYZYGY = 1.83686.. (6 letters)


• WOW = 1.83675.. (3 letters)


• SYZYGIUM = 1.73403.. (8 letters)


• WOODWORKING = 1.56381.. (11 letters)

(The normalization is to prefer everyday English words over IUPAC chemical names or 'Pneumonoultramicroscopicsilicovolcanoconiosis').

Letter values A:2 B:3 C:5 D:7 E:11 F:13 G:17 H:19 I:23 J:29 K:31 L:37 M:41 N:43 O:47 P:53 Q:59 R:61 S:67 T:71 U:73 V:79 W:83 X:89 Y:97 Z:101

Some Python example code:

import string
from math import log10

primes = { letter_prime[0]:letter_prime[1] for letter_prime in zip(string.ascii_uppercase, [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101])}
{'A': 2, 'B': 3, 'C': 5, 'D': 7, 'E': 11, 'F': 13, 'G': 17, 'H': 19, 'I': 23, 'J': 29, 'K': 31, 'L': 37, 'M': 41, 'N': 43, 'O': 47, 'P': 53, 'Q': 59, 'R': 61, 'S': 67, 'T': 71, 'U': 73, 'V': 79, 'W': 83, 'X': 89, 'Y': 97, 'Z': 101}

def letter_prime_score(word):
    prod = 1
    for letter in word:
        prod *= primes[letter]

    return log10(prod) / len(word)

letter_prime_score('SYZYGY')
1.8368600501100796

letter_prime_score('WOW')
1.8367513475626218

letter_prime_score('SYZYGIUM')
1.734027889906502


letter_prime_score('WOODWORKING')
1.5638076854903682

Answer

The word that I got was ZYZZYVA, a valid Scrabble word, and a type of snouted beetle, most famous for being the final word in the dictionary. The total points are $101^3 \times 97^2 \times 79 \times 2$, which is equal to $1.5316681 \times 10^{12}$. If we normalize this by taking the logarithm and dividing by the number of letters in the word, we get $\frac{\log 1.5316681 \times 10^{12}}{7} \approx 1.74073781098$.

In comparison, SYZYGIUM results in $\frac{\log 74511467904889}{8} \approx 1.73402788991$.

WOODWORKING results in $\frac{\log 159178548575947607}{11} \approx 1.56380768549$.

PNEUMONOULTRAMICROSCOPICSILICOVOLCANOCONIOSIS results in $\frac{\log 2.09 \times 10^{65}}{45} \approx 1.45155880636$.

Update On the other hand, my first suggestion punished long words as much as it favoured small words. Therefore:

1 letter: I results in $\frac{\log 23}{1} \approx 1.36172783602$.

2 letter: XU results in $\frac{\log 73 \times 89}{2} \approx 1.90635643338$.
Note: This word found by @Herb Wolfe! Go upvote their answer!! :D
3 letter: ZZZ results in $\frac{\log 1030301}{3} \approx 2.00432137378$.
Note: Check the link for ZZZ, it's in the Merriam Scrabble Dictionary!
4 letter: ZIZZ results in $\frac{\log 23696923}{4} \approx 1.84367298934$.
Note: Apparently ZIZZ is in the Merrriam Scrabble Dictionary too!
5 letter: ZZYZX results in $\frac{\log 8694710139}{5} \approx 1.98785102165$.
6 letter: ZYZZYX results in $\frac{\log 862775087701}{6} \approx 1.98931626609$.
Note: This word found by @peaceoutside! Go upvote their answer!! :D
7 letter: ZYZZYVA results in $\frac{\log 1.5316681 \times 10^{12}}{7} \approx 1.74073781098$.

8 letter: ZYZZYZUS results in $\frac{\log 4.7887992 \times 10^{15}}{8} \approx 1.96002832831$.