Number of A Word Out There [closed]

Let W1 be the word that is "the" most commonly appearing or used English word in all printed books and internet files. And {W2,W3,W4,W5,W6,W7,W8,W9,W10..Wnth} are the succeeding words in that order.

Which 3 of these succeeding words ,when combined, are equivalent to "W1" in terms of number of times it is written or typed in books and internet?

closed as off-topic by Dan Russell, IAmInPLS, Beastly Gerbil, Marius, AlconjaOct 24 '16 at 22:47

• This question does not appear to be about creation and solving of puzzles, within the scope defined in the help center.
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• note it is an empirical formulation using mathematical statistics and not exact. – TSLF Oct 23 '16 at 3:21
• Is this supposed to be a puzzle? Seems more like a reference search. – Dan Russell Oct 24 '16 at 13:50
• I'm voting to close this question as off-topic because it's not about the creation or solving of puzzles, and it is not a puzzle itself. – Dan Russell Oct 24 '16 at 13:51
• this may hint youtube.com/watch?v=fCn8zs912OE – TSLF Oct 30 '16 at 9:00

Based on comments, I'm assuming the question poser wanted us to use

Zipf's law: the frequency ranks of words are inversely proportional to their corresponding frequencies. That is, if the rank of a word is $r$ and its frequency is $f$ we have that for most texts (even in languages other than English) $f \approx \frac{C}{r^z}$. In particular, for English, $z$ is very close to 1.

Based on this,

We could say that W2, W3 and W6 have the same combined frequency as W1, because $1 = \frac12 + \frac13 + \frac16$.

However, it should be noted that the actual frequencies observed diverge from this statistical law near the extremes (which is what we're dealing with here, because the ranks are very small), which makes this result likely to be incorrect in practice. Even though word frequency data tends to vary depending on the corpus used, for an answer with actual data you can check Rubio`s answer.

I went with

The Rank size distribution calculation.
With the golden ratio of 1.618.
The number of words in Wn is about $1.618^n$ times the number of words in Wn+1
So we must find W1 = $1.618^x$ + $1.618^y$ + $1.618^z$ Unfortunately I'm out of time and really gotta go. So I can't finish now, but it seems point toward being. W2, W4 and W5

W2 = be
W3 = and
W4 = of
W5 = a
W6 = to
W7 = in
W8 = have
W9 = it
W10 = I
(frequency info based on the 450 million word Corpus of Contemporary American English.)

Based on statistical data on the frequency of word appearance in the written works in the corpus, the following 3 words appear to roughly equal "the" in their aggregate frequency of appearance:

W3 (and); W7 (in); and W8 (have).
W1 (the) 22038k occurrences ~= W3 (and) 10741k + W7 (in) 6996k + W8 (have) 4304k
It's not exact, but it's pretty close.

I used my copy of a word count file (a word-frequency profile of a corpus of Google Books data) which is linked to at http://norvig.com/mayzner.html. With that, you don't need 3 words: $W_2$, "of", and $W_3$, "and", have combined frequency slightly higher than that of $W_1$, "the". (In millions, the frequencies are $53097, 30966, 22632$.)