Last week I presented a "word" continuation puzzle. The algorithm used to create the list isn't too difficult to discover, applying English number words (one, two, three, ...) to the previous term (the zeroth assumed to be an empty string). Thus, one letter at a time, the second term from the first:
one +two
onet
onetw
netw
A letter gets added to the right if it doesn't already exist in the evolving string. It is deleted from the string if it does already exist. Thus the string will never contain more than one copy of any particular letter. If you noticed the double-comma near the end of my puzzle, that wasn't a typo: ourihten +onehundrednineteen results in an empty string, which +onehundredtwenty yields ohurweny, which +onehundredtwentyone yields one. Using Mathematica's built-in dictionary and ignoring already encountered words (such as one at index 121), here is a list of English found in a deep continuation:
1 one
21240 visaed
45660 fads
57242 ado
155868 woad
171524 aide
271966 ad
337664 waned
347660 audit
413700 and
423066 roads
507504 wained
537056 goads
557924 aid
615808 wad
619808 wade
635830 wand
1152766 mad
1250766 moaned
1272524 maid
1298168 made
2710904 maned
3526644 mashed
10984236 mawed
16170624 maiden
21730304 mated
67092006 mead
509056060 remands
540798800 moated
1000080796 boards
1000146526 bards
1000152766 bad
1000298168 bade
1000530740 baud
1000558076 broads
1000562062 brands
1000748080 bandit
1000750040 band
1000816952 bandy
2000710904 baned
...
Why would all of our subsequent English dictionary words appear at even indices?
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