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A180301
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Starting with 1, the next entry is the next higher integer whose spelling in US English comes lexicographically later.
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5
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1, 2, 200, 201, 202, 2000, 2001, 2002, 2200, 2201, 2202, 2000000000000, 2000000000001, 2000000000002, 2000000000200, 2000000000201, 2000000000202, 2000000002000, 2000000002001, 2000000002002, 2000000002200, 2000000002201, 2000000002202
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refs;
listen;
history;
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OFFSET
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1,2
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COMMENTS
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For example, "two" is after "one", but "three" is not alphabetically after "two" (and is thus not in the sequence); "two hundred" is after "two"; note that 2201 is spelled "two thousand two hundred one" and not "twenty two hundred and one".
The reverse alphabetical sequence starting at 1 consists only of the four terms 1, 4, 5, 8; no integer higher than 8 is alphabetically before "eight".
The comments regarding "and" above denote that US English is used.
Alphabetical order is with commas removed, but with spaces included, e.g., 8800 ("eight thousand eight hundred") would precede 8018 ("eight thousand eighteen").
In extending the sequence to large numbers, the "American system" (Weisstein link), also known as the "short scale" (Wikipedia link), was used.
a(12) = 2*10^12 ("two trillion"). The next term not displayed is a(24) = 2*10^36 ("two undecillion"). The highest known term is a(95) = 2*10^63 + 2*10^36 + 2*10^12 + 2202 ("two vigintillion two undecillion two trillion two thousand two hundred two"). See b-file and link to US English names of terms. (End)
If "and"'s were used, then a(8) = 2002 ("two thousand and two") and a(9..14) would be 2100 ("two thousand one hundred"), 2101, 2102, 2200, 2201, 2202. - Michael S. Branicky, Jul 14 2022
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LINKS
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FORMULA
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a(12+i) = 2*10^12 + a(i), for i in 1..11.
a(24+i) = 2*10^36 + a(i), for i in 1..23.
a(48+i) = 2*10^63 + a(i), for i in 1..47. (End)
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EXAMPLE
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1 ("one") is followed in lexicographic order first by 2 ("two"), and successively by 200 ("two hundred"), 201 ("two hundred one"), and so on.
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PROG
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(Python)
from num2words import num2words
def n2w(n):
return num2words(n).replace(" and", "").replace(chr(44), "")
def afind(limit, start=1):
last, t = start, start+1
print(start, end=", ")
while t <= limit:
target = n2w(last)
while n2w(t) <= target:
t += 1
if t > limit: return
last = t
print(t, end=", ")
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CROSSREFS
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KEYWORD
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easy,fini,nonn,word
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AUTHOR
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Dan Heisman (danheisman(AT)comcast.net), Aug 24 2010
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STATUS
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approved
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