A004740 -id:A004740 - cp's OEIS frontend

A108067 Integers in alphabetical order in U.S. English (another version of A004740).

8, 8000000000, 8000000008, 8000000018, 8018000000, 8018000008, 8018000018, 8018018000, 8018018008, 8018018018, 8018018800, 8018018808, 8018018818, 8018018880, 8018018888, 8018018885, 8018018884, 8018018889, 8018018881, 8018018887, 8018018886, 8018018883, 8018018882, 8018018811, 8018018815, 8018018850

Offset: 1

Lekraj Beedassy, May 31 2005

The last term of this infinite sequence is 0. - Lekraj Beedassy, Mar 09 2006
The penultimate term using the accepted naming convention would be TWO VIGINITILLION TWO UNDECILLION TWO TRILLION TWO THOUSAND TWO HUNDRED TWO. Because not all numbers have names, the sequence would not truly be infinite. - Matthew Goers, Jan 11 2012
The sequence does not count spaces and hyphens when alphabetizing. If spaces were counted, eight billion eight million would be after eight billion eight. - Charles Coker, Jul 29 2019
If spaces were counted, then eight billion eight hundred (and then many other terms, starting with eight billion eight hundred eight) would follow eight billion eight. These would all precede eight billion eight million (mentioned above). The current sequence's eight billion eighteen would not occur until after many more intervening terms. - Michael S. Branicky, Aug 17 2022 [clarified Oct 04 2023]

			EIGHT
EIGHT BILLION
EIGHT BILLION EIGHT
EIGHT BILLION EIGHTEEN
EIGHT BILLION EIGHTEEN MILLION
EIGHT BILLION EIGHTEEN MILLION EIGHT
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHTEEN
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED EIGHT
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED EIGHTEEN
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED EIGHTY
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED EIGHTY-EIGHT
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED EIGHTY-FIVE
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED EIGHTY-FOUR
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED EIGHTY-NINE
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED EIGHTY-ONE
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED EIGHTY-SEVEN
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED EIGHTY-SIX
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED EIGHTY-THREE
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED EIGHTY-TWO
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED ELEVEN
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED FIFTEEN
EIGHT BILLION EIGHTEEN MILLION EIGHTEEN THOUSAND EIGHT HUNDRED FIFTY

Cf. A004740.

Corrections and additions from Matthew Goers, Nov 02 2009

8018018811 and 8018018815 added in proper places and 801808818 corrected by Matthew Goers, Mar 22 2012

A019440 Integers in alphabetical order in British English.

Original entry on oeis.org

8, 8000000000000, 8000000000008, 8000000000018, 8000000000080, 8000000000088, 8000000000085

Offset: 1

N. J. A. Sloane, Neil Fernandez

			The list begins "eight", "eight billion", "eight billion and eight", ...

Cf. A004740, A026081.

A026081 Integers in reverse alphabetical order in U.S. English.

Original entry on oeis.org

0, 2202202202202, 2202202202222, 2202202202223, 2202202202226

Offset: 1

Neil Fernandez

This sequence is ill-defined from the 2nd term on. Proof: Assume that x is the term following the initial 0. Then the number x*10^(6n), for n sufficiently large, is spelled out: name(x) name(10^(6n)). This comes after name(x) in lexicographical order, and thus before x in this sequence. - M. F. Hasler, Nov 20 2009

			Zero, two trillion, two hundred and two billion, two hundred and two million, two hundred and two thousand, two hundred and two, ...

Cf. A004740, A019440.

See A127352 for another version. - N. J. A. Sloane, Dec 05 2009

A152611 Assign weights to the nonnegative integers as in A073327, then sort them by weight.

Original entry on oeis.org

1, 10, 5, 9, 8, 6, 3, 2, 4, 11, 0, 7, 15, 50, 18, 80, 40, 19, 12, 90, 16, 60, 13, 30, 51, 14, 20, 1000000000, 55, 59, 81, 100, 17, 70, 58, 5000000000, 9000000000, 85, 89, 500, 900, 1000000000000000000000000000000000, 41, 56, 91, 53, 8000000000, 88, 800, 52

Offset: 1

Matthew Goers, Oct 26 2009, Nov 02 2009, Nov 03 2009

We are using U.S. spelling, as in A108067 rather than A004740, and we ignore hyphens and spaces.

In the case of ties, sort by numerical value.

			Let wt(n) = A073327(n) denote the weight of n. We have wt(1) = 34 < wt(10) = 39 < wt(5) = 42 = wt(9) = 42 < wt(8) = 49 < ...

A127352 Integers less than 10^303 in reverse alphabetical order in U.S. English.

Original entry on oeis.org

0, 2000000000000000000000000002000000000000000000000002000000002202, 2000000000000000000000000002000000000000000000000002000000002222, 2000000000000000000000000002000000000000000000000002000000002223, 2000000000000000000000000002000000000000000000000002000000002226, 2000000000000000000000000002000000000000000000000002000000002227, 2000000000000000000000000002000000000000000000000002000000002221

Offset: 1

Michael B. Porter, Nov 24 2009

Since the use of alphabetic names is rare for numbers greater than 10^15, there is no universal agreement on the naming scheme for large integers, and there is some question whether this sequence would well-defined without the "less than 10^303" clause.
The Wikipedia article compares 8 dictionary sources and has names for the powers of 1000 up to 10^63 and for 10^303. These are also in the Mathworld link.
There are several conflicting schemes for extending the dictionary definitions. If we assume that the system of alphabetic names greater than 10^63 defines a word for every power of 1000 and that word comes before "vigintillion" alphabetically, the sequence can include all integers. However, many of the extension schemes listed do not meet that standard - some have multiple words and some have words that are alphabetically after "vigintillion".
For the powers of 1000 between 10^66 and 10^303, one source (http://www.mrob.com/pub/math/largenum.html) coins the name "vigintinonillion" for 10^90, but this format is inconsistent with other names listed in the same source, e.g. "duovigintillion", "sexoctogintillion". The name "novemvigintillion" seems to be more common. Otherwise, all sources have "vigintillion" as alphabetically last of all the powers of 1000 up to 10^303.
The terms are from Andrew Weimholt.

			zero,
two vigintillion two undecillion two trillion two thousand two hundred two,
two vigintillion two undecillion two trillion two thousand two hundred twenty two, etc.

Michael B. Porter, Table of n, a(n) for n = 1..15
Robert P. Munafo, Large numbers
Landon Curt Noll, How high can you count?
John J. G. Savard, Some notes on big numbers
Gerard Schildberger, Lists of Words for the Numbers in A060365 and A060366
Eric Weisstein's World of Mathematics, Large Number
Wikipedia, Names of large numbers

Cf. A004740, A019440, A060365, A060366.

See A026081 for another version.

A340671 a(n) is the number of values m such that, if the first n positive integers are arranged in alphabetical order in US English, the m-th term in the order is equal to m.

Original entry on oeis.org

1, 2, 1, 1, 0, 0, 0, 1, 0, 0, 2, 2, 3, 2, 0, 1, 0, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 3, 1, 2, 1, 2, 1, 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 0, 1, 1, 2, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 1, 0, 0, 1, 1, 2, 0, 1, 0, 1, 0

Offset: 1

Mikhail Soumar, Jan 15 2021

Nonnegative integers can be used instead of positive integers, since zero will always be the last element alphabetically and will not change the sequence of the other integers.
This sequence uses standard US English names for numbers. "and" is not used, e.g., 101 is rendered as "one hundred one" rather than "one hundred and one".
a(n) is equivalent to the number of terms in the n-th row of A124172 for which the term in the k-th column is equal to k.
For n < 100, a(n) + 2 = a(200 + n). This is because a(200) = 2, and the numbers starting with "two hundred" will follow all of 1-199 alphabetically, so the range [201, 200 + n] will be in the same order as [1, n]. Similarly, because a(2000) = 4, for n < 999, a(n) + 4 = a(2000 + n). [Editor's note: It is unclear how the author finds a(2000) = 4. Both versions mentioned below give a(2000) = 2. - M. F. Hasler, Jul 05 2024]
From Claudio Meller, Hans Havermann and Michael S. Branicky, Jul 03 2024: (Start)
A formalization of Philip Cohen's solution to "Alphabetizing the Integers" in (Eckler, p. 20).
When alphabetizing in the b-file and a-file, the space is assumed to precede any letter, so EIGHT HUNDRED comes before EIGHTEEN. No commas are used, but hyphens are used. (End)
At least two variants of this sequence are conceivable, depending on whether spaces and hyphens are considered or ignored, when sorting the English names of the numbers. If spaces are considered, "eight hundred" comes before "eighteen"; if they are ignored, "eighteen" comes only after all of "eight hundred ...". The two variants would not differ until a(815), where "eighteen" would be the only "fixed point" (i.e., listed at the 18th place) in the first variant, but not in the second variant (where it is listed in the 2nd place, after "eight"). - M. F. Hasler, Jul 05 2024

			a(1) = 1 ({one}, the 1st term is 1);
a(2) = 2 ({one, two}, the 1st term is 1 and the 2nd term is 2);
a(3) = 1 ({one, three, two}, the 1st term is 1);
a(4) = 1 ({four, one, three, two}, the 3rd term is 3);
a(11) = a(12) = 2 (the 4th term is 4 and the 7th term is 7);
a(13) = 3 (the 4th term is 4, the 7th term is 7, and the 12th term is 12).

Michael S. Branicky and Hans Havermann, Table of n, a(n) for n = 1..10000
Michael S. Branicky, Python program and utilities for A340671
A. Ross Eckler, Alphabetizing the Integers (Word Ways, 1981, Vol. 14, No. 1, pp. 18-20).
Hans Havermann and Michael S. Branicky, n, Set of Fixed Points for n = 1..10000

Cf. A124172, A005589, A004740.

apply( {A340671(n, cf=English)=sum(i=1, #n=vecsort([1..n], x->cf(x), 1), n[i]==i)}, [1..99]) \\ See A052360 for English(). To get the "ignore spaces and hyphens" variant, use "CF(x)=[c|c<-Vecsmall(English(x)), c>64]" as 2nd optional argument. To get the list of fixed points, replace "sum(i=1,(...))" by "[i|i<-[1..(...)]". - M. F. Hasler, Jul 05 2024

from num2words import num2words
def a(n):
    sorted_list = sorted([num2words(m) for m in range(1, n+1)])
    return sum(m == num2words(sorted_list.index(m)+1) for m in sorted_list)
print([a(n) for n in range(1, 101)]) # [Note: this program retains the "and" and commas. - Michael S. Branicky, Jul 05 2024]

# see link for faster version
from bisect import insort
from num2words import num2words
from itertools import count, islice
def n2w(n): # remove " and" and commas
    return num2words(n).replace(" and", "").replace(", ", " ")
def agen(): # generator of terms
    names = [] # a sorted list
    for n in count(1):
        insort(names, (n2w(n), n-1))
        fixed = [j+1 for j in range(n) if names[j][1] == j]
        yield len(fixed) # use "yield fixed" for list of fixed points
print(list(islice(agen(), 87))) # Michael S. Branicky, Jul 05 2024

A119796 Zero through ten in alphabetical order of English reverse spelling.

Original entry on oeis.org

3, 9, 1, 5, 10, 7, 0, 2, 4, 8, 6

Offset: 1

Jonathan Vos Post, Jul 30 2006

			a(1) = 3 because EERHT comes alphabetically first.
a(2) = 9 because ENIN comes alphabetically second.
a(3) = 1 because ENO comes alphabetically third.
a(4) = 5 because EVIF comes alphabetically fourth.

Cf. A000052, A004740, A019440, A026081, A001058, A057846, A108067, A072991.

A119898 1-digit numbers arranged in alphabetical order of English spelling reversed, then the 2-digit numbers so arranged, then the 3-digit numbers, etc.

Original entry on oeis.org

3, 9, 1, 5, 7, 0, 2, 4, 8, 6, 93, 53, 83, 73, 23, 33, 43, 63, 99, 59, 89, 79, 29, 39, 49, 69, 91, 51, 81, 71, 21, 31, 41, 61, 95, 55, 85, 75, 25, 35, 45, 65, 12, 19, 15, 18, 17, 13, 14, 16, 10, 11, 97, 57, 87, 77, 27, 37, 47, 67, 92, 52, 82, 72, 22, 32, 42, 62, 94, 54, 84, 74, 24

Offset: 1

Jonathan Vos Post, Aug 01 2006

Analog of A000052 where the alphabetical order is applied not to English name of integer but to the English name whose letters are then reversed.

			Eerht, Enin, Eno, Evif, Neves, Orez, Owt, Ruof, Thgie, Xis;
Eerhtytenin, Eerhtytfif, Eerhtythgie, Eerhtytneves, Eerhtytnewt, Eerhtytriht, Eerhtytrof, Eerhtytxis, Eninytenin, Eninytfif, Eninythgie, Eninytneves, Eninytnewt, Eninytriht, Eninytrof, Eninytxis, Enoytenin, Enoytfif, Enoythgie, Enoytneves, Enoytnewt, Enoytriht, Enoytrof, Enoytxis, Evifytenin, Evifytfif, ...

Cf. A000052, A004740, A019440, A026081, A001058, A057846, A108067, A072991.

f[a_, b_] := Sort@ Table[{ StringReverse[ StringReplace[ IntegerName[h, "Words"], {"\[Hyphen]" -> ""}]], h}, {h, a, b}]; Last /@ Join[f[0, 9], f[10, 99]] (* Giovanni Resta, Jun 13 2016 *)

Data and example corrected by Giovanni Resta, Jun 13 2016

A290483 Integers in alphabetical order in Spanish, using the long scale.

Original entry on oeis.org

14, 14000000000000, 14000000000014, 14000000014000, 14000000014014, 14000000014100, 14000000014114, 14000000014105, 14000000014150, 14000000014155, 14000000014154, 14000000014152, 14000000014159, 14000000014158, 14000000014156, 14000000014157, 14000000014153, 14000000014151, 14000000014140, 14000000014145

Offset: 1

Jose M. Arenas, Aug 03 2017

This sequence uses the long scale of numbers, so 10^9 is Thousand Millions (Mil Millones) and not One Billion (Un Billón).

			14 = "catorce".
14000000000000 = "catorce billones".
14000000000014 = "catorce billones catorce".
14000000014000 = "catorce billones catorce mil".
14000000014014 = "catorce billones catorce mil catorce".
14000000014100 = "catorce billones catorce mil cien".
14000000014114 = "catorce billones catorce mil ciento catorce".
14000000014105 = "catorce billones catorce mil ciento cinco".
14000000014150 = "catorce billones catorce mil ciento cincuenta".
14000000014155 = "catorce billones catorce mil ciento cincuenta y cinco".
14000000014154 = "catorce billones catorce mil ciento cincuenta y cuatro".
14000000014152 = "catorce billones catorce mil ciento cincuenta y dos".
14000000014159 = "catorce billones catorce mil ciento cincuenta y nueve".
14000000014158 = "catorce billones catorce mil ciento cincuenta y ocho".
14000000014156 = "catorce billones catorce mil ciento cincuenta y seis".
14000000014157 = "catorce billones catorce mil ciento cincuenta y siete".
14000000014153 = "catorce billones catorce mil ciento cincuenta y tres".
14000000014151 = "catorce billones catorce mil ciento cincuenta y uno".
14000000014140 = "catorce billones catorce mil ciento cuarenta".
14000000014145 = "catorce billones catorce mil ciento cuarenta y cinco".

Jose M. Arenas, Table of n, a(n) for n = 1..10000

Cf. A004740, A019440, A097960, A108067, A126686, A161390.

cp's OEIS Frontend

A108067 Integers in alphabetical order in U.S. English (another version of A004740).

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A019440 Integers in alphabetical order in British English.

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A026081 Integers in reverse alphabetical order in U.S. English.

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A152611 Assign weights to the nonnegative integers as in A073327, then sort them by weight.

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A127352 Integers less than 10^303 in reverse alphabetical order in U.S. English.

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A340671 a(n) is the number of values m such that, if the first n positive integers are arranged in alphabetical order in US English, the m-th term in the order is equal to m.

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Python

A119796 Zero through ten in alphabetical order of English reverse spelling.

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A119898 1-digit numbers arranged in alphabetical order of English spelling reversed, then the 2-digit numbers so arranged, then the 3-digit numbers, etc.

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Mathematica

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A290483 Integers in alphabetical order in Spanish, using the long scale.

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