A005589 -id:A005589 - cp's OEIS frontend

A109382 Levenshtein distance between successive English names of nonnegative integers, excluding spaces and hyphens.

4, 3, 4, 5, 3, 3, 4, 5, 4, 3, 4, 4, 6, 3, 3, 2, 4, 4, 3, 7, 3, 3, 4, 5, 3, 3, 4, 5, 4, 7, 3, 3, 4, 5, 3, 3, 4, 5, 4, 7, 3, 3, 4, 5, 3, 3, 4, 5, 4, 6, 3, 3, 4, 5, 3, 3, 4, 5, 4, 6, 3, 3, 4, 5, 3, 3, 4, 5, 4, 7, 3, 3, 4, 5, 3, 3, 4, 5, 4, 8, 3, 3, 4, 5, 3, 3, 4, 5, 4, 7, 3, 3, 4, 5, 3, 3, 4, 5, 4, 7, 3, 3, 4, 5, 3, 3

Offset: 0

Jonathan Vos Post, Aug 25 2005

			a(0) = 4 since LD(ZERO,ONE) requires 4 edits.
a(1) = 3 since LD(ONE,TWO) which requires 3 substitutions.
a(2) = 4 since LD(TWO,THREE) = requires 4 edits (leave the leftmost T unchanged), then 2 substitutions (W to H, O to R), then 2 insertions (E,E).
a(4) = 3 as LD(FOUR,FIVE) leaves the leftmost F unchanged, then requires 3 substitutions. From FIVE to SIX leaves the I unchanged. From SIX to SEVEN leaves the S unchanged. From TEN to ELEVEN leaves the EN unchanged. From ELEVEN to TWELVE leaves an E,L,V,E unchanged. From THIRTEEN to FOURTEEN leaves RTEEN unchanged. TWENTYNINE to THIRTY takes 7 edits. THIRTYNINE to FORTY takes 7 edits. SEVENTYNINE to EIGHTY takes 8 edits. EIGHTYNINE to NINETY takes 7 edits. NINETYNINE to ONEHUNDRED takes 7 edits.

Robert Israel, Table of n, a(n) for n = 0..10000
Michael Gilleland, Levenshtein Distance, in Three Flavors.
V. I. Levenshtein, Efficient reconstruction of sequences from their subsequences or supersequences, J. Combin. Theory Ser. A 93 (2001), no. 2, 310-332.
Landon Curt Noll, The English Name of a Number.
Robert G. Wilson v, American English names for the numbers from 0 to 100999 without spaces or hyphens.

Cf. A001477, A005589, A081355, A081356, A081230, A109809, A109811.

with(StringTools):
seq(Levenshtein(Select(IsAlpha, convert(n,english)),Select(IsAlpha,convert(n+1,english))),n=0..200); # Robert Israel, Jan 23 2018

(* First copy b109382.txt out of A109382 then *) levenshtein[s_List, t_List] := Module[{d, n = Length@s, m = Length@t}, Which[s === t, 0, n == 0, m, m == 0, n, s != t, d = Table[0, {m + 1}, {n + 1}]; d[[1, Range[n + 1]]] = Range[0, n]; d[[Range[m + 1], 1]] = Range[0, m]; Do[ d[[j + 1, i + 1]] = Min[d[[j, i + 1]] + 1, d[[j + 1, i]] + 1, d[[j, i]] + If[ s[[i]] === t[[j]], 0, 1]], {j, m}, {i, n}]; d[[ -1, -1]] ]]; f[x_] := Block[{str = ToString@ lst[[x]], len}, len = StringLength@ str; StringInsert[str, ",", Range[2, len]]]

a(n) = LD(nameof(n), nameof(n+1)).

More terms from Robert G. Wilson v, Jan 31 2006

Corrected by Robert Israel, Jan 23 2018

A139121 Total number of letters in the preceding terms spelled out in French.

Original entry on oeis.org

0, 4, 10, 13, 19, 26, 34, 46, 57, 70, 81, 94, 113, 123, 137, 151, 168, 184, 205, 217, 232, 250, 267, 287, 310, 322, 340, 357, 379, 403, 418, 435, 455, 478, 503, 516, 529, 546, 565, 585, 608, 619, 633, 651, 671, 692, 715, 729, 746, 765, 785, 808, 820, 833, 852, 873, 895, 920, 933, 952, 973, 995, 1020

Offset: 1

N. J. A. Sloane (based on Angelini's article), Jun 08 2008, Jun 15 2008

Form a sequence of French words as follows: look to the left, towards the beginning of the sequence and write down the number of letters you see; repeat; then replace the words by the corresponding numbers.
The sequence of words is: zero, quatre, dix, treize, dix-neuf, vingt-six, trente-quatre, quarante-six, cinquante-sept, ...
Hyphens, accents and spaces are not counted.
For an English version see A139097.

			The first word is "zero", because initially there are no letters to the left. The second word is "quatre" (and so a(2)=4), because at the end of the first word we can see four letters to the left. And so on.

E. Angelini, "Jeux de suites", in Dossier Pour La Science, pp. 32-35, Volume 59 (Jeux math'), April/June 2008, Paris.

Cf. A005589, A060403, A139097.

A139121(n)={ n>1 & return( #select(Vec(French(n=A139121(n-1))),x->x>"@")+n )}  /* see A167507 for French() */  \\ M. F. Hasler, Sep 29 2011

Offset and a(9) corrected (according to wording of example) and terms beyond a(9) from M. F. Hasler, Sep 29 2011

A381114 Triangle read by rows: T(n,k) is the number of the k-th eliminated person in a variation of the Josephus elimination process for n people, where the number of people skipped is equal to the number of letters in the previous number's English name.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 1, 2, 4, 3, 1, 5, 2, 4, 3, 1, 5, 4, 3, 6, 2, 1, 5, 3, 6, 2, 4, 7, 1, 5, 2, 3, 4, 7, 6, 8, 1, 5, 9, 8, 7, 2, 3, 6, 4, 1, 5, 9, 7, 4, 3, 2, 10, 8, 6, 1, 5, 9, 6, 2, 10, 7, 11, 8, 3, 4, 1, 5, 9, 4, 11, 7, 2, 3, 8, 12, 10, 6, 1, 5, 9, 3, 10, 4, 11, 8, 12, 13, 2, 6, 7, 1, 5, 9, 2, 8, 14, 7, 4, 3, 13, 12

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Feb 14 2025

In this variation of the Josephus elimination process, people numbered 1 through n are arranged in a circle. A pointer starts at person 1. Person 1 is eliminated, then three people are skipped because the number O-N-E has three letters. Then, the next person is eliminated. Next, three people are skipped because T-W-O has three letters, and the next person is eliminated. Then, five people are skipped because T-H-R-E-E has five letters, and so on. This repeats until no people remain. This sequence represents the triangle T(n, k), the order of elimination for the person numbered k in a circle of n people.
Every entry of the first column is 1.
In rows 5 and after, the second number is 5. In rows 9 and after, the third number is 9. In rows 15 and after, the fourth number is 15. In the limit, the numbers in a row form sequence A381128(k).

			Triangle begins:
 1;
 1, 2;
 1, 3, 2;
 1, 2, 4, 3;
 1, 5, 2, 4, 3;
 1, 5, 4, 3, 6, 2;
 1, 5, 3, 6, 2, 4, 7;
 ...
For n = 4, suppose four people are arranged in a circle. The first person is eliminated, then three people are skipped because O-N-E has three letters. The leftover people are now in order 2,3,4. Then, the next person (numbered 2) is eliminated. The leftover people are now ordered 3,4. The next three people are skipped, so the person eliminated is number 4. Then, 5 people are skipped, and the next person eliminated is number 3. Thus, the fourth row of the triangle is 1,2,4,3.

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248, A381127, A381128, A381129.

from num2words import num2words as n2w
def f(n): return sum(1 for c in n2w(n).replace(" and", "") if c.isalpha())
def row(n):
    c, i, J, out = 1, 0, list(range(1, n+1)), []
    while len(J) > 1:
        q = J.pop(i)
        out.append(q)
        i = (i + f(c))%len(J)
        c = c+1
    return out + [J[0]]
print([e for n in range(1, 14) for e in row(n)]) # Michael S. Branicky, Feb 16 2025

A381127 Triangle T(n,k) read by rows, where row n is a permutation of the numbers 1 through n, such that if a deck of n cards is prepared in this order, and Down-SpellUnder dealing is used, then the resulting cards will be dealt in increasing order.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 1, 2, 4, 3, 1, 3, 5, 4, 2, 1, 6, 4, 3, 2, 5, 1, 5, 3, 6, 2, 4, 7, 1, 3, 4, 5, 2, 7, 6, 8, 1, 6, 7, 9, 2, 8, 5, 4, 3, 1, 7, 6, 5, 2, 10, 4, 9, 3, 8, 1, 5, 10, 11, 2, 4, 7, 9, 3, 6, 8, 1, 7, 8, 4, 2, 12, 6, 9, 3, 11, 5, 10, 1, 11, 4, 6, 2, 12, 13, 8, 3, 5, 7, 9, 10, 1, 4, 9, 8, 2, 12, 7, 5, 3, 13, 14, 11, 10, 6

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Feb 14 2025

In Down-SpellUnder dealing, we deal the first card, then spell the positive integers starting from O-N-E, moving a card from the top of the deck underneath the deck for each letter in the English spelling of the number, followed by dealing or "putting down" the top card. So we start by dealing the first card, then putting 3 cards under because O-N-E has three letters, then we deal the next card. Then we put 3 cards under because T-W-O has three letters, then we deal a card. Then we put 5 cards under for T-H-R-E-E, and so on. This dealing sequence is highly irregular because it depends on English spelling. The dealing pattern starts: DUUUDUUUDUUUUUD, where each "U" corresponds to putting a card “under” and each "D" corresponds to dealing a card “down”.
This card dealing can be thought of as a generalized version of the Josephus problem. In this version of the Josephus problem, we start by executing the first person, then spell the positive integers in increasing order, each time skipping past 1 person for each letter and executing the next person. The card in row n and column k is x if and only if in the corresponding Josephus problem with n people, the person numbered x is the k-th person eliminated.
Equivalently, each row of the corresponding Josephus triangle A381114 is an inverse permutation of the corresponding row of this triangle. The first column contains only ones, since person number 1 always dies first in the corresponding Josephus problem. The index of the largest number in row n is A381129(n), corresponding to the index of the freed person in the corresponding Josephus problem. The number of card moves that we need to take if we start with n cards is A381128(n).

			Triangle begins:
 1;
 1, 2;
 1, 3, 2;
 1, 2, 4, 3;
 1, 3, 5, 4, 2;
 1, 6, 4, 3, 2, 5;
 1, 5, 3, 6, 2, 4, 7;
 1, 3, 4, 5, 2, 7, 6, 8;
  ...
For n = 4, suppose there are four cards arranged in the order 1,2,4,3. Card 1 is dealt, and then three cards go under the deck because O-N-E has three letters. Now, the deck is ordered 2,4,3. Card 2 is dealt, and three cards go under because T-W-O has three letters. Now, the leftover deck is ordered 3,4. Card 3 is dealt, and five cards go under because T-H-R-E-E has five letters. Then card 4 is dealt. The dealt cards are in numerical order. Thus, the fourth row of the triangle is 1, 2, 4, 3.

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248, A381114, A381128, A381129.

from num2words import num2words as n2w
def f(n): return sum(1 for c in n2w(n).replace(" and", "") if c.isalpha())
def row(n):
    c, i, J, out = 1, 0, list(range(1, n+1)), []
    while len(J) > 1:
        q = J.pop(i)
        out.append(q)
        i = (i + f(c))%len(J)
        c = c+1
    out.append(J[0])
    return [out.index(j)+1 for j in list(range(1, n+1))]
print([e for n in range(1, 15) for e in row(n)]) # Michael S. Branicky, Feb 16 2025

A381128 The number of card moves required to deal n cards using Down-SpellUnder dealing.

Original entry on oeis.org

1, 5, 9, 15, 20, 25, 29, 35, 41, 46, 50, 57, 64, 73, 82, 90, 98, 108, 117, 126, 133, 143, 153, 165, 176, 187, 197, 209, 221, 232, 239, 249, 259, 271, 282, 293, 303, 315, 327, 338, 344, 353, 362, 373, 383, 393, 402, 413, 424, 434, 440, 449, 458, 469, 479, 489, 498, 509, 520, 530, 536, 545, 554, 565, 575, 585, 594, 605

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Feb 14 2025

In Down-SpellUnder dealing, after dealing the i-th card we move a card from the top of the deck to the bottom for each letter in the English spelling of i. Then we deal the next card and proceed likewise. So we start by dealing card 1, then putting 3 cards under because O-N-E has three letters. Then, we deal the next card, and put three cards under 3 because T-W-O has three letters. We then deal again, put 5 under for T-H-R-E-E, and so on. This dealing sequence is highly irregular because it depends on English spelling. The dealing pattern starts: DUUUDUUUDUUUUUD, where D means 'deal', and U means 'under'.

			The dealing pattern to deal four cards is DUUUDUUUDUUUUUD. It contains 15 letters, so a(4) = 15.

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248, A381114, A381127, A381129.

from num2words import num2words as n2w
def spell(n):
    return sum(1 for c in n2w(n).replace(" and", "").replace(" ", "").replace(",", "").replace("-", ""))
moves_so_far = 0
l = []
for i in range(1, 41):
    moves_so_far += 1
    l += [str(moves_so_far)]
    moves_so_far += spell(i)
print(", ".join(l))

a(n) = A067278(n-1) + n.

A381129 A version of the Josephus problem: a(n) is the surviving integer under the spelling version of the elimination process, called Down-SpellUnder.

Original entry on oeis.org

1, 2, 2, 3, 3, 2, 7, 8, 4, 6, 4, 6, 7, 11, 10, 3, 14, 4, 17, 11, 3, 16, 7, 16, 7, 22, 2, 8, 24, 27, 7, 21, 13, 28, 30, 8, 3, 37, 12, 7, 8, 33, 7, 33, 44, 11, 32, 8, 6, 43, 2, 18, 49, 8, 32, 54, 26, 43, 44, 30, 40, 52, 26, 44, 8, 27, 60, 16, 11, 61, 70, 14, 58, 55

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Feb 14 2025

Arrange n people numbered 1, 2, 3, ..., n in a circle, increasing clockwise. Execute the person numbered 1, then spell the letters of O-N-E, moving one person clockwise for each letter. Once you are done, eliminate the next person. Then, spell the letters of T-W-O; in other words, skip three people and eliminate the next person. Following this, spell the letters of T-H-R-E-E; in other words, skip five people and eliminate the next person. Continue until one person remains. The number of this person is a(n).

			Consider n = 4 people. The first person eliminated is number 1. This leaves the remaining people in the order 2, 3, 4. The second person eliminated is number 2; the people left are in the order 3, 4. The next person eliminated is numbered 4, leaving only the person numbered 3. Thus, a(4) = 3.

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248, A381114, A381127, A381128.

from num2words import num2words as n2w
def f(n): return sum(1 for c in n2w(n).replace(" and", "") if c.isalpha())
def a(n):
    c, i, J = 1, 0, list(range(1, n+1))
    while len(J) > 1:
        q = J.pop(i)
        i = (i + f(c))%len(J)
        c = c+1
    return J[0]
print([a(n) for n in range(1, 75)]) # Michael S. Branicky, Feb 16 2025

a(22) and beyond from Michael S. Branicky, Feb 16 2025

A001161 Numbers containing an even number of letters.

Original entry on oeis.org

0, 4, 5, 9, 11, 12, 13, 14, 18, 19, 20, 24, 25, 29, 30, 34, 35, 39, 41, 42, 43, 46, 47, 48, 51, 52, 53, 56, 57, 58, 61, 62, 63, 66, 67, 68, 71, 72, 73, 76, 77, 78, 80, 84, 85, 89, 90, 94, 95, 99

Offset: 1

William P. Everts [ bille(AT)regenisys.com ]

Cf. A005589, complement of A001162.

A036448 Smallest positive number containing n e's when spelled out in US English.

Original entry on oeis.org

2, 1, 3, 11, 17, 111, 117, 317, 1317, 3317, 11317, 17317, 111317, 117317, 317317, 1317317, 3317317, 11317317, 17317317, 111317317, 117317317, 317317317, 1317317317, 3317317317, 11317317317, 17317317317, 111317317317, 117317317317, 317317317317, 1317317317317, 3317317317317

Offset: 0

Rodolfo Kurchan

From Michael S. Branicky, Oct 24 2020: (Start)
"US English" connotes that no "and" is used ("one hundred one") and, importantly here, that the names of large numbers follow the "American system" (Weisstein link), also known as the short scale (Wikipedia link). The previous a(8) and a(9) were based on "eleven hundred and seventeen" and "seventeen hundred and seventeen", which are less common written forms (Wikipedia English numbers link). To make the sequence precise, the common written form is adopted ("one thousand one hundred seventeen"; Wilson link; A000052 Example). Thus, a(n) is the least m such that A085513(m)=n.
The sequence follows the pattern of 1(317)^n, 3(317)^n, 11(317)^n, 17(317)^n, 111(317)^n, 117(317)^n, 317(317)^n for n = 0 through 7 and whenever the largest named power has no "e". a(50) > 10^21 = "one sextillion" which is the first power name that has an "e", breaking the pattern. In that case, a(50) = 1117(317)^6 and a(51) = 1(317)^7. Whenever the largest power has 1 "e" it follows this pattern. If it has m>1 "e"'s, the first block of three is shifted lower to a(7-m). See Wikipedia link for Names of large numbers for power names.
(End)

			One has 1 e.
Three has 2 e's.

Rodolfo Marcelo Kurchan, Problem 1882, Another Number Sequence, Journal of Recreational Mathematics, vol. 23, number 2, p. 141.

Eric Weisstein's World of Mathematics, Large Number
Wikipedia, English numerals
Wikipedia, Names of Large Numbers
Robert G. Wilson v, American English names for the numbers from 0 to 100999 without spaces or hyphens

Cf. A000027, A000052, A001477, A005589, A006933, A037196, A085513.

from num2words import num2words
def A036448(n):
    i = 1
    while num2words(i).count("e")!=n:
        i += 1
    return i
print([A036448(n) for n in range(1,12)]) # Michael S. Branicky, Oct 23 2020

a(8)-a(9) changed and a(11)-a(30) added by Michael S. Branicky, Oct 23 2020

a(0)=2 inserted by Sean A. Irvine, Nov 02 2020

A049528 Number of letters in n-th prime (in English).

Original entry on oeis.org

3, 5, 4, 5, 6, 8, 9, 8, 11, 10, 9, 11, 8, 10, 10, 10, 9, 8, 10, 10, 12, 11, 11, 10, 11, 13, 15, 15, 14, 18, 21, 19, 21, 20, 19, 18, 20, 20, 20, 22, 21, 19, 19, 21, 21, 20, 16, 21, 21, 20, 21, 20, 18, 18, 20, 20, 19, 20, 22, 19, 21, 21, 17, 18, 20, 21, 21, 23

Offset: 1

G. L. Honaker, Jr.

			a(13) = 8 because 'forty-one' contains 8 letters (not counting the hyphen).

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Cf. A000040, A005589, A052360.

from sympy import nextprime
from itertools import count, islice
from num2words import num2words as n2w
def f(n): return sum(1 for c in n2w(n).replace(" and", "") if c.isalpha())
def agen(p=2):
    while True: yield f(p); p = nextprime(p)
print(list(islice(agen(), 68))) # Michael S. Branicky, Jul 12 2022

a(n) = A005589(A000040(n)) or a(n) = A052360(A000040(n)) depending on whether hyphens and spaces are excluded or included. - Jonathan Vos Post, Oct 19 2007

a(n) = A005589(A000040(n)) since it does not count spaces or hyphens. - Michael S. Branicky, Jul 12 2022

a(66) and beyond from Michael S. Branicky, Jul 12 2022

A075826 n minus (number of letters in English name of n).

Original entry on oeis.org

-4, -2, -1, -2, 0, 1, 3, 2, 3, 5, 7, 5, 6, 5, 6, 8, 9, 8, 10, 11, 14, 12, 13, 12, 14, 15, 17, 16, 17, 19, 24, 22, 23, 22, 24, 25, 27, 26, 27, 29, 35, 33, 34, 33, 35, 36, 38, 37, 38, 40, 45, 43, 44, 43, 45, 46, 48, 47, 48, 50, 55, 53, 54, 53, 55, 56, 58, 57, 58, 60, 63

Offset: 0

Jon Perry, Oct 14 2002

a(n) < 0 for n < 4, a(n) = 0 for n = 4 and a(n) > 0 for n > 4. - Bernard Schott, Feb 11 2020

The French variant would be n - A167507(n) = (-4, -1, -2, -2, -2, 1, 3, 3, 4, 5, 7, 7, 7, 7, 6, 9, 11, ...). The German variant would be n - A007208(n) = (-3, -2, -1, 0, 1, 1, 1, 4, 5, 6, 8, 7, 5, 6, 7, 8, 9, 10, ...). - M. F. Hasler, Mar 02 2020

			Seven contains 5 letters, therefore a(7) = 7 - 5 = 2.

GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See pages 49 and 214.

Cf. A005589.

A075826(n)=n-A005589(n) \\ M. F. Hasler, Mar 02 2020

a(n) = n - A005589(n). - Michel Marcus, Feb 11 2020

Corrected and extended by David W. Wilson, Jul 04 2005

cp's OEIS Frontend

A109382 Levenshtein distance between successive English names of nonnegative integers, excluding spaces and hyphens.

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A139121 Total number of letters in the preceding terms spelled out in French.

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A381114 Triangle read by rows: T(n,k) is the number of the k-th eliminated person in a variation of the Josephus elimination process for n people, where the number of people skipped is equal to the number of letters in the previous number's English name.

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A381127 Triangle T(n,k) read by rows, where row n is a permutation of the numbers 1 through n, such that if a deck of n cards is prepared in this order, and Down-SpellUnder dealing is used, then the resulting cards will be dealt in increasing order.

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A381128 The number of card moves required to deal n cards using Down-SpellUnder dealing.

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A381129 A version of the Josephus problem: a(n) is the surviving integer under the spelling version of the elimination process, called Down-SpellUnder.

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A001161 Numbers containing an even number of letters.

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A036448 Smallest positive number containing n e's when spelled out in US English.

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A049528 Number of letters in n-th prime (in English).