A005589 -id:A005589 - cp's OEIS frontend

A281191 Number of holes in the (American) English name of n (as printed in lower case).

2, 2, 1, 2, 1, 1, 0, 2, 3, 1, 1, 3, 2, 2, 3, 2, 2, 4, 5, 3, 1, 3, 2, 3, 2, 2, 1, 3, 4, 2, 0, 2, 1, 2, 1, 1, 0, 2, 3, 1, 1, 3, 2, 3, 2, 2, 1, 3, 4, 2, 0, 2, 1, 2, 1, 1, 0, 2, 3, 1, 0, 2, 1, 2, 1, 1, 0, 2, 3, 1, 2, 4, 3, 4, 3, 3, 2, 4, 5, 3, 3, 5, 4, 5, 4, 4, 3, 5, 6, 4, 1, 3, 2, 3, 2, 2, 1, 3, 4, 2, 5, 7, 6, 7, 6

Offset: 0

Rick L. Shepherd, Jan 16 2017

For this sequence a font is used where a, b, d, e, o, p, and q each have one hole, g has two, and all other letters have no holes.

			The term a(101) = 7 because the name "one hundred one" contains seven total holes in these letters: o, e, d, e, d, o, and e.

Alois P. Heinz, Table of n, a(n) for n = 0..10000

Cf. A005589, A249572.

a:= n-> (s-> add((t-> `if`(t in {"a", "b", "d", "e", "o", "p", "q"}, 1,
        `if`(t="g", 2, 0)))(s[i]), i=1..length(s)))(convert(n, english)):
seq(a(n), n=0..104);  # Alois P. Heinz, Jul 30 2023

A305403 Number of Ukrainian letters in Ukrainian name of n.

Original entry on oeis.org

4, 4, 3, 3, 6, 4, 5, 3, 5, 6, 6, 10, 10, 10, 12, 10, 11, 10, 12, 12, 8, 12, 11, 11, 14, 12, 13, 11, 13, 14, 8, 12, 11, 11, 14, 12, 13, 11, 13, 14, 5, 9, 8, 8, 11, 9, 10, 8, 10, 11, 8, 12, 11, 11, 14, 12, 13, 11, 13, 14, 9, 13, 12, 12, 15, 13, 14, 12, 14, 15, 8, 12, 11, 11, 14, 12, 13, 11, 13, 14, 10, 14, 13

Offset: 0

Felix Fröhlich, May 31 2018

Apostrophes, when the name is written with Ukrainian letters, are not counted as letters. If they were, then, for example, a(5) would be 5, but here, a(5) = 4, because the apostrophe is ignored.

			Using the BGN/PCGN romanization system, the names of numbers are nul', odyn, dva, try, chotyry, .... Note that the number of letters in the romanized name of n does not necessarily coincide with the number of letters when the name of n is written in Ukrainian script.

E-Hern Lee, Table of n, a(n) for n = 0..10000
MyLanguages, Ukrainian numbers
Of Languages and Numbers, Counting in Ukranian
Wikipedia, List of numbers in various languages - Slavic languages
Wikipedia, Romanization of Ukrainian
Wikipedia, Ukrainian alphabet
Index entries for sequences related to names of numbers

Cf. A005589 (English), A007005 (French), A026858 (Italian), A006994 (Russian), A011762 (Spanish).

(* only works up to 999999 *)
upto20 = {0, 4, 3, 3, 6, 4, 5, 3, 5, 6, 6, 10, 10, 10, 12, 10, 11, 10, 12, 12, 8}
tens = {0, 6, 8, 8, 5, 8, 9, 8, 10, 9}
hundreds = {0, 3, 6, 6, 9, 6, 7, 6, 8, 8}
thousands = {0, 10, 9, 9, 12}
f[x_] := If[x == 0, 4, If[x > 999 && x < 5000, thousands[[Quotient[x, 1000] + 1]] + f[Mod[x, 1000]],
  If[
    x < 1000,
    If[Mod[x, 100] <= 20, upto20[[Mod[x, 100] + 1]], upto20[[Mod[x, 10] + 1]] + tens[[Mod[Quotient[x, 10], 10] + 1]]] + hundreds[[Mod[Quotient[x, 100], 10] + 1]] + thousands[[Mod[Quotient[x, 1000], 10] + 1]],
    5 + f[Quotient[x, 1000]] + If[Mod[x, 1000] == 0, 0, f[Mod[x, 1000]]]
  ]
]]
For[i = 0, i <= 10000, i++,
  j = 0;
  Print[i, " ", f[i]]
] (* E-Hern Lee, Jul 11 2018 *)

More terms from E-Hern Lee, Jul 11 2018

A309750 Number of letters in the English names of the months when the names are arranged in alphabetical order.

Original entry on oeis.org

4, 5, 6, 3, 3, 5, 3, 5, 4, 6, 3, 4

Offset: 1

Robert Dibley, Aug 15 2019

In alphabetical order, the months of the Gregorian calendar are: April (4), August (8), December (12), February (2), January (1), July (7), June (6), March (3), May (5), November (11), October (10), September (9).

Based originally on A260249 but with the additional layer of converting numbers to words and taking their lengths.

Cf. A005589, A260249.

Sort[Array[{DateValue[#, "MonthName"], StringLength@ IntegerName@ DateValue[#, "Month"]} &@ DateObject[{1, #, 1}] &, 12]][[All, -1]] (* Michael De Vlieger, Sep 04 2019 *)

a(n) = A005589(A260249(n)).

A356675 Lexicographically earliest infinite sequence satisfying a(1) > -1 and a(n-1) = A075826(a(n)).

Original entry on oeis.org

1, 5, 9, 16, 27, 38, 48, 58, 66, 76, 87, 98, 117, 136, 155, 177, 198, 215, 235, 254, 275, 295, 310, 333, 350, 372, 394, 411, 433, 452, 474, 495, 514, 535, 555, 576, 598, 615, 635, 650, 669, 689, 705, 728, 749, 773, 795, 810, 833, 850, 872, 894, 913, 934, 950, 973, 994, 1013, 1034, 1050, 1071, 1093

Offset: 1

Aidan Clarke, Aug 22 2022

There is at least one infinite sequence that satisfies the parameters of the sequence, because it is always true that A075826(k) < k for any k. (For the statement to be false, the number of letters must be less than or equal to 0.)

All known values in this sequence are guaranteed to be in any infinite sequence which can be produced, since iterating A075826(k) where the starting value of k is any number greater than 11126 converges at 6890 or larger. This proves any infinite sequence must contain 6890 and all values returned by iterating A075826 down to 0.

			A075826 yields the number of letters in US English spelling of each number subtracted from the number's value. For example, A075826(5) yields 1, because 5 - 4 (F-I-V-E) is 1. Our formula reveals that if 5 is in our sequence, it must come immediately after 1, which it does. However, 4, which must come immediately after 0, is not in our sequence, which we can prove because there exists no number k such that A075826(k) = 4. Each number less than the largest known value in this sequence produces a finite sequence unless it is in this sequence, because some number eventually is a dead end like 4.

Michel Marcus, Table of n, a(n) for n = 1..11623

Cf. A005589, A075826.

f(k) = k - A005589(k);
listd(nn) = {nn *= 2; my(vs = vector(nn)); my(list = List()); my(m=1, lbound = 0); listput(list, m); while (m < nn, if (vs[m] == 0, vs[m] = Vec(select(x->(x==m), vector(100, k, f(k+m-1)), 1)); if (#vs[m], vs[m] = apply(x->(x+m-1), vs[m]))); my(ok = 1, vc = vs[m]); if (! #vc, ok = 0, vc = select(x->(x>lbound), vc); if (! #vc, ok = 0);); if (!ok, lbound = list[#list]; listpop(list); if (! #list, return()); m = list[#list];, lbound = 0; m = vc[1]; listput(list, m););); Vec(select(x->(x<=nn/2), list));} \\ Michel Marcus, Aug 31 2022

a(n-1) = A075826(a(n)).

A362441 a(1) = 6; thereafter a(n) = smallest number with a(n-1) letters in British English.

Original entry on oeis.org

6, 11, 23, 124, 113323371373

Offset: 1

N. J. A. Sloane, Apr 22 2023

The GCHQ Puzzle Book gives an alternative for a(5).
a(6) is probably not well defined.
See A362442 for an American English version.

GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See pages 92 and 275.

Cf. A005589, A362123, A362442.

A362442 a(1) = 6; thereafter a(n) = smallest number with a(n-1) letters in American English.

Original entry on oeis.org

6, 11, 23, 323, 1103323373373373373373373373373

Offset: 1

N. J. A. Sloane, Apr 22 2023

a(5) should be findable, but a(6) will probably not be well defined.

See A362441 for a British English version.

			a(2) = 11 since "eleven" is the smallest number with 6 letters.
a(3) = 23 since "twenty three" is the smallest with 11 letters.

GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See pages 92 and 275.

Cf. A005589, A362123, A362441.

a(n) = A080777(a(n-1)) for n > 1. - Michael S. Branicky, Apr 22 2023

a(5) from Michael S. Branicky, Apr 22 2023 using A080777

A384131 Smallest positive number divisible by n that has n letters in US English, or 0 if none exists.

Original entry on oeis.org

6, 4, 40, 12, 70, 56, 36, 100, 33, 300, 1000000001, 406, 150, 112, 170, 162, 418, 11020, 336, 528, 828, 4800, 3300, 1404, 1620, 1512, 1218, 1770, 1147, 1344, 1353, 2788, 3325, 3888, 12728, 13376, 13338, 103360, 22878, 23478, 27778, 101728, 103725, 111734, 111578

Offset: 3

Jason Bard, May 20 2025

The GCHQ reference uses British English (including "and"). The disagreements are at a(14)-a(16), which the book lists as 70000000, 15000, 14000. Furthermore, the book lists a(13) as impossible.

GCHQ, The GCHQ Puzzle Book, Penguin, 2016 (see p. 47).

Michael S. Branicky, Table of n, a(n) for n = 3..100 (terms 3..69 from Jason Bard)

Cf. A005589, A134629.

mmax = 10^10; Do[m = n; While[StringLength[StringDelete[IntegerName[m, "Words"], {" ", "-", "\[Hyphen]", ","}]] != n, m += n; If[m > mmax, m = 0; Break[]]]; Print[m], {n, 3, 43}]

from num2words import num2words
from itertools import count, islice
def b(n): return sum(1 for c in num2words(n).replace(" and", "") if c.isalpha())
def a(n): return next(k for k in count(n, n) if b(k) == n)
print([a(n) for n in range(3, 13)]) # Michael S. Branicky, May 20 2025

a(n) >= A134629(n). - Michael S. Branicky, May 21 2025

A385513 The numbers of people in the "SpellUnder-Down" variant of the Josephus problem such that the last person is freed.

Original entry on oeis.org

1, 6, 7, 105, 181, 215, 821, 1907, 3176, 23388, 55058

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Jul 01 2025

In SpellUnder-Down dealing, we spell the number of the next card, putting a card under for each letter in the number, then we deal the next card. So we start with putting 3 cards under, for O-N-E, then deal, then 3 under for T-W-O, then deal, then 5 under for T-H-R-E-E, then deal. The dealing sequence is highly irregular because it depends on English spelling. The dealing pattern starts: UUUDUUUDUUUUUD. In the corresponding Josephus problem, we skip the next person for each under dealing, and eliminate the next person for each down dealing.
This sequence can be used in magic tricks with the SpellUnder-Down dealing pattern. The deck sizes in this sequence guarantee that after the dealing, the last card dealt is the one that was initially on the bottom.
The classical Josephus problem corresponds to under-down dealing. In this case, the last person is freed when the number of people is a power of 2 minus 1.
A naive probabilistic argument predicts the probability that A380204(k) = k is 1/k and expects this sequence to be infinite and distributed roughly as A002387. - Michael S. Branicky, Jul 24 2025

			Suppose there are 5 people in a circle. We start with skipping three people for O-N-E. After three people are skipped, the person number 4 is eliminated. The leftover people are 5,1,2,3 in order. Then we skip three people for T-W-O. The person number 3 eliminated, and the leftover people are 5,1,2 in order. Then we skip 5 people for T-H-R-E-E, and person number 2 is eliminated, and the leftover people are 5,1 in order. Then we skip four people for F-O-U-R. person number 5 is eliminated. Person 1 is freed. As person 1 is not last, 5 is NOT in this sequence.

Cf. A005589, A006257, A182459, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248, A385328.

{k | A380204(k) = k}. - Michael S. Branicky, Jul 24 2025

a(10)-a(11) from Michael S. Branicky, Jul 24 2025

A011752 Repeat n m times, where m is number of letters in English word for the number.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16

Offset: 0

Karl J. Schmerbauch (karl.j.schmerbauch(AT)boeing.com)

			Zero has four letters so sequence begins with four 0's, then three 1's since one has three letters, ...

Cf. A005589.

More terms from Kristen Bollmeier (s1168905(AT)cedarville.edu)

A051785 Number of letters in n-th Catalan number.

Original entry on oeis.org

2, 3, 4, 6, 4, 3, 3, 4, 3, 3, 4, 5, 6, 7, 6, 5, 6, 6, 5, 4, 7, 8, 9, 11, 9, 8, 8, 9, 8, 6, 8, 9, 10, 12, 10, 9, 9, 10, 9, 8, 10, 11, 12, 14, 12, 11, 11, 12, 11, 9, 11, 12, 13, 15, 13, 12, 12, 13, 12, 8, 10, 11, 12, 14, 12, 11, 11, 12, 11, 7, 9, 10, 11, 13, 11, 10, 10, 11, 10, 8, 10, 11

Offset: 1

N. J. A. Sloane and Richard Borcherds, Dec 09 1999

			In Catalan: un, dos, tres, quatre, cinc, sis, set, vuit, nou, deu, onze, dotze, tretze, catorze, quinze, setze, disset, divuit, dinou, vint, ...

J. Gili, Catalan Grammar, Dolphin. Oxford 1993, p. 39.
R. P. Stanley, Enumerative Combinatorics, vol. 2, problem 6.24.

Amer. Math. Monthly 103 (1996), see pages 538 and 577.
Catalan Institute of America, Home Page
Index entries for sequences related to number of letters in n

Cf. A000108, A005589.

More terms from F. Xavier Noria (fxn(AT)retemail.es) and Manuel Valdivia Prades (mvaldivia(AT)ugr.es), Dec 10 1999

cp's OEIS Frontend

A281191 Number of holes in the (American) English name of n (as printed in lower case).

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A305403 Number of Ukrainian letters in Ukrainian name of n.

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A309750 Number of letters in the English names of the months when the names are arranged in alphabetical order.

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A356675 Lexicographically earliest infinite sequence satisfying a(1) > -1 and a(n-1) = A075826(a(n)).

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A362441 a(1) = 6; thereafter a(n) = smallest number with a(n-1) letters in British English.

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A362442 a(1) = 6; thereafter a(n) = smallest number with a(n-1) letters in American English.

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A384131 Smallest positive number divisible by n that has n letters in US English, or 0 if none exists.

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A385513 The numbers of people in the "SpellUnder-Down" variant of the Josephus problem such that the last person is freed.

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