keyword:word - cp's OEIS frontend

A075774 Number of syllables in n in American English.

2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 2, 2, 2, 2, 3, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 3, 3, 2, 3, 3, 3, 3, 3, 3, 4, 3, 3, 2, 3, 3, 3, 3, 3, 3, 4, 3, 3, 2, 3, 3, 3, 3, 3, 3, 4, 3, 3, 2, 3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 4, 4, 2, 3, 3, 3, 3, 3, 3, 4, 3, 3, 2, 3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 4, 4, 4, 4

Offset: 0

Ethan B. Trewhitt, Oct 09 2002

Uses the convention of omitting a trailing 'and', so 101 is 'one hundred one' rather than 'one hundred and one.' - Eric W. Weisstein, May 11 2006
From Michael S. Branicky, May 28 2024: (Start)
The only numbers with a(n) = 1 are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12.
The only numbers with a(n) = 2 are 7, 13, 14, 15, 16, 18, 19, 20, 30, 40, 50, 60, 80, 90.
Those with a(n) = 3 and 4 are in A372807 and A180961, respectively. (End)

			a(76)=4 because seventy-six is split sev.en.ty.six, or four syllables.

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Number

Cf. A005589, A052360, A180961, A372807.

A075774(n, t=[10^9, 2, 10^6, 2, 1000, 2, 100, 2])={ n>99 && forstep( i=1, #t, 2, nA075774(n[1])+t[i+1]+if( n[2], A075774( n[2] )))); if( n<20, 1+!!setsearch(Set([0,7,13,14,15,16,18,19]),n) + 2*!!setsearch(Set([11,17]),n), 2+(n\10==7) + if(n%10, A075774(n%10)))}  \\ The "Set()" is not required in PARI v.2.6+ but we put it for downward compatibility. - M. F. Hasler, Nov 03 2013

def A075774(n):
    t = [(10**i, 2) for i in [12, 9, 6, 3, 2]]
    if n > 99:
        for ti, sti in t:
            if n >= ti:
                q, r = divmod(n, ti)
                return A075774(q) + sti + (A075774(r) if r else 0)
    if n < 20:
        return 1 + (n in {0, 7, 13, 14, 15, 16, 18, 19}) + 2*(n in {11, 17})
    else: return 2 + (n//10==7) + (A075774(n%10) if n%10 else 0)
print([A075774(n) for n in range(105)]) # Michael S. Branicky, Jun 27 2021 after M. F. Hasler

More terms from Eric W. Weisstein, May 11 2006

A000861 Numbers ending with a vowel in American English.

Original entry on oeis.org

0, 1, 2, 3, 5, 9, 12, 21, 22, 23, 25, 29, 31, 32, 33, 35, 39, 41, 42, 43, 45, 49, 51, 52, 53, 55, 59, 61, 62, 63, 65, 69, 71, 72, 73, 75, 79, 81, 82, 83, 85, 89, 91, 92, 93, 95, 99, 101, 102, 103, 105, 109, 112, 121, 122, 123, 125, 129, 131, 132, 133, 135, 139, 141, 142

Offset: 1

N. J. A. Sloane

The German name of n ends in a vowel iff n==2 or n==3 (mod 100), so the German analog of this sequence would be {2, 3, 102, 103, 202, 203, ...}. - M. F. Hasler, Sep 20 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-1).

Cf. A247359 (complement), A152592 (characteristic fct of the complement), A059437 (a variant of the latter: vowels or "y").

Cf. A247361 (French analog) and A247360 (complement).

print1(0",");for(n=1,3,for(k=10^(n-1),10^n-1,m100=k%100;m10=k%10;if(m100==12 || ((m100>20 || m100<10) && ((m10==1)||(m10==2)||(m10==3)||(m10==5)||(m10==9))),print1(k",")))) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), Nov 20 2006

is(n,S=[1,2,3,5,9,12])=setsearch(S,if(n%100>19,n%10,n%100))||!n \\ Add 0 in S (and optionally remove trailing ||...) for the A059437 variant. Add "Set()" around [...] for PARI versions < 2.6. - M. F. Hasler, Sep 14 2014

From Chai Wah Wu, Apr 18 2024: (Start)
a(n) = a(n-1) + a(n-46) - a(n-47) for n > 48.
G.f.: x^2*(x^46 + 4*x^45 + 2*x^44 + x^43 + x^42 + 2*x^41 + 4*x^40 + 2*x^39 + x^38 + x^37 + 2*x^36 + 4*x^35 + 2*x^34 + x^33 + x^32 + 2*x^31 + 4*x^30 + 2*x^29 + x^28 + x^27 + 2*x^26 + 4*x^25 + 2*x^24 + x^23 + x^22 + 2*x^21 + 4*x^20 + 2*x^19 + x^18 + x^17 + 2*x^16 + 4*x^15 + 2*x^14 + x^13 + x^12 + 2*x^11 + 4*x^10 + 2*x^9 + x^8 + x^7 + 9*x^6 + 3*x^5 + 4*x^4 + 2*x^3 + x^2 + x + 1)/(x^47 - x^46 - x + 1). (End)

Corrected by Logan K. Young (lkyfella(AT)yahoo.com), Mar 19 2005

More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Nov 20 2006

A164790 a(n) is the smallest nonnegative number whose American English name has the letter "e" in the n-th position.

Original entry on oeis.org

8, 0, 1, 3, 3, 12, 13, 17, 21, 23, 23, 73, 101, 103, 103, 112, 113, 117, 121, 123, 123, 173, 323, 373, 1103, 1103, 1112, 1113, 1117, 1121, 1123, 1123, 1173, 1323, 1373, 3323, 3373, 11373, 13323, 13373, 17373, 23323, 23373, 73373, 101373, 103323, 103373, 111373

Offset: 1

Claudio Meller, Aug 26 2009

			a(1)=8 ("Eight"), a(2)=0 ("zEro"), a(3)=1 ("onE"), a(4)=3 ("thrEe").

GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See page 70.

See A362120-A362122 for other versions.

Cf. A164789 ("o"), A164791 ("n"), A164792 ("t"), A164793 ("i"), A164794 ("f"), A164795 ("h"), A164796 ("r"), A164797 ("u").

from num2words import num2words
from itertools import count, islice
def n2w(n):
    return "".join(c for c in num2words(n).replace(" and ", "") if c.isalpha())
def A164790(n, t="e", i0=0): # t is target letter, i0 is start
    return next(i for i in count(i0) if len(w:=n2w(i))>=n and w[n-1]==t)
print([A164790(n) for n in range(1, 38)]) # Michael S. Branicky, Apr 21 2023

# faster for initial segment of sequence; uses n2w/imports above
def A164790gen(t="e", i0=0, offset=1): # generator of terms w
    adict, n = dict(), offset
    for i in count(i0):
        w = n2w(i)
        if t in w:
            locs = [i+1 for i, c in enumerate(w) if w[i] == t]
            for v in locs:
                if v not in adict: adict[v] = i
        while n in adict: yield adict[n]; n += 1
print(list(islice(A164790gen(), 50))) # Michael S. Branicky, Apr 21 2023

a(25) and beyond from Michael S. Branicky, Mar 25 2021

Name edited by N. J. A. Sloane, Apr 20 2023

A380201 Triangle T(n,k) read by rows, where row n is a permutation of numbers 1 through n, such that if a deck of n cards is prepared in this order, and SpellUnder-Down dealing is used, then the resulting cards are put down in increasing order.

Original entry on oeis.org

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

Offset: 1

Tanya Khovanova and the MIT PRIMES STEP junior group, Jan 16 2025

In Spell Under-Down dealing, we spell the positive integers starting from O-N-E, moving 1 card from the top of the deck underneath the deck for each letter, followed by dealing or "putting down" the top card. So we start by putting 3 cards under for O-N-E, then we deal a card. Then we put 3 cards under for T-W-O, then we deal a card. Then we put 5 cards under for T-H-R-E-E, and subsequently deal a card. This dealing sequence is highly irregular because it depends on English spelling. The dealing pattern starts: UUUDUUUDUUUUUD, 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 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 A380247 is an inverse permutation of the corresponding row of this triangle. The first column is A380246, the order of elimination of the first person in the corresponding Josephus problem. The index of the largest number in row n is A380204(n), corresponding to the index of the freed person in the corresponding Josephus problem. The number of card moves if we start with n cards is A380202 = A067278(n) + n.

			Triangle begins:
  1;
  2, 1;
  1, 3, 2;
  2, 4, 3, 1;
  5, 3, 2, 1, 4;
  4, 2, 5, 1, 3, 6;
  2, 3, 4, 1, 6, 5, 7;
  5, 6, 8, 1, 7, 4, 3, 2;
  ...
For n = 4 suppose there are four cards arranged in order 2, 4, 3, 1. Three cards go under for each letter in O-N-E, then 1 is dealt. Now the deck is ordered 2,4,3. Three cards go under for each letter in T-W-O, then card 2 is dealt. Now the leftover deck is ordered 4,3. Five cards go under for each letter in T-H-R-E-E, then card 3 is dealt. Finally, card 4 is dealt. The dealt cards are in numerical order. Thus, the fourth row of the triangle is 2, 4, 3, 1.

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

from num2words import num2words as n2w
def spell(n):
    return sum(1 for c in n2w(n).replace(" and", "").replace(" ", "").replace(",","").replace("-", ""))
def nthRow(n):
    l = []
    for i in range(0,n):
        l.append(0)
    zp = 0
    for j in range(1,n+1):
        zc = 0
        while zc <= spell(j):
            if l[zp] == 0:
                zc += 1
            zp += 1
            zp = zp % n
        l[zp-1] = str(j)
    return l
l = []
for i in range(1,20):
    l += nthRow(i)
print(", ".join(l))

A380204 A version of the Josephus problem: a(n) is the surviving integer under the spelling version of the elimination process.

Original entry on oeis.org

1, 1, 2, 2, 1, 6, 7, 3, 5, 3, 5, 6, 10, 9, 2, 13, 3, 16, 10, 2, 15, 6, 15, 6, 21, 1, 7, 23, 26, 6, 20, 12, 27, 29, 7, 2, 36, 11, 6, 7, 32, 6, 32, 43, 10, 31, 7, 5, 42, 1, 17, 48, 7, 31, 53, 25, 42, 43, 29, 39, 51, 25, 43, 7, 26, 59, 15, 10, 60, 69, 13, 57, 54, 66, 57, 30, 9, 35, 64, 9, 65, 1, 15, 3, 79, 47, 86, 7

Offset: 1

Tanya Khovanova and the MIT PRIMES STEP junior group, Jan 16 2025

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

Michael S. Branicky, Table of n, a(n) for n = 1..20000

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

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:
        i = (i + f(c))%len(J)
        q = J.pop(i)
        c = c+1
    return J[0]
print([a(n) for n in range(1, 89)]) # Michael S. Branicky, Jan 26 2025

Terms a(22) and beyond corrected by Michael S. Branicky, Feb 15 2025

A380246 Elimination order of the first person in a variation of the Josephus problem, where the number of skipped people correspond to the number of letters in consecutive numbers, called SpellUnder-Down.

Original entry on oeis.org

1, 2, 1, 2, 5, 4, 2, 5, 6, 4, 6, 10, 3, 12, 6, 8, 15, 4, 13, 19, 14, 17, 5, 22, 18, 26, 6, 20, 13, 17, 19, 23, 7, 25, 21, 31, 22, 32, 8, 31, 38, 20, 29, 9, 27, 18, 43, 10, 15, 50, 37, 20, 16, 41, 11, 21, 39, 36, 34, 32, 29, 12, 36, 50, 27, 53, 35, 19, 45, 67, 13, 20, 70, 59, 74, 26, 21, 40, 65, 14, 49, 82, 33, 43, 28, 34, 53, 15

Offset: 1

Tanya Khovanova and the MIT PRIMES STEP junior group, Jan 17 2025

Arrange n people numbered 1,2,3,...,n in a circle, increasing clockwise. Starting with the person numbered 1, 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. a(n) is the order of elimination of the first person.

			Consider n = 4 people. The first person eliminated is number 4. This leaves the remaining people in order 1, 2, 3. The second person eliminated is number 1. Thus, person number 1 is eliminated in the second round, implying that a(4) = 2.

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

from num2words import num2words as n2w
def spell(n):
    return sum(1 for c in n2w(n).replace(" and", "").replace(" ", "").replace(chr(44), "").replace("-", ""))
def nthRow(n):
    l = []
    for i in range(0,n):
        l.append(0)
    zp = 0
    for j in range(1,n+1):
        zc = 0
        while zc <= spell(j):
            if l[zp] == 0:
                zc += 1
            zp += 1
            zp = zp % n
        l[zp-1] = str(j)
    return l
l = []
for i in range(1,89):
    l += [nthRow(i)[0]]
print(l)

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) > 0:
        i = (i + f(c))%len(J)
        q = J.pop(i)
        if q == 1: return c
        c = c+1
print([a(n) for n in range(1, 89)]) # Michael S. Branicky, Feb 15 2025

A008520 Numbers whose American English name contains the letter 'e'.

Original entry on oeis.org

0, 1, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 33, 35, 37, 38, 39, 41, 43, 45, 47, 48, 49, 51, 53, 55, 57, 58, 59, 61, 63, 65, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90

Offset: 1

N. J. A. Sloane

A085513(a(n)) > 0. - Reinhard Zumkeller, Jan 23 2015

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Roel and Bas van Dijk, Numerals package, Hackage (Haskell packages)
Index entries for sequences related to number of letters in n

Cf. A006933 (complement), A085513.

Cf. A008519 (o), A008522 (t), A008536 (n), A008538 (s), A008540 (f), A008553 (y).

import Data.Maybe (fromJust)
import Data.Text (Text); import qualified Data.Text as T (any)
import Text.Numeral.Grammar.Reified (defaultInflection)
import qualified Text.Numeral.Language.EN as EN  -- see link
a008520 n = a008520_list !! (n-1)
a008520_list = filter (T.any (== 'e') . numeral) [0..] where
   numeral :: Integer -> Text
   numeral = fromJust . EN.gb_cardinal defaultInflection
-- Reinhard Zumkeller, Jan 23 2015

A008520Q[n_]:=StringContainsQ[IntegerName[n,"Words"],"e"];Select[Range[0,200],A008520Q] (* Paolo Xausa, Aug 11 2023 *)

Name edited by Michael De Vlieger, Aug 11 2023

A380202 Number of card moves to deal n cards using the SpellUnder-Down dealing.

Original entry on oeis.org

4, 8, 14, 19, 24, 28, 34, 40, 45, 49, 56, 63, 72, 81, 89, 97, 107, 116, 125, 136, 146, 156, 168, 179, 190, 200, 212, 224, 235, 246, 256, 266, 278, 289, 300, 310, 322, 334, 345, 355, 364, 373, 384, 394, 404, 413, 424, 435, 445, 455, 464, 473, 484, 494, 504, 513, 524, 535, 545, 555, 564, 573, 584, 594, 604, 613, 624, 635, 645

Offset: 1

Tanya Khovanova and the MIT PRIMES STEP junior group, Jan 16 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.

			The dealing pattern to deal three cards is UUUDUUUDUUUUUD. It contains 14 letters, thus, a(3) = 14.

Cf. A005589, A006257, A067278, A225381, A321298, A378635, A380201, A380204, A380246, A380247, A380248.

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

A380247 Triangle read by rows: T(n,k) is the number of the k-th eliminated person in the variation of the Josephus elimination process for n people, where the number of people skipped correspond to the number of letters in the next number in English alphabet.

Original entry on oeis.org

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

Offset: 1

Tanya Khovanova and the MIT PRIMES STEP junior group, Jan 17 2025

In this variation of the Josephus elimination process, the numbers 1 through n are arranged in a circle. A pointer starts at position 1. Then three people are skipped because 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 numbers remain. This sequence represents the triangle T(n, k), where n is the number of people in the circle, and T(n, k) is the elimination order of the k-th person in the circle.

In rows 4 and after, the first number is 4. In rows 8 and after, the second number is 8. In rows 14 and after, the third number is 14. In the limit the numbers form sequence A380202.

			Triangle begins:
  1;
  2, 1;
  1, 3, 2;
  4, 1, 3, 2;
  4, 3, 2, 5, 1;
  4, 2, 5, 1, 3, 6;
  4, 1, 2, 3, 6, 5, 7;
  ...
For n = 4 suppose four people are arranged in a circle corresponding to the fourth row of the triangle. Three people are skipped for each letter in O-N-E; then the 4th person is eliminated. This means the row starts with 4. The next three people are skipped, and the person eliminated is number 1. Thus, the next element in the row is 1. Then, 5 people are skipped, and the next person eliminated is number 3. Similarly, the last person eliminated is number 2. Thus, the fourth row of this triangle is 4, 1, 3, 2.

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

from num2words import num2words as n2w
def spell(n):
    return sum(1 for c in n2w(n).replace(" and", "").replace(" ", "").replace(chr(44), "").replace("-", ""))
def inverse_permutation(p):
    inv = [0] * len(p)
    for i, x in enumerate(p):
        inv[x-1] = i +1
    return inv
def nthRow(n):
    l = []
    for i in range(0,n):
        l.append(0)
    zp = 0
    for j in range(1,n+1):
        zc = 0
        while zc <= spell(j):
            if l[zp] == 0:
                zc += 1
            zp += 1
            zp = zp % n
        l[zp-1] = j
    return l
l = []
for i in range(1,15):
    l += inverse_permutation(nthRow(i))
print(l)

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 = 1, 0, list(range(1, n+1))
    out = []
    while len(J) > 1:
        i = (i + f(c))%len(J)
        q = J.pop(i)
        out.append(q)
        c = c+1
    out.append(J[0])
    return out
print([e for n in range(1, 15) for e in row(n)]) # Michael S. Branicky, Feb 15 2025

A007208 Number of letters in German name of n.

Original entry on oeis.org

4, 4, 4, 4, 4, 4, 5, 6, 4, 4, 4, 3, 5, 8, 8, 8, 8, 8, 8, 8, 7, 13, 14, 14, 14, 14, 15, 16, 14, 14, 7, 13, 14, 14, 14, 14, 15, 16, 14, 14, 7, 13, 14, 14, 14, 14, 15, 16, 14, 14, 7, 13, 14, 14, 14, 14, 15, 16, 14, 14, 7, 13, 14, 14, 14, 14, 15, 16, 14, 14, 7

Offset: 0

N. J. A. Sloane

Standard German orthography; a letter with an umlaut or ß is counted as a single letter: e.g., 30 maps to length("dreißig") = 7.
There are ambiguities from n=100 on, since both, "hundert" and "einhundert" are equally valid and common. The same applies for 1000 with "tausend" or "eintausend". - M. F. Hasler, Nov 03 2013
In contrast to English (A005589 vs A052360) and French (A007005 vs A167507), there are no spaces or other punctuation in German names for numbers, until 10^6 = "eine Million". - M. F. Hasler, Sep 20 2014
There also appears to be an ambiguity on whether there is an 's' in the middle of 101*10^3, "(ein)hundertein(s)tausend". - M. F. Hasler, Apr 08 2023

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Cf. A005589 and A052360 (English analog).

Cf. A007005 and A167507 (French analog).

/* Because names with ä, ö, ü or ß can't be entered directly as a string in the GP interface, we use a separate list for the names, for efficiency and readability of the main function. Note that the default lexicographical order is that of ISO 8859-1 character codes ("z" < "ß" < "ä"). In applications where this is not suitable, the special characters below can be replaced, e.g., with "ae, oe, ue, ss" or "a, o, u, s". [M. F. Hasler, Jul 05 2024] */
{deutsch = ["eins", "zwei", "drei", "vier", Str("f"Strchr(252)"nf"), "sechs", "sieben", "acht", "neun", "zehn", "elf", Str("zw"Strchr(246)"lf"),  "dreizehn", "vierzehn", Str("f"Strchr(252)"nfzehn"), "sechzehn", "siebzehn", "achtzehn", "neunzehn", "zwanzig", Str("drei"Strchr(223)"ig"), "vierzig", Str("f"Strchr(252)"nfzig"), "sechzig", "siebzig", "achtzig", "neunzig"]}
German(n, e="eins", power=0, name="")={ if(power /* internal helper function */
  , n = divrem(n, power); Str(German(n[1], e) name, if(n[2], German(n[2]), ""))
  , n < 20, if(n>1, deutsch[n], n, e, "null")
  , n < 100, Str(if(n%10, Str(German(n%10, "ein") "und"), "") deutsch[n\10+18])
  , n < 1000, German(n, "ein", 100, "hundert") \\ replace "ein" with "" to get
  , n < 10^6, German(n, "ein", 1000, "tausend")\\ hundert/tausend without "ein-"
  , my(t=3); while(n>=10^t, t+=3); German(n, "ein", 10^t-=3, strprintf(
      if(n\10^t>1, " %sen", t%2, "e %se", "e %s")  if(n%10^t, " ", ""),
      Str(["M", "B", "Tr", "Quadr", "Quint", "Sext", "Sept", "Oct", "Non",
           "Dez", "Undez" /* etc. */][t\6], "illi", ["on", "ard"][t%2+1])))
  )} \\ updated Mar 03 2020, Apr 08 2023, Jul 05 2024
A007208 = n -> #German(n) \\ M. F. Hasler, Nov 01 2013
A007208(n) = vecsum([c>32|c<-Vecsmall(German(n))]) \\ To exclude spaces; irrelevant for n < 10^6. - M. F. Hasler, Jul 05 2024

Corrected by Markus Stausberg (markus(AT)polomi.de), Aug 08 2004

Initial term a(0) = 4 = #"null" added by M. F. Hasler, Nov 01 2013

cp's OEIS Frontend

A075774 Number of syllables in n in American English.

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A000861 Numbers ending with a vowel in American English.

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A164790 a(n) is the smallest nonnegative number whose American English name has the letter "e" in the n-th position.

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A380201 Triangle T(n,k) read by rows, where row n is a permutation of numbers 1 through n, such that if a deck of n cards is prepared in this order, and SpellUnder-Down dealing is used, then the resulting cards are put down in increasing order.

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A380204 A version of the Josephus problem: a(n) is the surviving integer under the spelling version of the elimination process.

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A380246 Elimination order of the first person in a variation of the Josephus problem, where the number of skipped people correspond to the number of letters in consecutive numbers, called SpellUnder-Down.

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A008520 Numbers whose American English name contains the letter 'e'.

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A380202 Number of card moves to deal n cards using the SpellUnder-Down dealing.

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