A139097 -id:A139097 - cp's OEIS frontend

A060403 Each term is the previous term plus the number of letters in the previous number, as conventionally spelled out in American English.

1, 4, 8, 13, 21, 30, 36, 45, 54, 63, 73, 85, 95, 105, 119, 137, 158, 178, 200, 210, 223, 244, 263, 283, 304, 320, 338, 361, 381, 402, 416, 434, 455, 475, 497, 519, 538, 560, 576, 597, 619, 637, 658, 678, 700, 712, 730, 748, 770, 789, 811, 829, 851, 871, 893

Offset: 1

Kevin Langdon (kevin.langdon(AT)polymath-systems.com), Apr 05 2001

Increases more slowly than A160395 since American English does not use 'and' to separate hundreds from the rest of the number. E.g., 619 = "six hundred nineteen" in American English but "six hundred and nineteen" in British English. - Carl R. White, May 12 2009

			a(2)=4 because a(1)=1 and 4 is 1 plus the number of letters in "one," 3.

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

Michael De Vlieger, Table of n, a(n) for n = 1..10001

Cf. A005589 See A139097 for another version.

For British English see A160395. - Carl R. White, May 12 2009

NestList[#+Length[Select[Characters[IntegerName[#,"Words"]],LetterQ ]]&,1,54] (* James C. McMahon, Jul 30 2024 *)

More terms from Carl R. White, May 12 2009

A329447 Start with a(0)=0; thereafter, look left and identify the least frequent digit d so far (in case of a tie, choose the smallest d): after then a(n) = 10c + d, where c > 0 is the number of times d has appeared so far.

Original entry on oeis.org

0, 10, 11, 20, 12, 22, 30, 13, 23, 33, 40, 14, 24, 34, 44, 50, 15, 25, 35, 45, 55, 60, 16, 26, 36, 46, 56, 66, 70, 17, 27, 37, 47, 57, 67, 77, 80, 18, 28, 38, 48, 58, 68, 78, 88, 90, 19, 29, 39, 49, 59, 69, 79, 89, 99, 100, 112, 113, 114, 115, 116, 117, 118, 119, 120, 123, 124, 125, 126, 127, 128, 129

Offset: 0

Eric Angelini and M. F. Hasler, Nov 14 2019

The term "10c + d" are to be read "c digits d have appeared so far", as in the "look and see" sequences llike A045918.
It follows immediately from the definition that all terms are distinct. For a sorted list of the terms, see A376779. For a tabular method of computing a(n), see the triangle in A377905. - N. J. A. Sloane, Nov 11 2024
An analogous sequence may be obtained for any initial term a(0). Sequence A329448 lists the starting values that will appear a second time later in the respective sequence.

M. F. Hasler, Table of n, a(n) for n = 0..10000 (independently computed by _Jean-Marc Falcoz_).
Eric Angelini, Look left and say, Cinquante signes Blog Post, Nov 14 2019.
Eric Angelini, Look left and say, Cinquante signes Blog Post, Nov 14 2019. [Local copy, with permission]

Cf. A045918 (describe n), A005150 (the classic "Say What You See"), A139282 (count vowels so far), A139097 (count letters so far).

A376779 gives terms in increasing order. See also A377905.

a[0]:= 0;
S[0]:= 1:
for i from 1 to 9 do S[i]:= 0 od:
for n from 1 to 100 do
  a[n]:= min(select(`>=`,[seq(10*S[i]+i, i=0..9)],10));
  L:= convert(a[n],base,10);
  for d from 0 to 9 do S[d]:= S[d] + numboccur(d,L) od;
od:
seq(a[n],n=0..100); # Robert Israel, Nov 14 2019

A329447_vec(N)={my(c=Vec(1,10),t); vector(N,i, for(j=1, #i=vecsort(c,,1), if(c[i[j]], i=i[j];break)); for(j=1, #i=digits(t=c[i]*10+i-1), c[i[j]+1]++);t)} \\ Returns the vector a(1..N)

from itertools import islice
def agen():  # generator of terms
    counts = [1] + [0 for i in range(1, 10)]
    yield 0
    while True:
        m = float('inf')
        for i in range(10):
            if counts[i] and counts[i] < m:
                m, argm = counts[i], i
        an = 10*m + argm
        yield an
        for d in str(an): counts[int(d)] += 1
print(list(islice(agen(), 72))) # Michael S. Branicky, Nov 11 2024

Edited by N. J. A. Sloane, Nov 10 2024

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

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A060403 Each term is the previous term plus the number of letters in the previous number, as conventionally spelled out in American English.

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A329447 Start with a(0)=0; thereafter, look left and identify the least frequent digit d so far (in case of a tie, choose the smallest d): after then a(n) = 10c + d, where c > 0 is the number of times d has appeared so far.

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

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Author

Keywords

Comments

Examples

References

Crossrefs

Programs

PARI

Extensions