A134629 Least nonnegative integer which requires n letters to spell in English, excluding spaces and hyphens. A right inverse of A005589.

1, 0, 3, 11, 15, 13, 17, 24, 23, 73, 101, 104, 103, 111, 115, 113, 117, 124, 123, 173, 323, 373, 1104, 1103, 1111, 1115, 1113, 1117, 1124, 1123, 1173, 1323, 1373, 3323, 3373, 11373, 13323, 13373, 17373, 23323, 23373, 73373, 101373, 103323, 103373, 111373, 113323, 113373, 117373, 123323, 123373

Offset: 3

Author

Robert G. Wilson v, Nov 04 2007

Keywords

• nonn
• word

Comments

Variant of A080777. - R. J. Mathar, Dec 13 2008

This is one of many possible right inverses of A005589, i.e., A005589 o A134629 = id (of course on the domain of this sequence, [3 .. oo)). It does not satisfy A134629 o A005589 = id. - M. F. Hasler, Feb 25 2018

Examples

			a(3) = 1: "one", a(4) = 0: "zero", a(5) = 3: "three", a(6) = 11: "eleven", a(7) = 15: "fifteen", etc.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 3..608 (terms < 10^54)
• Hans Havermann, Table giving n and American English name for n, for 0 <= n <= 100999, without spaces or hyphens
• Landon Curt Noll, The English Name of a Number.
• Eric Weisstein's World of Mathematics, Number.

Crossrefs

Cf. A005589, A052363, A080777.

Programs

Mathematica

(* first load Hans Havermann's text file a001477.txt and then *)
f[n_] := Length@ Characters@ ToString@ lst[[n]]; g[n_] :=  Block[{k = 1}, While[ f[k] != n, k++]; k]; Array[g, 41, 3] - 1 (* Robert G. Wilson v, May 26 2013 *)
f[n_] := Length@ StringPartition[ StringReplace[ IntegerName[n, "Words"], ", " | " " | "\[Hyphen]" -> ""], 1] (* after Giovanni Resta in A005589 *); t[] := 0; k = 1; While[k < 174000, a = f@k; If[t[a] == 0, t[a] = k]; k++]; t[4] = 0 (* only {0, 4, 5 & 9} have just four letters *); t@# & /@ Range[3, 54] (* _Robert G. Wilson v, May 25 2018 *)

PARI

A134629(n)=for(k=1,oo,A005589(k)==n&&return(k)) \\ M. F. Hasler, Feb 25 2018

Python

from num2words import num2words as n2w
from itertools import count, islice
def f(n): return sum(1 for c in n2w(n).replace(" and", "") if c.isalpha())
def agen(): # generator of terms
    adict, n = dict(), 3
    for k in count(0):
        v = f(k)
        if v not in adict: adict[v] = k
        while n in adict: yield adict[n]; n += 1
print(list(islice(agen(), 40))) # Michael S. Branicky, Feb 19 2023

Extensions

More terms and reworded name from M. F. Hasler, Feb 25 2018