A134629 - OEIS

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

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OFFSET

3,3

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

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.

EXAMPLE

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

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 *)

PROG

(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