A303701 Number of distinct letters in the (American) English name of n, excluding spaces and hyphens.

4, 3, 3, 4, 4, 4, 3, 4, 5, 3, 3, 4, 5, 6, 7, 5, 6, 5, 6, 4, 5, 6, 6, 7, 9, 8, 8, 7, 8, 6, 5, 8, 7, 6, 8, 8, 7, 9, 7, 7, 5, 7, 6, 7, 6, 8, 8, 9, 9, 8, 4, 7, 6, 7, 7, 6, 6, 8, 7, 6, 5, 8, 7, 8, 9, 8, 5, 8, 8, 7, 6, 7, 8, 8, 10, 8, 8, 6, 9, 7, 6, 8, 8, 7, 10, 8, 8, 9, 6, 7, 5, 6, 7, 7, 9, 7, 7, 7

Offset

0,1

Comments

First occurrence of k=3..23: 1, 0, 8, 13, 14, 25, 24, 74, 112, 127, 125, 165, 265, 1265, 2568, 12468, 1002568, 1001002568, 1000001001002568, 1000000001000001001002568, 1001000000001000001001002568.

Since there are no number words consisting of just one or two letters, nor any number words which contain the letters "j", "k" or "z"; the above list is complete.

Links

Table of n, a(n) for n=0..97.

Formula

a(n) <= A005589(n).

Example

a(7) = 4 since "seven" has 4 distinct letters. (Cf. A005589(7) = 5.)

Mathematica

f[n_] := Length@ Union@ StringPartition[ StringReplace[ IntegerName[n, "Words"], ", " | " " | "\[Hyphen]" -> ""], 1]; Array[f, 98, 0]

Prog

(Python)
from num2words import num2words
def n2w(n):
  map = {ord(c): None for c in "-, "}
  return num2words(n).replace(" and", "").translate(map)
def a(n): return len(set(n2w(n)))
print([a(n) for n in range(98)]) # Michael S. Branicky, Apr 03 2021

Crossrefs

Cf. A005589.

Sequence in context: A362330 A129344 A048853 * A179845 A180217 A270651

Adjacent sequences: A303698 A303699 A303700 * A303702 A303703 A303704

Keyword

nonn,word

Author

Robert G. Wilson v, Apr 29 2018

Status

approved