A354523 Number of distinct letters in the English word for n that can also be found in the English word for n+1.

2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 2, 4, 4, 4, 4, 3, 4, 3, 5, 6, 5, 6, 6, 6, 6, 5, 6, 3, 5, 6, 5, 5, 6, 5, 6, 6, 6, 3, 5, 5, 5, 5, 5, 6, 6, 6, 7, 4, 4, 5, 4, 5, 4, 4, 5, 5, 5, 3, 5, 6, 5, 6, 6, 5, 5, 6, 6, 5, 6, 7, 6, 7, 7, 7, 6, 6, 7, 4, 6, 7, 6, 7, 7, 6, 7, 6, 6, 5, 5, 6, 5, 6, 6, 5, 6, 5, 5, 2, 7

Offset

0,1

Comments

US English is assumed (i.e., 101 = 'one hundred one' instead of 'one hundred and one').

There are no zero values since min{a(k) | 0 <= k < 1000} = 1 and "nine" and "one" share common letters whenever the initial power name changes. - Michael S. Branicky, Aug 19 2022

Links

Table of n, a(n) for n=0..100.

Example

a(0) = 2 since the letters 'e' and 'o' in 'zero' can also be found in 'one'.
a(11)= 3 since the letters 'e', 'l' and 'v' in 'eleven' can also be found in 'twelve'.

Mathematica

a[n_]:= Length[Intersection[Characters[IntegerName[n]], Characters[IntegerName[n+1]], CharacterRange["a", "z"]]]; Array[a, 101, 0] (* Stefano Spezia, Aug 19 2022 *)

Prog

(Python)
from num2words import num2words as n2w
def b(n): return set(c for c in n2w(n).replace(" and", "") if c.isalpha())
def a(n): return len(b(n) & b(n+1))
print([a(n) for n in range(101)]) # Michael S. Branicky, Aug 19 2022

Crossrefs

Sequence in context: A375732 A147753 A333355 * A116531 A375039 A101871

Adjacent sequences: A354520 A354521 A354522 * A354524 A354525 A354526

Keyword

nonn,easy,word

Author

Ray G. Opao, Aug 16 2022

Status

approved