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A345866 Nonnegative integers ordered by lowest number value per letter (in American English) and then, in case of ties, numerically. 0
0, 1, 3, 2, 4, 5, 7, 8, 13, 14, 11, 17, 6, 12, 23, 15, 9, 18, 16, 21, 19, 24, 22, 27, 25, 28, 26, 29, 33, 10, 20, 37, 34, 31, 38, 35, 32, 39, 36, 43, 47, 48, 44, 30, 45, 41, 42, 53, 49, 57, 46, 58, 123, 54, 127, 73, 128, 55, 117, 124, 125, 113, 63, 114, 133 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
Uses the convention of omitting a trailing 'and', so 101 is 'one hundred one' rather than 'one hundred and one.'
LINKS
EXAMPLE
17 has a value per letter of 17/9, which is more than 11's value of 11/6 but less than 6's value of 2.
From Jon E. Schoenfield, Jun 27 2021: (Start)
Table begins:
letter count
n a(n) name A005589(a(n)) a(n)/A005589(a(n))
-- ---- ------------ ------------- ------------------
1 0 zero 4 0/4 = 0
2 1 one 3 1/3 = 0.33333...
3 3 three 5 3/5 = 0.6
4 2 two 3 2/3 = 0.66666...
5 4 four 4 4/4 = 1
6 5 five 4 5/4 = 1.25
7 7 seven 5 7/5 = 1.4
8 8 eight 5 8/5 = 1.6
9 13 thirteen 8 13/8 = 1.625
10 14 fourteen 8 14/8 = 1.75
11 11 eleven 6 11/6 = 1.83333...
12 17 seventeen 9 17/9 = 1.88888...
13 6 six 3 6/3 = 2
14 12 twelve 6 12/6 = 2
15 23 twenty-three 11 23/11 = 2.09090...
(End)
PROG
(Python)
from num2words import num2words as n2w
def A005589(n): return sum(c.isalpha() for c in n2w(n).replace(" and", ""))
def aseq(N): return sorted(range(10*N), key=lambda x: (x/A005589(x), x))[:N]
print(aseq(65)) # Michael S. Branicky, Jun 27 2021
CROSSREFS
Cf. A005589.
Sequence in context: A234027 A364839 A165279 * A125060 A039882 A370803
KEYWORD
nonn,word
AUTHOR
Jack Zilinskas, Jun 26 2021
STATUS
approved

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Last modified April 24 05:26 EDT 2024. Contains 371918 sequences. (Running on oeis4.)