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A157899
a(n) is the smallest number whose name in US English contains n consonants.
1
1, 2, 3, 12, 13, 22, 23, 102, 103, 112, 113, 122, 123, 223, 323, 1112, 1113, 1122, 1123, 1223, 1323, 2323, 3323, 12323, 13323, 22323, 23323, 102323, 103323, 112323, 113323, 122323, 123323, 223323, 323323, 1113323, 1122323, 1123323, 1223323, 1323323, 2323323, 3323323
OFFSET
1,2
COMMENTS
In US English, "102" is written as "one hundred two".
Vowels are {a, e, i, o, u, y} here. If y is a consonant, then a(5)..a(7) = 21, 22, 23 are the first of subsequent differences. - Michael S. Branicky, Jan 29 2022
LINKS
Michael S. Branicky, Table of n, a(n) for n = 1..2478 (all terms < 1000 centillion)
Wiktionary, one hundred one (US).
Wiktionary, one hundred and one (UK).
Michael S. Branicky, Python program
EXAMPLE
"oNe" has 1 consonant, "TWo" has 2, "THRee" has 3, "TWeLVe" has 4, "THiRTeeN" has 5, "TWeNTyTWo" has 6, "TWeNTyTHRee" has 7, "oNe HuNDReD TWo" has 8, "oNe HuNDReD THRee" has 9, "oNe HuNDReD TWeLVe" has 10.
PROG
(Python) # see linked program for going to large numbers
from num2words import num2words
def consonants(n):
w = num2words(n).replace(" and", "")
return sum(1 for c in w if c in "bcdfghjklmnpqrstvwxz")
def aupton(nn):
k, adict, alst = 1, {}, []
while len(alst) < nn:
c = consonants(k)
if c not in adict:
adict[c] = k
alst = []
for i in range(1, max(adict)+1):
if i not in adict:
break
alst.append(adict[i])
k += 1
return alst
print(aupton(27)) # Michael S. Branicky, Jan 29 2022
CROSSREFS
Sequence in context: A039565 A032805 A239017 * A157900 A157902 A102660
KEYWORD
nonn,word
AUTHOR
Rodolfo Kurchan, Mar 08 2009
EXTENSIONS
Edited by Jon E. Schoenfield, Sep 29 2018
STATUS
approved