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A085513
Number of "e"s in n (in English).
6
1, 1, 0, 2, 0, 1, 0, 2, 1, 1, 1, 3, 2, 2, 2, 2, 2, 4, 3, 3, 1, 2, 1, 3, 1, 2, 1, 3, 2, 2, 0, 1, 0, 2, 0, 1, 0, 2, 1, 1, 0, 1, 0, 2, 0, 1, 0, 2, 1, 1, 0, 1, 0, 2, 0, 1, 0, 2, 1, 1, 0, 1, 0, 2, 0, 1, 0, 2, 1, 1, 2, 3, 2, 4, 2, 3, 2, 4, 3, 3, 1, 2, 1, 3, 1, 2, 1, 3, 2, 2, 1, 2, 1, 3, 1, 2, 1, 3, 2, 2
OFFSET
0,4
COMMENTS
a(A006933(n)) = 0; a(A008520(n)) > 0. - Reinhard Zumkeller, Jan 23 2015
a(A006933(n)) = 0; a(A008520(n)) > 0; a(A121065(n)) = n and a(m) != n for m < A121065(n). - Reinhard Zumkeller, Jan 24 2015
EXAMPLE
a(123) = 5 because "onE hundrEd twEnty-thrEE" has 5 e's.
MATHEMATICA
StringCount[IntegerName/@Range[0, 99], "e"] (* Ivan N. Ianakiev, Mar 25 2017 *)
PROG
(Haskell)
import Data.Maybe (fromJust)
import Data.Text (Text); import qualified Data.Text as T (unpack))
import Text.Numeral.Grammar.Reified (defaultInflection)
import qualified Text.Numeral.Language.EN as EN -- see link
a085513 = length . filter (== 'e') . T.unpack . numeral where
numeral :: Integer -> Text
numeral = fromJust . EN.gb_cardinal defaultInflection
-- Reinhard Zumkeller, Jan 23 2015
(Python)
from num2words import num2words
def A085513(n):
return num2words(n).count('e') # Chai Wah Wu, Dec 20 2019
CROSSREFS
Cf. A005589.
Sequence in context: A364916 A365923 A089650 * A362913 A259965 A117054
KEYWORD
easy,nonn,word
AUTHOR
Samuel Thompson (samuelt(AT)ugcs.caltech.edu), Jul 03 2003
STATUS
approved