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Numbers with digits in alphabetical order (in English).
19

%I #20 Aug 17 2022 16:11:40

%S 0,1,2,3,4,5,6,7,8,9,10,11,12,13,16,17,20,22,30,32,33,40,41,42,43,44,

%T 46,47,49,50,51,52,53,54,55,56,57,59,60,62,63,66,70,72,73,76,77,80,81,

%U 82,83,84,85,86,87,88,89,90,91,92,93,96,97,99,100

%N Numbers with digits in alphabetical order (in English).

%C a(142447) = A053433(1023) = 8549176320 is the greatest term not containing any repeating digits. - _Reinhard Zumkeller_, Oct 05 2014

%H Reinhard Zumkeller, <a href="/A053432/b053432.txt">Table of n, a(n) for n = 1..10000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/WordSequence.html">Word Sequence</a>

%H Wikipedia, <a href="http://de.wikipedia.org/wiki/Zahlen_in_unterschiedlichen_Sprachen#0_bis_10">Zahlen in unterschiedlichen Sprachen</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/List_of_numbers_in_various_languages">List of numbers in various languages</a>

%o (Haskell)

%o import Data.IntSet (fromList, deleteFindMin, union)

%o a053432 n = a053432_list !! (n-1)

%o a053432_list = 0 : f (fromList [1..9]) where

%o f s = x : f (s' `union`

%o fromList (map (+ 10 * x) $ dropWhile (/= mod x 10) digs))

%o where (x, s') = deleteFindMin s

%o digs = [8, 5, 4, 9, 1, 7, 6, 3, 2, 0]

%o -- _Reinhard Zumkeller_, Oct 05 2014.

%o (Python)

%o from itertools import count, islice, combinations_with_replacement as cwr

%o def agen(): # generator of terms

%o for d in count(1):

%o out = sorted(int("".join(t)) for t in cwr("8549176320", d))

%o yield from out[1-int(d==1):] # remove extra 0's

%o print(list(islice(agen(), 65))) # _Michael S. Branicky_, Aug 17 2022

%Y Cf. A247750 (Czech), A247751 (Danish), A247752 (Dutch), A247753 (Finnish), A247754 (French), A247755 (German), A247756 (Hungarian), A247757 (Italian), A247758 (Latin), A247759 (Norwegian), A247760 (Polish), A247757 (Portuguese), A247761 (Russian), A247762 (Slovak), A161390 (Spanish), A247759 (Swedish), A247764 (Turkish).

%K nonn,base,word,easy

%O 1,3

%A _G. L. Honaker, Jr._, Jan 10 2000