A009994 Numbers with digits in nondecreasing order.
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 23, 24, 25, 26, 27, 28, 29, 33, 34, 35, 36, 37, 38, 39, 44, 45, 46, 47, 48, 49, 55, 56, 57, 58, 59, 66, 67, 68, 69, 77, 78, 79, 88, 89, 99, 111, 112, 113, 114, 115, 116, 117, 118, 119, 122
Offset: 1
References
- Amarnath Murthy and Robert J. Clarke, Some Properties of Staircase sequence, Mathematics & Informatics Quarterly, Volume 11, No. 4, November 2001.
- Frank A. Hanna, The Compilation of Manufacturing Statistics. U.S. Department of Commerce, Bureau of the Census, 1959.
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Reinhard Zumkeller)
- David Applegate, Marc LeBrun, N. J. A. Sloane, Dismal Arithmetic, J. Int. Seq. 14 (2011) # 11.9.8.
- David Radcliffe and Brendan McKay, Re: A009994, SeqFan list, Jul 29 2019
- Eric Weisstein's World of Mathematics, Digit.
- Index entries for 10-automatic sequences.
Crossrefs
Programs
-
Haskell
import Data.Set (fromList, deleteFindMin, insert) a009994 n = a009994_list !! n a009994_list = 0 : f (fromList [1..9]) where f s = m : f (foldl (flip insert) s' $ map (10*m +) [m `mod` 10 ..9]) where (m,s') = deleteFindMin s -- Reinhard Zumkeller, Aug 10 2011
-
Magma
[k:k in [0..122]|Sort(Intseq(k)) eq Reverse(Intseq(k))]; // Marius A. Burtea, Jul 28 2019
-
Maple
A[0]:= [0]: A[1]:= [$1..9]: for d from 2 to 4 do A[d]:= map(t -> seq(10*t+i,i=(t mod 10) .. 9), A[d-1]): od: seq(op(A[d]),d=0..4); # Robert Israel, Jul 28 2019
-
Mathematica
Select[Range[0, 125], LessEqual@@IntegerDigits[#] &] (* Ray Chandler, Oct 25 2011 *)
-
PARI
is(n)=n=digits(n);n==vecsort(n) \\ Charles R Greathouse IV, Dec 03 2013
-
Python
from itertools import combinations_with_replacement def A009994generator(): yield 0 l = 1 while True: for i in combinations_with_replacement('123456789',l): yield int(''.join(i)) l += 1 # Chai Wah Wu, Nov 11 2015
-
Scala
def hasDigitsSorted(n: Int): Boolean = { val digSort = Integer.parseInt(n.toString.toCharArray.sorted.mkString) n == digSort } (0 to 200).filter(hasDigitsSorted()) // _Alonso del Arte, Apr 20 2020
Formula
a(n) >> exp(n^(1/10)). - Charles R Greathouse IV, Mar 15 2014
a(n) ~ 10^((9! n)^(1/9) - 5), since 10^(d - 1) <= a(n) < 10^d for binomial(d + 8, 9) < n <= binomial(d + 9, 9) = (d + 5 - epsilon)^9 / 9!. Using epsilon = 10/(3n) + o(1/n) gives more precise estimate. [Following Radcliffe and McKay, cf. SeqFan list.] - M. F. Hasler, Jul 30 2019
Comments