A221714 -id:A221714 - cp's OEIS frontend

A004086 Read n backwards (referred to as R(n) in many sequences).

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 21, 31, 41, 51, 61, 71, 81, 91, 2, 12, 22, 32, 42, 52, 62, 72, 82, 92, 3, 13, 23, 33, 43, 53, 63, 73, 83, 93, 4, 14, 24, 34, 44, 54, 64, 74, 84, 94, 5, 15, 25, 35, 45, 55, 65, 75, 85, 95, 6, 16, 26, 36, 46, 56, 66, 76, 86, 96, 7, 17, 27, 37, 47

Offset: 0

N. J. A. Sloane

Also called digit reversal of n.

Leading zeros (after the reversal has taken place) are omitted. - N. J. A. Sloane, Jan 23 2017

Indranil Ghosh, Table of n, a(n) for n = 0..50000 (first 1001 terms from Franklin T. Adams-Watters)
Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, Finding structure in sequences of real numbers via graph theory: a problem list, arXiv:2012.04625 [math.CO], Dec 08 2020.
Michael Penn, A digit moving number puzzle., YouTube video, 2022.

Cf. A004185, A004186, A009996, A073138, A188649, A221714.

Cf. A030101, A030102, A030103, A030104, A030105, A030107, A030108.

a004086 = read . reverse . show  -- Reinhard Zumkeller, Apr 11 2011

|.&.": i.@- 1e5 NB. Stephen Makdisi, May 14 2018

read transforms; A004086 := digrev; #cf "Transforms" link at bottom of page
A004086:=proc(n) local s,t; if n<10 then n else s:=irem(n,10,'t'); while t>9 do s:=s*10+irem(t,10,'t') od: s*10+t fi end; # M. F. Hasler, Jan 29 2012

Table[FromDigits[Reverse[IntegerDigits[n]]], {n, 0, 75}]
IntegerReverse[Range[0,80]](* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 13 2018 *)

dig(n) = {local(m=n,r=[]); while(m>0,r=concat(m%10,r);m=floor(m/10));r}
A004086(n) = {local(b,m,r);r=0;b=1;m=dig(n);for(i=1,matsize(m)[2],r=r+b*m[i];b=b*10);r} \\ Michael B. Porter, Oct 16 2009

A004086(n)=fromdigits(Vecrev(digits(n))) \\ M. F. Hasler, Nov 11 2010, updated May 11 2015, Sep 13 2019

def A004086(n):
    return int(str(n)[::-1]) # Chai Wah Wu, Aug 30 2014

For n > 0, a(a(n)) = n iff n mod 10 != 0. - Reinhard Zumkeller, Mar 10 2002
a(n) = d(n,0) with d(n,r) = r if n=0, otherwise d(floor(n/10), r*10+(n mod 10)). - Reinhard Zumkeller, Mar 04 2010
a(10*n+x) = x*10^m + a(n) if 10^(m-1) <= n < 10^m and 0 <= x <= 9. - Robert Israel, Jun 11 2015

Extended by Ray Chandler, Dec 30 2004

A004185 Arrange digits of n in increasing order, then (for n > 0) omit the zeros.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2, 12, 22, 23, 24, 25, 26, 27, 28, 29, 3, 13, 23, 33, 34, 35, 36, 37, 38, 39, 4, 14, 24, 34, 44, 45, 46, 47, 48, 49, 5, 15, 25, 35, 45, 55, 56, 57, 58, 59, 6, 16, 26, 36, 46, 56, 66, 67, 68, 69, 7, 17, 27, 37, 47

Offset: 0

N. J. A. Sloane

Record values: A009994. - Reinhard Zumkeller, Dec 05 2009
If we define "sortable primes" as prime numbers that remain prime when their digits are sorted in increasing order, then all absolute primes (A003459) are sortable primes but not all sortable primes are absolute primes. For example, 311 is both sortable and absolute, and 271 is sortable but not absolute, since its digits can be permuted to 217 = 7 * 31 or 712 = 2^3 * 89, etc. - Alonso del Arte, Oct 05 2013
The above mentioned "sortable primes" are listed in A211654, the nontrivial ones (with digits not in nondecreasing order) in A086042. - M. F. Hasler, Jul 30 2019

			a(19) = 19 because the digits are already in increasing order.
a(20) = 2 because the digits of 20 are 2 and 0, which in increasing order are 0 and 2, but since zero-padding is not allowed on the left, the zero digit is dropped and we are left with 2.
a(21) = 12 because the digits of 21 are 2 and 1, which in increasing order are 1 and 2.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A004719, A004151, A004086, A004186, A009996, A221714, A073138, A193581, A193582, A065641.

Cf. A211654 (sortable primes) and subsequence A086042 (nontrivial solutions).

import Data.List (sort)
a004185 n = read $ sort $ show n :: Integer
-- Reinhard Zumkeller, Aug 10 2011

A004185:=func; [n eq 0 select 0 else A004185(n): n in [0..57]]; // Bruno Berselli, Apr 03 2012

A004185 := proc(n)
    local dgs;
    convert(n,base,10) ;
    dgs := sort(%,`>`) ;
    add( op(i,dgs)*10^(i-1),i=1..nops(dgs)) ;
end proc:
seq(A004185(n),n=0..20) ; # R. J. Mathar, Jul 26 2015

FromDigits[Sort[DeleteCases[IntegerDigits[#], 0]]]&/@Range[0, 60] (* Harvey P. Dale, Nov 29 2011 *)

a(n)=fromdigits(vecsort(digits(n))) \\ Charles R Greathouse IV, Feb 06 2017

def A004185(n):
    return int(''.join(sorted(str(n))).replace('0','')) if n > 0 else 0 # Chai Wah Wu, Nov 10 2015

A073138 Largest number having in its binary representation the same number of 0's and 1's as n.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 6, 7, 8, 12, 12, 14, 12, 14, 14, 15, 16, 24, 24, 28, 24, 28, 28, 30, 24, 28, 28, 30, 28, 30, 30, 31, 32, 48, 48, 56, 48, 56, 56, 60, 48, 56, 56, 60, 56, 60, 60, 62, 48, 56, 56, 60, 56, 60, 60, 62, 56, 60, 60, 62, 60, 62, 62, 63, 64, 96, 96, 112, 96, 112, 112

Offset: 0

Reinhard Zumkeller, Jul 16 2002

From Trevor Green (green(AT)snoopy.usask.ca), Nov 26 2003: (Start)
a(n)/n has an accumulation point at x exactly when x is in the interval [1, 2]. Proof: Clearly n <= a(n) < 2n. Let b(n) = a(n)/n, then b(n) must always lie in [1,2) and all the accumulation points of the sequence must lie in [1,2]. We shall show that every such number is an accumulation point.
First, consider any d-bit integer n. Suppose that z of these bits are 0. Let n' be the (d+z)-bit integer whose first d bits are the same as those of n and whose remaining bits are all 1. Then a(n') will have to be the (d+z)-bit integer whose first d bits are all 1 and whose last z bits are all 0.
Thus n' = (n+1)*2^z-1; a(n') = (2^d-1)2^z; and b(n') = (2^d-1)/(n+1) + epsilon, where 0 < epsilon < 2^(1-d). So to get an accumulation point x, we just choose n(d) to be the d-bit integer such that (2^d-1)/(n(d)+1) < x <= (2^d-1)/n(d), or equivalently, n(d) = floor((2^d-1)/x). If x lies in [1,2), then n(d) will always be a d-bit number for sufficiently large d.
Then n'(d) yields an increasing subsequence of the integers for which b(n'(d)) converges to x. For x = 2, choose n(d) = 2^(d-1), which is always a d-bit number; then b(n'(d)) = (2^d-1)/(2^(d-1)+1) + epsilon = 2 + epsilon', where epsilon' also heads for 0 as d blows up. This proves the claim.
(End)

			a(20)=24, as 20='10100' and 24 is the greatest number having two 1's and three 0's: 17='10001', 18='10010', 20='10100' and 24='11000'.

Seiichi Manyama, Table of n, a(n) for n = 0..8191 (terms 0..1023 from T. D. Noe)
Index entries for sequences related to binary expansion of n

Cf. A007088, A073137, A000523, A073139, A073140, A073141, A319650.
Cf. A030109.
Cf. A038573.
Decimal equivalent of A221714. - N. J. A. Sloane, Jan 26 2013

a073138 n = a038573 n * a080100 n  -- Reinhard Zumkeller, Jan 16 2012

a:= n-> Bits[Join](sort(Bits[Split](n))):
seq(a(n), n=0..100);  # Alois P. Heinz, Jun 26 2021

f[n_] := Module[{idn=IntegerDigits[n, 2], o, l}, l=Length[idn]; o=Count[idn, 1]; FromDigits[Join[Table[1, {o}], Table[0, {l-o}]], 2]]; Table[f[i], {i, 0, 70}]
ln[n_] := Module[{idn=IntegerDigits[n, 2], len, zer}, len=Length[idn]; zer=Count[idn, 0]; FromDigits[Join[Table[1, {len-zer}], Table[0, {zer}]], 2]]; Table[ln[i], {i, 0, 70}]
a[z_] := 2^(Floor[Log[2, z]] + 1) * (1 - 2^(-Sum[k, {k, IntegerDigits[n, 2]}])) Column[Table[a[p], {p, 500}], Right] (* Trevor G. Hyde (thyde12(AT)amherst.edu), Jul 14 2008 *)
Table[FromDigits[ReverseSort[IntegerDigits[n,2]],2],{n,0,70}] (* Harvey P. Dale, Mar 13 2023 *)

a(n) = fromdigits(vecsort(binary(n),,4), 2); \\ Michel Marcus, Sep 26 2018

def a(n): return int("".join(sorted(bin(n)[2:], reverse=True)), 2)
print([a(n) for n in range(71)]) # Michael S. Branicky, Jun 27 2021

def A073138(n): return (m:=1<>n.bit_count()) # Chai Wah Wu, Aug 18 2025

a(n+1) = a(floor(n/2))*2 + (n mod 2)*(2^floor(log_2(n)) - a(floor(n/2))); a(0)=0.
A023416(a(n)) = A023416(n), A000120(a(n)) = A000120(n).
a(0)=0, a(1)=1, a(2n) = 2a(n), a(2n+1) = a(n) + 2^floor(log_2(n)). - Ralf Stephan, Oct 05 2003
a(n) = 2^(floor(log_2(n)) + 1) * (1 - 2^(-d(n))) where d(n) = digit sum of base-2 expansion of n. - Trevor G. Hyde (thyde12(AT)amherst.edu), Jul 14 2008
a(n) = A038573(n) * A080100(n). - Reinhard Zumkeller, Jan 16 2012
n <= a(n) < 2n. - Charles R Greathouse IV, Aug 07 2024

cp's OEIS Frontend

A004086 Read n backwards (referred to as R(n) in many sequences).

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A004185 Arrange digits of n in increasing order, then (for n > 0) omit the zeros.

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Haskell

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A073138 Largest number having in its binary representation the same number of 0's and 1's as n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Python

Python

Formula