A004185 -id:A004185 - cp's OEIS frontend

A237671 Let m_n denote the number which is obtained from n-base representation of m if its digits are written in nondecreasing order; then a(n) is the smallest period of the sequence which is defined by the recurrence b(0)=0, b(1)=1, b(k)=(b(k-1) + b(k-2))_n, for k>=2, or a(n)=0, if there is no such period.

1, 3, 16, 6, 20, 24, 16, 36, 120, 300, 20, 288, 28, 192, 200, 552, 180, 192, 180, 1380, 224, 60, 1728, 912, 3800, 756, 576, 1776, 4102, 15480, 3540, 1344, 10800, 14328, 800, 2304, 1520, 1890, 1232, 11280, 9040, 31152, 49544, 3660, 6360, 3696, 13248, 21408

Offset: 2

Vladimir Shevelev and Peter J. C. Moses, Feb 11 2014

We conjecture that the sequence b is always eventually periodic, and so a(n)>0.

			For n=5, b-sequence begins 0,1,1,2,3,1,4,1,1,2,... It has period {1,1,2,3,1,4} of length 6. So a(5)=6.
a(10) = 120, because the eventual period of A069638 is 120.

Pontus von Brömssen, Table of n, a(n) for n = 2..250 (terms 2..100 from Giovanni Resta)

Cf. A069638, A237568, A004185, A306773.

import sympy,functools
def digits2int(x,b):
  return functools.reduce(lambda n,d:b*n+d,x,0)
def A237671(n):
  return next(sympy.cycle_length(lambda x:(x[1],digits2int(sorted(sympy.ntheory.factor_.digits(sum(x),n)[1:]),n)),(0,1)))[0] # Pontus von Brömssen, Aug 28 2020

A272918 Fibonacci numbers with the base 10 digits sorted into increasing order.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 13, 12, 34, 55, 89, 144, 233, 377, 16, 789, 1579, 2458, 1148, 5667, 1469, 11177, 25678, 34668, 2557, 112339, 114689, 111378, 122459, 2348, 1234669, 123789, 2345578, 257788, 2245679, 1233459, 11245778, 1368899, 23456689, 11233455, 11145568

Offset: 0

Alonso del Arte, May 10 2016

Leading zeros are omitted, of course.

			a(8) = 12 because F(8) = 21, so the digits in ascending order become 12.
a(9) = 34 = F(9), the digits are already in ascending order.

G. C. Greubel, Table of n, a(n) for n = 0..500

Cf. A000045, A004185.

Table[FromDigits[Sort[IntegerDigits[Fibonacci[n]]]], {n, 0, 49}]

from gmpy2 import fib
for n in range(500):
   print(''.join(sorted(list(str(fib(n))))),end=', ')
# Soumil Mandal, May 14 2016

a(n) = A004185(A000045(n)). - Michel Marcus, May 17 2016

A354049 The smallest number that includes all the digits of n but does not equal n.

Original entry on oeis.org

10, 10, 12, 13, 14, 15, 16, 17, 18, 19, 100, 101, 21, 31, 41, 51, 61, 71, 81, 91, 102, 12, 122, 32, 42, 52, 62, 72, 82, 92, 103, 13, 23, 133, 43, 53, 63, 73, 83, 93, 104, 14, 24, 34, 144, 54, 64, 74, 84, 94, 105, 15, 25, 35, 45, 155, 65, 75, 85, 95, 106, 16, 26, 36, 46, 56, 166, 76, 86, 96, 107

Offset: 0

Scott R. Shannon and Zach J. Shannon, May 16 2022

The terms cannot start with a leading zero so any number including a zero must have at least one digit greater than zero as its first digit. See the examples below.

			a(9) = 19 as there is no smaller number that includes the digit 9 but does not equal 9.
a(10) = 100 as there is no smaller number that includes the digits 1 and 0 but does not equal 10. Note that '01' = 1 is not allowed.
a(20) = 102 as there is no smaller number that includes the digits 2 and 0 but does not equal 20. Note that '02' = 2 is not allowed.
a(22) = 122 as there is no smaller number that includes two 2 digits but does not equal 22.
a(200) = 1002 as there is no smaller number that includes two 0 digits and the digit 2 but does not equal 200.

Scott R. Shannon, Table of n, a(n) for n = 0..9999
Scott R. Shannon, Image of the first 100000 terms. The green line is y = n.

Cf. A004185, A004185, A337227.

vd(n) = my(d=if (n, digits(n), [0])); vector(10, k, #select(x->(x==k-1), d));
isok(k, n, d) = if (k!=n, my(dd=vd(k)); for (i=1, #d, if (dd[i] < d[i], return(0))); return(1));
a(n) = my(k=0, d=vd(n)); while(!isok(k, n, d), k++); k; \\ Michel Marcus, May 17 2022

def ok(k, n):
    if k == n: return False
    sk, sn = str(k), str(n)
    return all(sk.count(d) >= sn.count(d) for d in set(sn))
def a(n):
    k = 0
    while not ok(k, n): k += 1
    return k
print([a(n) for n in range(71)]) # Michael S. Branicky, May 23 2022

A032553 Arrange digits of cubes in ascending order.

Original entry on oeis.org

0, 1, 8, 27, 46, 125, 126, 334, 125, 279, 1, 1133, 1278, 1279, 2447, 3357, 469, 1349, 2358, 5689, 8, 1269, 1468, 11267, 12348, 12556, 15677, 13689, 12259, 23489, 27, 12799, 23678, 33579, 3349, 24578, 45666, 3556, 24578, 13599, 46

Offset: 0

Patrick De Geest, Apr 15 1998

			Leading zeros discarded (e.g., 40^3 = 64000 = 00046 becomes 46).

Enrique Pérez Herrero, Table of n, a(n) for n = 0..1000

Cf. A032554.
Cf. A028907, A028908.
Cf. A004186, A004185.

Table[FromDigits[Sort[IntegerDigits[n^3]]],{n,0,40}] (* Enrique Pérez Herrero, Sep 29 2013 *)

A068636 a(n) = Min(n, R(n)), where R(n) (A004086) = digit reversal of n.

Original entry on oeis.org

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, 57

Offset: 1

Amarnath Murthy, Feb 27 2002

a(n) = A004185(n) for n <= 100. - Reinhard Zumkeller, Apr 03 2015

			a(12) = min(12,21) = 12. a(34632) = min(34632,23643) = 23643.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A068637, A004086, A004185.

a068636 n = min n $ a004086 n  -- Reinhard Zumkeller, Apr 03 2015

a:= n-> min(n,(s-> parse(cat(seq(s[-i], i=1..length(s)))))(""||n)):
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 22 2015

Table[Min[n,IntegerReverse[n]],{n,80}] (* The program uses the IntegerReverse function from Mathematica version 10 *) (* Harvey P. Dale, Nov 27 2015 *)

def A068636(n): return min(n,int(str(n)[::-1])) # Chai Wah Wu, Jun 26 2025

A096089 Let f(n) = largest number formed using digits of n, g(n) = smallest number formed using digits of n; then a(n) = floor(f(n)/g(n)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 10, 1, 1, 2, 2, 3, 3, 4, 4, 4, 10, 1, 1, 1, 1, 2, 2, 2, 2, 3, 10, 2, 1, 1, 1, 1, 1, 1, 2, 2, 10, 2, 1, 1, 1, 1, 1, 1, 1, 1, 10, 3, 2, 1, 1, 1, 1, 1, 1, 1, 10, 3, 2, 1, 1, 1, 1, 1, 1, 1, 10, 4, 2, 1, 1, 1, 1, 1, 1, 1, 10, 4, 2, 2, 1, 1, 1, 1, 1, 1, 10, 4, 3, 2, 1, 1, 1, 1, 1, 1, 100, 10, 17, 23, 29, 34

Offset: 1

Amarnath Murthy, Jun 22 2004

			a(12324) = floor(43221/12234) = 3.
a(1098) = floor(9810/0189) = 51.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A004186, A004185, A096090, A096091.

A096089 := proc(n)
    floor( A004186(n)/A004185(n)) ;
end proc: # R. J. Mathar, Jul 26 2015

a(n) = d = digits(n); fromdigits(vecsort(d, , 4)) \ fromdigits(vecsort(d))

More terms from Sam Handler (sam_5_5_5_0(AT)yahoo.com), Jul 19 2004

a(1)..a(9) prepended by David A. Corneth, Jan 21 2019

A057169 Least integer with the same nonzero decimal digits as n and one more 0 digit.

Original entry on oeis.org

10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 200, 102, 202, 203, 204, 205, 206, 207, 208, 209, 300, 103, 203, 303, 304, 305, 306, 307, 308, 309, 400, 104, 204, 304, 404, 405, 406, 407, 408, 409, 500, 105

Offset: 1

N. J. A. Sloane, Sep 15 2000

Robert Israel, Table of n, a(n) for n = 1..10000

Decimal analog of A057168. Cf. A004185.

f:= proc(n) local L,m,p;
L:= convert(n,base,10);
m:= nops(L);
L:= sort(subs(0=NULL,L));
p:= nops(L);
10^m*L[1]+add(L[j]*10^(p-j),j=2..p)
end proc:
map(f, [$1..100]); # Robert Israel, May 06 2018

Definition corrected by Robert Israel, May 06 2018

A135374 Mersenne numbers with digits sorted in increasing order and zeros suppressed.

Original entry on oeis.org

1, 3, 7, 15, 13, 36, 127, 255, 115, 123, 247, 459, 1189, 13368, 23677, 35556, 11137, 122346, 224578, 145578, 112579, 133449, 367888, 11256777, 13334455, 1366788, 112234777, 234455568, 11356789, 112334778, 1234446778, 2244567999

Offset: 1

Jonathan Vos Post, Dec 09 2007

This is to A000225 as A078726 is to A000215. a(n) is prime for n = 2, 3, 5, 7, 15, 27, 29 and which other values?

Cf. A000215, A000225, A004185, A078726.

a(n) = A004185(2^n - 1) = 2^n - 1, base 10, with digits sorted in increasing order and zeros suppressed.

Offset corrected by Arkadiusz Wesolowski, Jun 02 2011

A346296 a(0) = 1; thereafter a(n) = 2*a(n-1) + 1, with digits rearranged into nondecreasing order.

Original entry on oeis.org

1, 3, 7, 15, 13, 27, 55, 111, 223, 447, 589, 1179, 2359, 1479, 2599, 1599, 1399, 2799, 5599, 11199, 22399, 44799, 58999, 117999, 235999, 147999, 259999, 159999, 139999, 279999, 559999, 1119999, 2239999, 4479999, 5899999, 11799999, 23599999, 14799999, 25999999

Offset: 0

Ctibor O. Zizka, Jul 13 2021

			a(3) = A004185(2*7+1) = A004185(15) = 15.
a(4) = A004185(2*15+1) = A004185(31) = 13.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,100,-100).

Cf. A004185, A010888, A057615, A069638, A321542.

a[0] = 1; a[n_] := a[n] = FromDigits @ Sort @ IntegerDigits[2*a[n - 1] + 1]; Array[a, 45, 0] (* Amiram Eldar, Jul 13 2021 *)
NestList[FromDigits[Sort[IntegerDigits[2#+1]]]&,1,40] (* Harvey P. Dale, Oct 01 2023 *)

lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = fromdigits(vecsort(digits(2*va[n-1]+1)));); va;} \\ Michel Marcus, Aug 31 2021

from itertools import accumulate
def atis(anm1, _): return int("".join(sorted(str(2*anm1+1))))
print(list(accumulate([1]*39, atis))) # Michael S. Branicky, Aug 31 2021

a(n) = A004185(2*a(n-1)+1).
For k >= 1;
a(12*k-9) = 100^(k-1) *   16 - 1;
a(12*k-8) = 100^(k-1) *   14 - 1;
a(12*k-7) = 100^(k-1) *   28 - 1;
a(12*k-6) = 100^(k-1) *   56 - 1;
a(12*k-5) = 100^(k-1) *  112 - 1;
a(12*k-4) = 100^(k-1) *  224 - 1;
a(12*k-3) = 100^(k-1) *  448 - 1;
a(12*k-2) = 100^(k-1) *  590 - 1;
a(12*k-1) = 100^(k-1) * 1180 - 1;
a(12*k)   = 100^(k-1) * 2360 - 1;
a(12*k+1) = 100^(k-1) * 1480 - 1;
a(12*k+2) = 100^(k-1) * 2600 - 1.
G.f.: -(1800*x^15 -720*x^14 +1080*x^13 -1080*x^12 -590*x^11 -142*x^10 -224*x^9 -112*x^8 -56*x^7 -28*x^6 -14*x^5 +2*x^4 -8*x^3 -4*x^2 -2*x -1) / ((x-1)*(10*x^6-1)*(10*x^6+1)). - Alois P. Heinz, Aug 02 2021
a(n) = 100*a(n-12) + 99 for n >= 15. - Pontus von Brömssen, Sep 01 2021

A348287 Arrange nonzero digits of n in increasing order then append the zeros.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 12, 22, 23, 24, 25, 26, 27, 28, 29, 30, 13, 23, 33, 34, 35, 36, 37, 38, 39, 40, 14, 24, 34, 44, 45, 46, 47, 48, 49, 50, 15, 25, 35, 45, 55, 56, 57, 58, 59, 60, 16, 26, 36, 46, 56, 66

Offset: 0

Simon Strandgaard, Oct 10 2021

Shares 101 initial terms with A328447. First difference is A328447(101)=101 vs A348287(101)=110.

			a(1010) = 1100,
a(1234567890) = 1234567890,
a(9876543210) = 1234567890.

Cf. A004185, A328447.

a[n_] := 10^DigitCount[n, 10, 0] * FromDigits[Sort[IntegerDigits[n]]]; Array[a, 100, 0] (* Amiram Eldar, Oct 10 2021 *)

a(n) = fromdigits(vecsort(digits(n), x->if(x,x,10)));

def a(n): s = str(n); return int("".join(sorted(s)))*10**s.count('0')
print([a(n) for n in range(68)]) # Michael S. Branicky, Oct 10 2021

p Array.new(40) { |n|
  n.to_s.gsub('0','a').split(//).sort.join.gsub('a','0').to_i
}

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A272918 Fibonacci numbers with the base 10 digits sorted into increasing order.

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A354049 The smallest number that includes all the digits of n but does not equal n.

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A032553 Arrange digits of cubes in ascending order.

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A068636 a(n) = Min(n, R(n)), where R(n) (A004086) = digit reversal of n.

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A096089 Let f(n) = largest number formed using digits of n, g(n) = smallest number formed using digits of n; then a(n) = floor(f(n)/g(n)).

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A057169 Least integer with the same nonzero decimal digits as n and one more 0 digit.

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A135374 Mersenne numbers with digits sorted in increasing order and zeros suppressed.

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