A151949 -id:A151949 - cp's OEIS frontend

A151956 a(1) = 1002; thereafter a(n) = (a(n-1) with digits sorted into descending order) - (a(n-1) with digits sorted into ascending order) (see the Kaprekar map, A151949).

1002, 2088, 8532, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174, 6174

Offset: 1

Harvey P. Dale, Aug 18 2009

For the list of fixed points see A099009.

Index entries for the Kaprekar map

Cf. A099009, A151947, A151949, A151950, A069746, A151955, A151951, A151957.

A010785 Repdigit numbers, or numbers whose digits are all equal.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 111, 222, 333, 444, 555, 666, 777, 888, 999, 1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999, 11111, 22222, 33333, 44444, 55555, 66666, 77777, 88888, 99999, 111111, 222222, 333333, 444444, 555555, 666666

Offset: 0

N. J. A. Sloane

Complement of A139819. - David Wasserman, May 21 2008
Subsequence of A134336 and of A178403. - Reinhard Zumkeller, May 27 2010
Subsequence of A193460. - Reinhard Zumkeller, Jul 26 2011
Intersection of A009994 and A009996. - David F. Marrs, Sep 29 2018
Beiler (1964) called these numbers "monodigit numbers". The term "repdigit numbers" was used by Trigg (1974). - Amiram Eldar, Jan 21 2022

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, p. 83.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Eric F. Bravo, Carlos A. Gómez and Florian Luca, Product of Consecutive Tribonacci Numbers With Only One Distinct Digit, J. Int. Seq., Vol. 22 (2019), Article 19.6.3.
Eric Fernando Bravo, On concatenations of Padovan and Perrin numbers, Math. Commun. (2023) Vol 28, 105-119.
Mahadi Ddamulira, Repdigits as sums of three balancing numbers, Mathematica Slovaca, (2019), hal-02405969.
Mahadi Ddamulira, Padovan numbers that are concatenations of two distinct repdigits, arXiv:2003.10705 [math.NT], 2020.
Mahadi Ddamulira, Tribonacci numbers that are concatenations of two repdigits, hal-02547159, Mathematics [math] / Number Theory [math.NT], 2020.
Mahadi Ddamulira, Padovan numbers that are concatenations of two distinct repdigits, Mathematica Slovaca, Vol. 71, No. 2 (2021), pp. 275-284.
Bart Goddard and Jeremy Rouse, Sum of two repdigits a square, arXiv:1607.06681 [math.NT], 2016. Mentions this sequence.
Bir Kafle, Florian Luca and Alain Togbé, Triangular Repblocks, Fibonacci Quart., Vol. 56, No. 4 (2018), pp. 325-328.
Bir Kafle, Florian Luca and Alain Togbé, Pentagonal and heptagonal repdigits, Annales Mathematicae et Informaticae, Vol. 52 (2020), pp. 137-145.
Benedict Vasco Normenyo, Bir Kafle, and Alain Togbé, Repdigits as Sums of Two Fibonacci Numbers and Two Lucas Numbers, Integers, Vol. 19 (2019), Article A55.
Salah Eddine Rihane and Alain Togbé, Repdigits as products of consecutive Padovan or Perrin numbers, Arab. J. Math., Vol. 10 (2021), pp. 469-480.
Charles W. Trigg, Infinite sequences of palindromic triangular numbers, The Fibonacci Quarterly, Vol. 12, No. 2 (1974), pp. 209-212.
Eric Weisstein's World of Mathematics, Repdigit.
Wikipedia, Repdigit.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,-10).

Cf. A000005, A009994, A009996, A010879, A037904, A047842, A059995, A065444, A134336, A139819, A151949, A178401, A178403, A180410, A193459, A193460, A202022, A227362, A244112.

a010785 n = a010785_list !! n
a010785_list = 0 : r [1..9] where
   r (x:xs) = x : r (xs ++ [10*x + x `mod` 10])
-- Reinhard Zumkeller, Jul 26 2011

[(n-9*Floor((n-1)/9))*(10^Floor((n+8)/9)-1)/9: n in [0..50]]; // Vincenzo Librandi, Nov 10 2014

A010785 := proc(n)
    (n-9*floor(((n-1)/9)))*((10^(floor(((n+8)/9)))-1)/9) ;
end proc:
seq(A010785(n), n = 0 .. 100); # Robert Israel, Nov 09 2014

fQ[n_]:=Module[{id=IntegerDigits[n]}, Length[Union[id]]==1]; Select[Range[0,10000], fQ] (* Vladimir Joseph Stephan Orlovsky, Dec 29 2010 *)
Union[FromDigits/@Flatten[Table[PadRight[{},i,n],{n,0,9},{i,6}],1]] (* or *) LinearRecurrence[{0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,-10}, {0,1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88},40] (* Harvey P. Dale, Dec 28 2011 *)
Union@ Flatten@ Table[k (10^n - 1)/9, {k, 0, 9}, {n, 6}] (* Robert G. Wilson v, Oct 09 2014 *)
Table[(n - 9 Floor[(n-1)/9]) (10^Floor[(n+8)/9] - 1)/9, {n, 0, 50}] (* José de Jesús Camacho Medina, Nov 06 2014 *)

a(n)=10^((n+8)\9)\9*((n-1)%9+1) \\ Charles R Greathouse IV, Jun 15 2011

nxt(n,t=n%10)=if(t<9,n*(t+1),n*10+9)\t \\ Yields the term a(k+1) following a given term a(k)=n. M. F. Hasler, Jun 24 2016

is(n)={1==#Set(digits(n))}
inv(n) = 9*#Str(n) + n%10 - 9 \\ David A. Corneth, Jun 24 2016

def a(n): return 0 if n == 0 else int(str((n-1)%9+1)*((n-1)//9+1))
print([a(n) for n in range(55)]) # Michael S. Branicky, Dec 29 2021

print([0]+[int(d*r) for r in range(1, 7) for d in "123456789"]) # Michael S. Branicky, Dec 29 2021

# without string operations
def a(n): return 0 if n == 0 else (10**((n-1)//9+1)-1)//9*((n-1)%9+1)
print([a(n) for n in range(55)]) # Michael S. Branicky, Nov 03 2023

A037904(a(n)) = 0. - Reinhard Zumkeller, Dec 14 2007
A178401(a(n)) > 0. - Reinhard Zumkeller, May 27 2010
From Reinhard Zumkeller, Jul 26 2011: (Start)
For n > 0: A193459(a(n)) = A000005(a(n)).
for n > 10: a(n) mod 10 = floor(a(n)/10) mod 10.
A010879(n) = A010879(A059995(n)). (End)
A202022(a(n)) = 1. - Reinhard Zumkeller, Dec 09 2011
a(0)=0, a(1)=1, a(2)=2, a(3)=3, a(4)=4, a(5)=5, a(6)=6, a(7)=7, a(8)=8, a(9)=9, a(10)=11, a(11)=22, a(12)=33, a(13)=44, a(14)=55, a(15)=66, a(16)=77, a(17)=88, a(n) = 11*a(n-9) - 10*a(n-18). - Harvey P. Dale, Dec 28 2011
A151949(a(n)) = 0; A180410(a(n)) = A227362(a(n)). - Reinhard Zumkeller, Jul 09 2013
a(n) = (n - 9*floor((n-1)/9))*(10^floor((n+8)/9) - 1)/9. - José de Jesús Camacho Medina, Nov 06 2014
G.f.: x*(1+2*x+3*x^2+4*x^3+5*x^4+6*x^5+7*x^6+8*x^7+9*x^8)/((1-x^9)*(1-10*x^9)). - Robert Israel, Nov 09 2014
A047842(a(n)) = A244112(a(n)). - Reinhard Zumkeller, Nov 11 2014
Sum_{n>=1} 1/a(n) = (7129/2520) * A065444 = 3.11446261209177581335... - Amiram Eldar, Jan 21 2022

Name clarified by Jon E. Schoenfield, Nov 10 2023

A003558 Least number m > 0 such that 2^m == +-1 (mod 2n + 1).

Original entry on oeis.org

1, 1, 2, 3, 3, 5, 6, 4, 4, 9, 6, 11, 10, 9, 14, 5, 5, 12, 18, 12, 10, 7, 12, 23, 21, 8, 26, 20, 9, 29, 30, 6, 6, 33, 22, 35, 9, 20, 30, 39, 27, 41, 8, 28, 11, 12, 10, 36, 24, 15, 50, 51, 12, 53, 18, 36, 14, 44, 12, 24, 55, 20, 50, 7, 7, 65, 18, 36, 34, 69, 46

Offset: 0

N. J. A. Sloane

nonn

Multiplicative suborder of 2 (mod 2n+1) (or sord(2, 2n+1)).
This is called quasi-order of 2 mod b, with b = 2*n+1, for n >= 1, in the Hilton/Pederson reference.
For the complexity of computing this, see A002326.
Also, the order of the so-called "milk shuffle" of a deck of n cards, which maps cards (1,2,...,n) to (1,n,2,n-1,3,n-2,...).  See the paper of Lévy. - Jeffrey Shallit, Jun 09 2019
It appears that under iteration of the base-n Kaprekar map, for even n > 2 (A165012, A165051, A165090, A151949 in bases 4, 6, 8, 10), almost all cycles are of length a(n/2 - 1); proved under the additional constraint that the cycle contains at least one element satisfying "number of digits (n-1) - number of digits 0 = o(total number of digits)". - Joseph Myers, Sep 05 2009
From Gary W. Adamson, Sep 20 2011: (Start)
a(n) can be determined by the cycle lengths of iterates using x^2 - 2, seed 2*cos(2*Pi/N); as shown in the A065941 comment of Sep 06 2011. The iterative map of the logistic equation 4x*(1-x) is likewise chaotic with the same cycle lengths but initiating the trajectory with sin^2(2*Pi/N), N = 2n+1 [Kappraff & Adamson, 2004]. Chaotic terms with the identical cycle lengths can be obtained by applying Newton's method to i = sqrt(-1) [Strang, also Kappraff and Adamson, 2003], resulting in the morphism for the cot(2*Pi/N) trajectory: (x^2-1)/2x. (End)
From Gary W. Adamson, Sep 11 2019: (Start)
Using x^2 - 2 with seed 2*cos(Pi/7), we obtain the period-three trajectory 1.8019377...-> 1.24697...-> -0.445041... For an odd prime N, the trajectory terms represent diagonal lengths of regular star 2N-gons, with edge the shortest value (0.445... in this case.) (Cf. "Polygons and Chaos", p. 9, Fig 4.) We can normalize such lengths by dividing through with the lowest value, giving 3 diagonals of the 14-gon: (1, 2.801937..., 4.048917...). Label the terms ranked in magnitude with odd integers (1, 3, 5), and we find that the diagonal lengths are in agreement with the diagonal formula (sin(j*Pi)/14)/(sin(Pi/14)), with j = (1,3,5). (End)
Roots of signed n-th row A054142 polynomials are chaotic with respect to the operation (-2, x^2), with cycle lengths a(n). Example: starting with a root to x^3 - 5x^2 + 6x - 1 = 0; (2 + 2*cos(2*Pi/N) = 3.24697...); we obtain the trajectory (3.24697...-> 1.55495...-> 0.198062...); the roots to the polynomial with cycle length 3 matching a(3) = 3. - Gary W. Adamson, Sep 21 2011
From Juhani Heino, Oct 26 2015: (Start)
Start a sequence with numbers 1 and n. For next numbers, add previous numbers going backwards until the sum is even. Then the new number is sum/2. I conjecture that the sequence returns to 1,n and a(n) is the cycle length.
For example:
  1,7,4,2,1,7,... so a(7) = 4.
  1,6,3,5,4,2,1,6,... so a(6) = 6. (End)
From Juhani Heino, Nov 06 2015: (Start)
Proof of the above conjecture: Let n = -1/2; thus 2n + 1 = 0, so operations are performed mod (2n + 1). When the member is even, it is divided by 2. When it is odd, multiply by n, so effectively divide by -2. This is all well-defined in the sense that new members m are 1 <= m <= n. Now see what happens starting from an odd member m. The next member is -m/2. As long as there are even members, divide by 2 and end up with an odd -m/(2^k). Now add all the members starting with m. The sum is m/(2^k). It's divided by 2, so the next member is m/(2^(k+1)). That is the same as (-m/(2^k))/(-2), as with the definition.
So actually start from 1 and always divide by 2, although the sign sometimes changes. Eventually 1 is reached again. The chain can be traversed backwards and then 2^(cycle length) == +-1 (mod 2n + 1).
To conclude, we take care of a(0): sequence 1,0 continues with zeros and never returns to 1. So let us declare that cycle length 0 means unavailable. (End)
From Gary W. Adamson, Aug 20 2019: (Start)
Terms in the sequence can be obtained by applying the doubling sequence mod (2n + 1), then counting the terms until the next term is == +1 (mod 2n + 1).  Example: given 25, the trajectory is (1, 2, 4, 8, 16, 7, 14, 3, 6, 12).
The cycle ends since the next term is 24 == -1 (mod 25) and has a period of 10. (End)
From Gary W. Adamson, Sep 04 2019: (Start)
Conjecture of Kappraff and Adamson in "Polygons and Chaos", p. 13 Section 7, "Chaos and Number": Given the cycle length for N = 2n + 1, the same cycle length is present in bases 4, 9, 16, 25, ..., m^2, for the expansion of 1/N.
Examples: The cycle length for 7 is 3, likewise for 1/7 in base 4: 0.021021021.... In base 9 the expansion of 1/7 is 0.125125125... Check: The first few terms are 1/9 + 2/81 + 5/729 = 104/279 = 0.1426611... (close to 1/7 = 0.142857...). (End)
From Gary W. Adamson, Sep 24 2019: (Start)
An exception to the rule for 1/N in bases m^2: (when N divides m^2 as in 1/7 in base 49, = 7/49, rational). When all terms in the cycle are the same, the identity reduces to 1/N in (some bases) = .a, a, a, .... The minimal values of "a" for 1/N are provided as examples, with the generalization 1/N in base (N-1)^2 = .a, a, a, ... for N odd:
   1/3  in base  4  =  .1, 1, 1, ...
   1/5  in base 16  =  .3, 3, 3, ...
   1/7  in base 36  =  .5, 5, 5, ...
   1/9  in base 64  =  .7, 7, 7, ...
   1/11 in base 100 =  .9, 9, 9, ... (Check: the first three terms are 9/100 +9/(100^2) + 9/(100^3) = 0.090909 where 1/11 = 0.09090909...). (End)
For N = 2n+1, the corresponding entry is equal to the degree of the polynomial for N shown in (Lang, Table 2, p. 46). As shown, x^3 - 3x - 1 is the minimal polynomial for N = 9, with roots (1.87938..., -1.53208..., 0.347296...); matching the (abs) values of the 2*cos(Pi/9) trajectory using x^2 - 2. Thus, a(4) = 3. If N is prime, the polynomials shown in Table 2 are the same as those for the same N in A065941. If different, the minimal polynomials shown in Table 2 are factors of those in A065941. - Gary W. Adamson, Oct 01 2019
The terms in the 2*cos(Pi/N) trajectory (roots to the minimal polynomials in A187360 and (Lang)), are quickly obtained from the doubling trajectory (mod N) by using the operation L(m) 2*cos(x)--> 2*cos(m*x), where L(2), the second degree Lucas polynomial (A034807) is x^2 - 2. Relating to the heptagon and using seed 2*cos(Pi/7), we obtain the trajectory 1.8019..., 1.24697..., and 0.445041....; cyclic with period 3. All such roots can be derived from the N-th roots of Unity and can be mapped on the Vesica Piscis. Given the roots of Unity (Polar 1Angle(k*2*Pi/N), k = 1, 2, ..., (N-1)/2) the Vesica Piscis maps these points on the left (L) circle to the (R) circle by adding 1A(0) or (a + b*I) = (1 + 0i). But this operation is the same as vector addition in which the resultant vector is 1 + 1A(k*(2*Pi/N)). Example: given the radius at 2*Pi/7 on the left circle, this maps to (1 + 1A(2*Pi/7)) on the right circle; or 1A(2*Pi/7) --> (1.8019377...A(Pi/7).  Similarly, 1 + 1A((2)*2*Pi/7)) maps to (1.24697...A (2*Pi/7); and 1 + 1A(3*2*Pi/7) maps to (.0445041...A(3*Pi/7). - Gary W. Adamson, Oct 23 2019
From Gary W. Adamson, Dec 01 2021: (Start)
As to segregating the two sets: (A014659 terms are those N = (2*n+1), N divides (2^m - 1), and (A014657 terms are those N that divide (2^m + 1)); it appears that the following criteria apply: Given IcoS(N, 1) (cf. Lang link "On the Equivalence...", p. 16, Definition 20), if the number of odd terms is odd, then N belongs to A014659, otherwise A014657. In IcoS(11, 1): (1, 2, 4, 3, 5), three odd terms indicate that 11 is a term in A014657. IcoS(15, 1) has the orbit (1, 2, 4, 7) with two odd terms indicating that 15 is a term in A014659.
It appears that if sin(2^m * Pi/N) has a negative sign, then N is in A014659; otherwise N is in A014657. With N = 15, m is 4 and sin(16 * Pi/15) is -0.2079116... If N is 11, m is 5 and sin(32 * Pi/11) is 0.2817325. (End)
On the iterative map using x^2 - 2, (Devaney, p. 126) states that we must find the function that takes 2*cos(Pi) -> 2*cos(2*Pi). "However, we may write 2*cos(2*Pi) = 2*(2*cos^2(Pi) - 1) = (2*cos(Pi))^2 - 2. So the required function is x^2 - 2." On the period 3 implies chaos theorem of James Yorke and T.Y. Li, proved in 1975; Devaney (p. 133) states that if F is continuous and we find a cycle of period 3, there are infinitely many other cycles for this map with every possible period. Check: The x^2 - 2 orbit for 7 has a period of 3, so this entry has periodic points of all other periods. - Gary W. Adamson, Jan 04 2023
It appears that a(n) is the length of the cycle starting at 2/(2*n+1) for the map x->1 - abs(2*x-1). - Michel Marcus, Jul 16 2025

			a(3) = 3 since f(x) = x^2 - 2 has a period of 3 using seed 2*cos(2*Pi/7), where 7 = 2*3 + 1.
a(15) = 5 since the iterative map of the logistic equation 4x*(1-x) has a period 5 using seed sin^2(2*Pi)/N; N = 31 = 2*15 + 1.

Peter Hilton and Jean Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, 2010, pp. 261-264.
Carl Schick, Trigonometrie und unterhaltsame Zahlentheorie, Bokos Druck, Zürich, 2003 (ISBN 3-9522917-0-6).
Robert L. Devaney, A First Course in Chaotic Dynamical Systems, Theory and Experiment; Perseus Books Publishing, 1992, pp. 121-126.

T. D. Noe, Table of n, a(n) for n = 0..1000
R. Bekes, J. Pedersen, and B. Shao, Mad tea party cyclic partitions, Coll. Math. J. 43 (1) (2012) 25-36, Jstor.
William Q. Erickson, Daniel Herden, Jonathan Meddaugh, Mark R. Sepanski, Mitchell Minyard, and Kyle Rosengartner, Limits and Periodicity of Metamour 2-Distance Graphs, arXiv:2409.02306 [math.CO], 2024. See p. 9.
Daniel Gabric and Jeffrey Shallit, Borders, Palindrome Prefixes, and Square Prefixes, arXiv:1906.03689 [cs.DM], 2019.
P. Hilton and J. Pedersen, On Factoring 2^k+-1, The Math. Educ. 5 (1) (1994) 29-31.
Jay Kappraff and Gary W. Adamson, Polygons and Chaos, Bridges: Mathematical Connections in Art, Music, and Science, 2001, pages 67-80.
Anthony Kay and Katrina Downes-Ward, Fixed Points and Cycles of the Kaprekar Transformation: 1. Odd Bases, J. Int. Seq., Vol. 25 (2022), Article 22.6.7.
Anthony Kay and Katrina Downes-Ward, Fixed Points and Cycles of the Kaprekar Transformation: 2. Even bases, arXiv:2408.12257 [math.CO], 2024. See p. 6.
Torleiv Klove, On covering sets for limited-magnitude errors, Cryptogr. Commun. 8 (3) (2016) 415-433
Wolfdieter Lang, The Field Q(2Cos Pi/N), its Galois group and length ratios in the regular n-gon, arXiv:1210.1018 [math.GR], 2012-2017.
Wolfdieter Lang, On the Equivalence of Three Complete Cyclic Systems of Integers, arXiv:2008.04300 [math.NT], 2020.
Paul Lévy, Sur quelques classes de permutations, Compositio Mathematica, 8 (1951), 1-48.
Olivier Martin, Andrew M. Odlyzko, and Stephen Wolfram, Algebraic Properties of Cellular Automata, Communications in Mathematical Physics, volume 93, number 2, June 1984, pages 219-258, appendix B part C sord_N(2). Also third author's copy (and text).
Harry J. Smith, XICalc - Extra Precision Integer Calculator. [Broken link]
Gilbert Strang, A Chaotic Search for i, College Mathematics Journal 22, 3-12, (1991) [JSTOR].
Eric Weisstein's World of Mathematics, Multiplicative Order.
Eric Weisstein's World of Mathematics, Suborder Function.

Cf. A054142, A065941, A085478, A160657, A179480, A135303 (coach numbers), A216371 (odd primes with one coach), A000215 (Fermat numbers).
A216066 is an essentially identical sequence apart from the offset.
Cf. A329593, A332433 (signs).
Cf. A014657, A014659.

A003558 := proc(n)
    local m,mo ;
    if n = 0 then
        return 0 ;
    end if;
    for m from 1 do
        mo := modp(2^m,2*n+1) ;
        if mo in {1,2*n} then
            return m;
        end if;
    end do:
end proc:
seq(A003558(n),n=0..20) ; # R. J. Mathar, Dec 01 2014
f:= proc(n) local t;
      t:= numtheory:-mlog(-1,2,n);
      if t = FAIL then numtheory:-order(2,n) else t fi
end proc:
0, seq(f(2*k+1),k=1..1000); # Robert Israel, Oct 26 2015

Suborder[a_,n_]:=If[n>1&&GCD[a,n]==1,Min[MultiplicativeOrder[a,n,{-1,1}]],0];
Join[{1},Table[Suborder[2,2n+1],{n,100}]] (* T. D. Noe, Aug 02 2006 *) (* revised by Vincenzo Librandi, Apr 11 2020 *)

a(n) = {m=1; while(m, if( (2^m) % (2*n+1) == 1 || (2^m) % (2*n+1) == 2*n, return(m)); m++)} \\ Altug Alkan, Nov 06 2015

isok(m, n) = my(md = Mod(2, 2*n+1)^m); (md==1) || (md==-1);
A003558(n) = my(m=1); while(!isok(m,n) , m++); m; \\ Michel Marcus, May 06 2020

def A003558(n):
    m, k = 1, 2 % (c:=(r:=n<<1)+1)
    while not (k==1 or k==r):
        k = 2*k%c
        m += 1
    return m # Chai Wah Wu, Oct 09 2023

a(n) = log_2(A160657(n) + 2) - 1. - Nathaniel Johnston, May 22 2009
a(n-1) = card {cos((2^k)*Pi/(2*n-1)): k in N} for n >= 1 (see A216066, an essentially identical sequence, for more information). - Roman Witula, Sep 01 2012
a(n) <= n. - Charles R Greathouse IV, Sep 15 2012 [For n >= 1]
a(n) = min{k > 0 | q_k = q_0} where q_0 = 1 and q_k = |2*n+1 - 2*q_{k-1}| (cf. [Schick, p. 4]; q_k=1 for n=1; q_k=A010684(k) for n=2; q_k=A130794(k) for n=3; q_k=|A154870(k-1)| for n=4; q_k=|A135449(k)| for n=5.) - Jonathan Skowera, Jun 29 2013
2^(a(n)) == A332433(n) (mod (2*n+1)), and (2^(a(n)) - A332433(n))/(2*n+1) = A329593(n), for n >= 0. - Wolfdieter Lang, Apr 09 2020

More terms from Harry J. Smith, Feb 11 2005

Entry revised by N. J. A. Sloane, Aug 02 2006 and again Dec 10 2017

A099009 Fixed points of the Kaprekar mapping f(n) = n' - n'', where in n' the digits of n are arranged in descending, in n'' in ascending order.

Original entry on oeis.org

0, 495, 6174, 549945, 631764, 63317664, 97508421, 554999445, 864197532, 6333176664, 9753086421, 9975084201, 86431976532, 555499994445, 633331766664, 975330866421, 997530864201, 999750842001, 8643319766532, 63333317666664

Offset: 1

Klaus Brockhaus, Sep 22 2004

There are no seven-digit fixed points.
Let d(n) denote n repetitions of the digit d. The sequence includes the following for all n>=0: 5(n)499(n)4(n)5, 63(n)176(n)4, 8643(n)1976(n)532. - Jens Kruse Andersen, Oct 04 2004
0's in n giving leading 0's in n'' is allowed.
For every natural number n let n' and n" be the numbers obtained by arranging the digits of n into decreasing and increasing order, and let f(n)=n'-n". It is known that the number 6174 is invariant under this transformation and that applying f a certain number of times to a number n with four digits the numbers 0, 495 or 6174 are always reached. - Vincenzo Librandi, Nov 17 2010
Each term of A055162(n) corresponds to A099009(n+1), with its digits being reordered in the ascending manner. - Alexander R. Povolotsky, Apr 27 2012
All terms of this sequence are divisible by nine, a(n)/9 = A132155(n). - Alexander R. Povolotsky, Apr 29 2012
A055160 differs from this sequence only at the positions of two terms in it: 554999445 and 555499994445. - Alexander R. Povolotsky, May 01 2012
The union of the sequences A214555, A214556, A214557, A214558, A214559 and the element 0 gives the sequence A099009. - Syed Iddi Hasan, Jul 24 2012
The comment made by Jens Kruse Andersen is missing one more family of terms (which starts with one or more digits "9" and ends with the digit "1"): 97508421, 9753086421, 9975084201, 975330866421, 997530864201, 999750842001, ... This family could be generalized (using the same method as in Andersen's comment) and it is actually covered by Syed Iddi Hasan in A214559. Also A214557 and A214558 (both - by Syed Iddi Hasan) are variants of Andersen's 8643(n)1976(n)532. - Alexander R. Povolotsky, Mar 14 2015
Fixed points of A151949. - Reinhard Zumkeller, Mar 23 2015

			6174 is a fixed point of the mapping and hence a term: 6174 -> 7641 - 1467 = 6174.

Syed Iddi Hasan, Table of n, a(n) for n = 1..8924
Mauro Fiorentini, Kaprekar (costante di) (in Italian)
Manuj Mishra, Illustration of first 8923 terms, with each digit in a different color
Manuj Mishra, Illustration as above but only including terms of even length
Manuj Mishra, Illustration as above but only including terms of odd length
Joseph Myers, List of cycles under Kaprekar map (all numbers with <= 60 digits; cycles are represented by their smallest value)
Conrad Roche, Kaprekar Series Generator.
Eric Weisstein's World of Mathematics, Kaprekar Routine
Index entries for the Kaprekar map

Cf. A090429, A069746, A099010, A151959, A055162, A132155, A055160, A214557, A214558, A214559
In other bases: A163205 (base 2), A164997 (base 3), A165016 (base 4), A165036 (base 5), A165055 (base 6), A165075 (base 7), A165094 (base 8), A165114 (base 9).
Cf. A151949, A004185, A004186.

a099009 n = a099009_list !! (n-1)
a099009_list = [x | x <- [0..], a151949 x == x]
-- Reinhard Zumkeller, Mar 23 2015

a:=func; [k:k in [0..10^7]|a(k)]; // Marius A. Burtea, Sep 12 2019

f[n_] := Block[{d = IntegerDigits@ n, a, b}, a = FromDigits@ Sort@ d; b = FromDigits@ Reverse@ Sort@ d; n == b - a]; Select[Range@ 1000000, f] (* Michael De Vlieger, Mar 20 2015 *)

# (version 2.4) from Tim Peters
def extend(base, start, n):
    if n == 0:
        yield base
        return
    for i in range(start, 10):
        for x in extend(base + str(i), i, n-1):
            yield x
def drive(n):
    result = []
    for lo in extend("", 0, n):
        ilo = int(lo)
        if ilo == 0 and n > 1:
            continue
        hi = lo[::-1]
        diff = str(int(hi) - ilo)
        diff = "0" * (n - len(diff)) + diff
        if sorted(diff) == list(lo):
            result.append(diff)
    return sorted(result)
for n in range(1, 17):
    # print("Length", n)
    # print('-' * 40)
    for r in drive(n):
        print(r, end=', ')

More terms from Jens Kruse Andersen and Tim Peters (tim(AT)python.org), Oct 04 2004

Corrected by Jens Kruse Andersen, Oct 25 2004

A165012 a(n) = image of n under the base-4 Kaprekar map n -> (n with digits sorted into descending order) - (n with digits sorted into ascending order).

Original entry on oeis.org

0, 0, 0, 0, 3, 0, 3, 6, 6, 3, 0, 3, 9, 6, 3, 0, 15, 15, 30, 45, 15, 0, 15, 30, 30, 15, 15, 30, 45, 30, 30, 30, 30, 30, 30, 45, 30, 15, 15, 30, 30, 15, 0, 15, 45, 30, 15, 15, 45, 45, 45, 45, 45, 30, 30, 30, 45, 30, 15, 15, 45, 30, 15, 0, 63, 75, 138, 201, 75, 63, 126, 189, 138, 126

Offset: 0

Joseph Myers, Sep 04 2009

			For n = 11, 11_10 = 23_4. So, a(11) = 32_4 - 23_4 = 14 - 11 = 3. - _Indranil Ghosh_, Feb 02 2017

Indranil Ghosh, Table of n, a(n) for n = 0..16384 (terms 0..1024 from Joseph Myers)
Index entries for the Kaprekar map

Cf. A165013.

In other bases: A164884 (base 2), A164993 (base 3), A165032 (base 5), A165051 (base 6), A165071 (base 7), A165090 (base 8), A165110 (base 9), A151949 (base 10).

b4km[n_]:=Module[{idn4=Sort[IntegerDigits[n,4]]},FromDigits[ Reverse[ idn4],4]-FromDigits[idn4,4]]; Array[b4km,80,0]

cons(m) = {local(b, r); r=0; b=1; for(i=1, matsize(m)[2], r=r+b*m[i]; b=b*4); r}
A165012(n) = {local(m, r); r=[]; m=n; while(m>0, r=concat(m%4, r); m=floor(m/4)); cons(vecsort(r,,0))-cons(vecsort(r,,4))} \\ Michael B. Porter, Nov 05 2009

A165051 a(n) = image of n under the base-6 Kaprekar map n -> (n with digits sorted into descending order) - (n with digits sorted into ascending order).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 5, 0, 5, 10, 15, 20, 10, 5, 0, 5, 10, 15, 15, 10, 5, 0, 5, 10, 20, 15, 10, 5, 0, 5, 25, 20, 15, 10, 5, 0, 35, 35, 70, 105, 140, 175, 35, 0, 35, 70, 105, 140, 70, 35, 35, 70, 105, 140, 105, 70, 70, 70, 105, 140, 140, 105, 105, 105, 105, 140, 175, 140, 140, 140

Offset: 0

Joseph Myers, Sep 04 2009

			For n = 10, 10_10 = 14_6. So, a(10) = 41_6 - 14_6 = 25 - 10 = 15. - _Indranil Ghosh_, Feb 02 2017

Indranil Ghosh, Table of n, a(n) for n = 0..46656 (terms 0..1296 from Joseph Myers)
Index entries for the Kaprekar map

Cf. A165052.

In other bases: A164884 (base 2), A164993 (base 3), A165012 (base 4), A165032 (base 5), A165071 (base 7), A165090 (base 8), A165110 (base 9), A151949 (base 10).

a[n_] := With[{dd = IntegerDigits[n, 6]}, FromDigits[ReverseSort[dd], 6] - FromDigits[Sort[dd], 6]];
a /@ Range[0, 100] (* Jean-François Alcover, Jan 08 2020 *)

A165090 a(n) is the image of n under the base-8 Kaprekar map n -> (n with digits sorted into descending order) - (n with digits sorted into ascending order).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 7, 0, 7, 14, 21, 28, 35, 42, 14, 7, 0, 7, 14, 21, 28, 35, 21, 14, 7, 0, 7, 14, 21, 28, 28, 21, 14, 7, 0, 7, 14, 21, 35, 28, 21, 14, 7, 0, 7, 14, 42, 35, 28, 21, 14, 7, 0, 7, 49, 42, 35, 28, 21, 14, 7, 0, 63, 63, 126, 189, 252, 315, 378, 441, 63, 0, 63, 126, 189

Offset: 0

Joseph Myers, Sep 04 2009

			For n = 11, 11_10 = 13_8. So, a(11) = 31_8 - 13_8 = 25 - 11 = 14. - _Indranil Ghosh_, Feb 02 2017

Indranil Ghosh, Table of n, a(n) for n = 0..32768 (terms 0..4096 from Joseph Myers)
Index entries for the Kaprekar map

Cf. A165091.

In other bases: A164884 (base 2), A164993 (base 3), A165012 (base 4), A165032 (base 5), A165051 (base 6), A165071 (base 7), A165110 (base 9), A151949 (base 10).

a[n_] := With[{dd = IntegerDigits[n, 8]}, FromDigits[ReverseSort[dd], 8] - FromDigits[Sort[dd], 8]];
a /@ Range[0, 100] (* Jean-François Alcover, Jan 08 2020 *)

def A165090(n):
    if n==0:return 0
    return int("".join(sorted(oct(n)[2:],reverse=True)),8)-int("".join(sorted(oct(n)[2:])),8) # Indranil Ghosh, Feb 02 2017

A165110 a(n) = image of n under the base-9 Kaprekar map n -> (n with digits sorted into descending order) - (n with digits sorted into ascending order).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0, 8, 16, 24, 32, 40, 48, 56, 16, 8, 0, 8, 16, 24, 32, 40, 48, 24, 16, 8, 0, 8, 16, 24, 32, 40, 32, 24, 16, 8, 0, 8, 16, 24, 32, 40, 32, 24, 16, 8, 0, 8, 16, 24, 48, 40, 32, 24, 16, 8, 0, 8, 16, 56, 48, 40, 32, 24, 16, 8, 0, 8, 64, 56, 48, 40, 32, 24, 16, 8

Offset: 0

Joseph Myers, Sep 04 2009

			For n = 11, 11_10 = 12_9. So, a(11) = 21_9 - 12_9 = 19 - 11 = 8. - _Indranil Ghosh_, Feb 02 2017

Indranil Ghosh, Table of n, a(n) for n = 0..59049 (terms 0..6561 from Joseph Myers)
Index entries for the Kaprekar map

Cf. A165111.

In other bases: A164884 (base 2), A164993 (base 3), A165012 (base 4), A165032 (base 5), A165051 (base 6), A165071 (base 7), A165090 (base 8), A151949 (base 10).

a[n_] := With[{dd = IntegerDigits[n, 9]}, FromDigits[ReverseSort[dd], 9] - FromDigits[Sort[dd], 9]];
a /@ Range[0, 100] (* Jean-François Alcover, Jan 08 2020 *)

A165032 a(n) = image of n under the base-5 Kaprekar map n -> (n with digits sorted into descending order) - (n with digits sorted into ascending order).

Original entry on oeis.org

0, 0, 0, 0, 0, 4, 0, 4, 8, 12, 8, 4, 0, 4, 8, 12, 8, 4, 0, 4, 16, 12, 8, 4, 0, 24, 24, 48, 72, 96, 24, 0, 24, 48, 72, 48, 24, 24, 48, 72, 72, 48, 48, 48, 72, 96, 72, 72, 72, 72, 48, 48, 48, 72, 96, 48, 24, 24, 48, 72, 48, 24, 0, 24, 48, 72, 48, 24, 24, 48, 96, 72, 48, 48, 48, 72, 72, 72

Offset: 0

Joseph Myers, Sep 04 2009

			For n = 10, 10_10 = 20_5. So, a(10) = 20_5 - 2_5 = 10 - 2 = 8. - _Indranil Ghosh_, Feb 02 2017

Indranil Ghosh, Table of n, a(n) for n = 0..15625 (terms 0..3125 from Joseph Myers)
Index entries for the Kaprekar map

Cf. A165033.

In other bases: A164884 (base 2), A164993 (base 3), A165012 (base 4), A165051 (base 6), A165071 (base 7), A165090 (base 8), A165110 (base 9), A151949 (base 10).

a[n_] := With[{dd = IntegerDigits[n, 5]}, FromDigits[ReverseSort[dd], 5] - FromDigits[Sort[dd], 5]];
a /@ Range[0, 100] (* Jean-François Alcover, Jan 08 2020 *)

A165071 a(n) = image of n under the base-7 Kaprekar map n -> (n with digits sorted into descending order) - (n with digits sorted into ascending order).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 6, 0, 6, 12, 18, 24, 30, 12, 6, 0, 6, 12, 18, 24, 18, 12, 6, 0, 6, 12, 18, 24, 18, 12, 6, 0, 6, 12, 30, 24, 18, 12, 6, 0, 6, 36, 30, 24, 18, 12, 6, 0, 48, 48, 96, 144, 192, 240, 288, 48, 0, 48, 96, 144, 192, 240, 96, 48, 48, 96, 144, 192, 240, 144, 96, 96, 96

Offset: 0

Joseph Myers, Sep 04 2009

			For n = 11, 11_10 = 14_7. So, a(11) = 41_7 - 14_7 = 29 - 11 = 18. - _Indranil Ghosh_, Feb 02 2017

Indranil Ghosh, Table of n, a(n) for n = 0..16807 (terms 0..2401 from Joseph Myers)
Index entries for the Kaprekar map

Cf. A165072.

In other bases: A164884 (base 2), A164993 (base 3), A165012 (base 4), A165032 (base 5), A165051 (base 6), A165090 (base 8), A165110 (base 9), A151949 (base 10).

a[n_] := With[{dd = IntegerDigits[n, 7]}, FromDigits[ReverseSort[dd], 7] - FromDigits[Sort[dd], 7]];
a /@ Range[0, 100] (* Jean-François Alcover, Jan 08 2020 *)

cp's OEIS Frontend

A151956 a(1) = 1002; thereafter a(n) = (a(n-1) with digits sorted into descending order) - (a(n-1) with digits sorted into ascending order) (see the Kaprekar map, A151949).

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A010785 Repdigit numbers, or numbers whose digits are all equal.

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A003558 Least number m > 0 such that 2^m == +-1 (mod 2n + 1).

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A099009 Fixed points of the Kaprekar mapping f(n) = n' - n'', where in n' the digits of n are arranged in descending, in n'' in ascending order.

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A165012 a(n) = image of n under the base-4 Kaprekar map n -> (n with digits sorted into descending order) - (n with digits sorted into ascending order).

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A165051 a(n) = image of n under the base-6 Kaprekar map n -> (n with digits sorted into descending order) - (n with digits sorted into ascending order).

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A165090 a(n) is the image of n under the base-8 Kaprekar map n -> (n with digits sorted into descending order) - (n with digits sorted into ascending order).