A002283 -id:A002283 - cp's OEIS frontend

A099676 Partial sums of repdigits of A002283.

9, 108, 1107, 11106, 111105, 1111104, 11111103, 111111102, 1111111101, 11111111100, 111111111099, 1111111111098, 11111111111097, 111111111111096, 1111111111111095, 11111111111111094, 111111111111111093, 1111111111111111092, 11111111111111111091

Offset: 1

Labos Elemer, Nov 17 2004

a(n) is the maximal positive integer k such that the sequence 1, 2, 3, 4, ..., k-1, k has a total of n*k digits. - Bui Quang Tuan, Mar 12 2015

			9 + 99 + 999 + 9999 + 99999 = a(5) = 111105.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (12,-21,10).

Cf. A057932, A002275-A002283, A099669-A099674.

[(10/9)*(10^n-1)-n: n in [1..20]]; // Vincenzo Librandi, Mar 14 2014

a:=n->sum((10^(n-j)-1^(n-j)), j=0..n): seq(a(n), n=1..17); # Zerinvary Lajos, Jan 15 2007

<Vincenzo Librandi, Mar 14 2014 *)
LinearRecurrence[{12,-21,10},{9,108,1107},20] (* Harvey P. Dale, Apr 18 2015 *)

Vec(-9*x/((x-1)^2*(10*x-1)) + O(x^100)) \\ Colin Barker, Mar 12 2014

[gaussian_binomial(n,1,10)-n for n in range(2,19)] # Zerinvary Lajos, May 29 2009

a(n) = (10/9)*(10^n-1) - n. - R. Piyo (nagoya314(AT)yahoo.com), Dec 10 2004
From Colin Barker, Mar 12 2014: (Start)
a(n) = 12*a(n-1)-21*a(n-2)+10*a(n-3).
G.f.: -9*x / ((x-1)^2*(10*x-1)). (End)
E.g.f.: exp(x)*(10*(exp(9*x) - 1) - 9*x)/9. - Stefano Spezia, Sep 13 2023

A093851 a(n) = A002283(n-1) + floor(A052268(n)/(1+n)).

Original entry on oeis.org

4, 39, 324, 2799, 24999, 228570, 2124999, 19999999, 189999999, 1818181817, 17499999999, 169230769229, 1642857142856, 15999999999999, 156249999999999, 1529411764705881, 14999999999999999, 147368421052631577, 1449999999999999999, 14285714285714285713

Offset: 1

Amarnath Murthy, Apr 18 2004

The first column r=1 of a triangle defined by T(n,r) = 10^(n-1) -1 + r*floor(9*10^(n-1)/(n+1)).

A row starts with a (virtual) 0th column of a rep-9-digit and fills the remainder with n+1 numbers in arithmetic progression with the largest step such that all numbers in the n-th row are n-digit numbers.

			The triangle starts in row n=1 as
4 9 # -1, -1+5, -1+2*5
39 69 99 # 9,9+30,9+2*30
324 549 774 999 # 99, 99+225, 99+2*225, 99+3*225
2799 4599 6399 8199 9999 # 999, 999+1800, 999+2*1800,..
...
The sequence contains the first column.

G. C. Greubel, Table of n, a(n) for n = 1..995

Cf. A061772, A093846, A093847, A093450, A093552.

[10^(n-1) -1 +Floor(9*10^(n-1)/(n+1)): n in [1..20]]; // G. C. Greubel, Apr 02 2019

A093851 := proc(n) 10^(n-1)-1+floor(9*10^(n-1)/(n+1)) ; end proc: seq(A093851(n),n=1..20) ; # R. J. Mathar, Oct 14 2010

Table[10^(n-1) -1 +Floor[9*10^(n-1)/(n+1)], {n, 1, 20}] (* G. C. Greubel, Apr 02 2019 *)

{a(n) = 10^(n-1) -1 +floor(9*10^(n-1)/(n+1))}; \\ G. C. Greubel, Apr 02 2019

[10^(n-1) -1 +floor(9*10^(n-1)/(n+1)) for n in (1..20)] # G. C. Greubel, Apr 02 2019

a(n) = 10^(n-1) -1 + floor(9*10^(n-1)/(n+1)). - G. C. Greubel, Apr 02 2019

More terms from R. J. Mathar, Oct 14 2010

A002275 Repunits: (10^n - 1)/9. Often denoted by R_n.

Original entry on oeis.org

0, 1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111, 1111111111, 11111111111, 111111111111, 1111111111111, 11111111111111, 111111111111111, 1111111111111111, 11111111111111111, 111111111111111111, 1111111111111111111, 11111111111111111111

Offset: 0

N. J. A. Sloane

R_n is a string of n 1's.
Base-4 representation of Jacobsthal bisection sequence A002450. E.g., a(4)= 1111 because A002450(4)= 85 (in base 10) = 64 + 16 + 4 + 1 = 1*(4^3) + 1*(4^2) + 1*(4^1) + 1. - Paul Barry, Mar 12 2004
Except for the first two terms, these numbers cannot be perfect squares, because x^2 != 11 (mod 100). - Zak Seidov, Dec 05 2008
For n >= 0: a(n) = (A000225(n) written in base 2). - Jaroslav Krizek, Jul 27 2009, edited by M. F. Hasler, Jul 03 2020
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=10, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Except 0, 1 and 11, all these integers are Brazilian numbers, A125134. - Bernard Schott, Dec 24 2012
Numbers n such that 11...111 = R_n = (10^n - 1)/9 is prime are in A004023. - Bernard Schott, Dec 24 2012
The terms 0 and 1 are the only squares in this sequence, as a(n) == 3 (mod 4) for n>=2. - Nehul Yadav, Sep 26 2013
For n>=2 the multiplicative order of 10 modulo the a(n) is n. - Robert G. Wilson v, Aug 20 2014
The above is a special case of the statement that the order of z modulo (z^n-1)/(z-1) is n, here for z=10. - Joerg Arndt, Aug 21 2014
From Peter Bala, Sep 20 2015: (Start)
Let d be a divisor of a(n). Let m*d be any multiple of d. Split the decimal expansion of m*d into 2 blocks of contiguous digits a and b, so we have m*d = 10^k*a + b for some k, where 0 <= k < number of decimal digits of m*d. Then d divides a^n - (-b)^n (see McGough). For example, 271 divides a(5) and we find 2^5 + 71^5 = 11*73*271*8291 and 27^5 + 1^5 = 2^2*7*31*61*271 are both divisible by 271. Similarly, 4*271 = 1084 and 10^5 + 84^5 = 2^5*31*47*271*331 while 108^5 + 4^5 = 2^12*7*31*61*271 are again both divisible by 271. (End)
Starting with the second term this sequence is the binary representation of the n-th iteration of the Rule 220 and 252 elementary cellular automaton starting with a single ON (black) cell. - Robert Price, Feb 21 2016
If p > 5 is a prime, then p divides a(p-1). - Thomas Ordowski, Apr 10 2016
0, 1 and 11 are only terms that are of the form x^2 + y^2 + z^2 where x, y, z are integers. In other words, a(n) is a member of A004215 for all n > 2. - Altug Alkan, May 08 2016
Except for the initial terms, the binary representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 737", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Mar 17 2017
The term "repunit" was coined by Albert H. Beiler in 1964. - Amiram Eldar, Nov 13 2020
q-integers for q = 10. - John Keith, Apr 12 2021
Binomial transform of A001019 with leading zero. - Jules Beauchamp, Jan 04 2022

Albert H. Beiler, Recreations in the Theory of Numbers: The Queen of Mathematics Entertains, New York: Dover Publications, 1964, chapter XI, p. 83.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 235-237.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 197-198.
Samuel Yates, Peculiar Properties of Repunits, J. Recr. Math. 2, 139-146, 1969.
Samuel Yates, Prime Divisors of Repunits, J. Recr. Math. 8, 33-38, 1975.

David Wasserman, Table of n, a(n) for n = 0..1000
Eudes Antonio Costa and Fernando Soares Carvalho, On repunit polynomials sequence, Braz. Elec. J. Math. (2024). See pp. 2, 15.
Eudes Antonio Costa, Douglas Catulio Santos, Paula Maria Machado Cruz Catarino, and Elen Viviani Pereira Spreafico, On Gaussian and Quaternion Repunit Numbers, Rev. Mat. UFOP (Brazil, 2024) Vol. 2. See p. 2.
Eudes Antonio Costa, Paula Maria Machado Cruz Catarino, and Douglas Catulio Santos, A Study of the Symmetry of the Tricomplex Repunit Sequence with Repunit Sequence, Symmetry (2024) Vol. 17, No. 1, 28.
Dmytro S. Inosov and Emil Vlasák, Cryptarithmically unique terms in integer sequences, arXiv:2410.21427 [math.NT], 2024. See pp. 3, 18.
Makoto Kamada, Factorizations of 11...11 (Repunit).
Douglas Catulio Santos, Eudes Antonio Costa, and Paula Maria Machado Cruz Catarino, On Gersenne Sequence: A Study of One Family in the Horadam-Type Sequence, Axioms 14, 203, (2025). See p. 4.
W. M. Snyder, Factoring Repunits, Am. Math. Monthly, Vol. 89, No. 7 (1982), pp. 462-466.
Amelia Carolina Sparavigna, On Repunits, Politecnico di Torino (Italy, 2019).
Amelia Carolina Sparavigna, Composition Operations of Generalized Entropies Applied to the Study of Numbers, International Journal of Sciences (2019) Vol. 8, No. 4, 87-92.
Amelia Carolina Sparavigna, The groupoids of Mersenne, Fermat, Cullen, Woodall and other Numbers and their representations by means of integer sequences, Politecnico di Torino, Italy (2019), [math.NT].
Amelia Carolina Sparavigna, Some Groupoids and their Representations by Means of Integer Sequences, International Journal of Sciences (2019) Vol. 8, No. 10.
Eric Weisstein's World of Mathematics, Repunit.
Eric Weisstein's World of Mathematics, Demlo Number.
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton.
Wikipedia, Repunit.
Amin Witno, A Family of Sequences Generating Smith Numbers, J. Int. Seq. 16 (2013) #13.4.6.
Stephen Wolfram, A New Kind of Science.
Samuel Yates, The Mystique of Repunits, Math. Mag., Vol. 51, No. 1 (1978), pp. 22-28.
Index to Elementary Cellular Automata.
Index entries for 10-automatic sequences.
Index entries for sequences related to cellular automata.
Index entries for linear recurrences with constant coefficients, signature (11,-10).
Index entries for "core" sequences.

Partial sums of 10^n (A011557). Factors: A003020, A067063.
Bisections give A099814, A100706.
Cf. A000042, A002276, A002277, A002278, A002279, A002280, A002281, A002282, A002283, A004023, A046053, A059988, A065444, A075412, A075415, A083278, A095370, A102380, A125134, A178635, A204845, A204846, A204847, A204848, A206244.
Numbers having multiplicative digital roots 0-9: A034048, A002275, A034049, A034050, A034051, A034052, A034053, A034054, A034055, A034056.
Cf. A000225, A001019, A001045, A002450, A004215, A015565, A125118, A242614, A242622.
Cf. A010785, A017173, A105279.

a002275 = (`div` 9) . subtract 1 . (10 ^)
a002275_list = iterate ((+ 1) . (* 10)) 0
-- Reinhard Zumkeller, Jul 05 2013, Feb 05 2012

[(10^n-1)/9: n in [0..25]]; // Vincenzo Librandi, Nov 06 2014

seq((10^k - 1)/9, k=0..30); # Wesley Ivan Hurt, Sep 28 2013

Table[(10^n - 1)/9, {n, 0, 19}] (* Alonso del Arte, Nov 15 2011 *)
Join[{0},Table[FromDigits[PadRight[{},n,1]],{n,20}]] (* Harvey P. Dale, Mar 04 2012 *)

a[0]:0$
a[1]:1$
a[n]:=11*a[n-1]-10*a[n-2]$
A002275(n):=a[n]$
makelist(A002275(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

a(n)=(10^n-1)/9; \\ Michael B. Porter, Oct 26 2009

my(x='x+O('x^30)); concat(0, Vec(x/((1-10*x)*(1-x)))) \\ Altug Alkan, Apr 10 2016

print([(10**n-1)//9 for n in range(100)]) # Michael S. Branicky, Apr 30 2022

[lucas_number1(n, 11, 10) for n in range(21)]  # Zerinvary Lajos, Apr 27 2009

a(n) = 10*a(n-1) + 1, a(0)=0.
a(n) = A000042(n) for n >= 1.
Second binomial transform of Jacobsthal trisection A001045(3n)/3 (A015565). - Paul Barry, Mar 24 2004
G.f.: x/((1-10*x)*(1-x)). Regarded as base b numbers, g.f. x/((1-b*x)*(1-x)). - Franklin T. Adams-Watters, Jun 15 2006
a(n) = 11*a(n-1) - 10*a(n-2), a(0)=0, a(1)=1. - Lekraj Beedassy, Jun 07 2006
a(n) = A125118(n,9) for n>8. - Reinhard Zumkeller, Nov 21 2006
a(n) = A075412(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 10^(n-1) with a(0)=0. - Vincenzo Librandi, Jul 22 2010
a(n) = A242614(n,A242622(n)). - Reinhard Zumkeller, Jul 17 2014
E.g.f.: (exp(9*x) - 1)*exp(x)/9. - Ilya Gutkovskiy, May 11 2016
a(n) = Sum_{k=0..n-1} 10^k. - Torlach Rush, Nov 03 2020
Sum_{n>=1} 1/a(n) = A065444. - Amiram Eldar, Nov 13 2020
From Elmo R. Oliveira, Aug 02 2025: (Start)
a(n) = A002283(n)/9 = A105279(n)/10.
a(n) = A010785(A017173(n-1)) for n >= 1. (End)

A000533 a(0)=1; a(n) = 10^n + 1, n >= 1.

Original entry on oeis.org

1, 11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001, 100000000001, 1000000000001, 10000000000001, 100000000000001, 1000000000000001, 10000000000000001

Offset: 0

N. J. A. Sloane

Also, b^n+1 written in base b, for any base b >= 2.
Also, A083318 written in base 2. - Omar E. Pol, Feb 24 2008, Dec 30 2008
Also, palindromes formed from the reflected decimal expansion of the concatenation of 1 and infinite 0's. - Omar E. Pol, Dec 14 2008
It seems that the sequence gives 'all' positive integers m such that m^4 is a palindrome. Note that a(0)^4 = 1 is a palindrome and for n > 0, a(n)^4 = (10^n + 1)^4 = 10^(4n) + 4*10^(3n) + 6*10^(2n) + 4*10^(n) + 1 is a palindrome. - Farideh Firoozbakht, Oct 28 2014
a(n)^2 starts with a(n)+1 for n >= 1. - Dhilan Lahoti, Aug 31 2015
From Peter Bala, Sep 25 2015: (Start)
The simple continued fraction expansion of sqrt(a(2*n)) = [10^n; 2*10^n, 2*10^n, ...] has period 1.
The simple continued fraction expansion of sqrt(a(6*n))/a(2*n) = [10^n - 1; 1, 10^n - 1, 10^n - 1, 1, 2*(10^n - 1), ...] has period 5.
The simple continued fraction expansion of sqrt(a(10*n))/a(2*n) = [10^(3*n) - 10^n; 10^n, 10^n, 2*(10^(3*n) - 10^n), ...] has period 3.
As n increases, these expansions have large partial quotients.
A theorem of Kuzmin in the measure theory of continued fractions says that large partial quotients are the exception in continued fraction expansions.
Empirically, we also see exceptionally large partial quotients in the continued fraction expansions of the m-th root of the numbers a(m*n), for m >= 3. For example, it appears that the continued fraction expansion of a(3*n)^(1/3) begins [10^n; 3*10^(2*n), 10^n, 4.5*10^(2*n), 0.8*10^n, ( 9*10^(2*n + 2) - 144 + 24*(2^mod(n,3) - 1) )/168, ...]. As n increases, the expansion begins with 6 large partial quotients. An example is given below. Cf. A002283, A066138 and A168624.
(End)
a(1) and a(2) are the only prime terms up to n=100000. - Daniel Arribas, Jun 04 2016
Based on factors from A001271, the first abundant number in this sequence should occur in the first M terms, where M is the double factorial M=7607!!. Is any abundant number known in this sequence? - Sergio Pimentel, Oct 04 2019
The (3^5 * 5^2 * 7^2 * 11^2 * 13^2 * 17 * 19 * 23 * 29 * 31 * 37 * 41 * 43 * 47 * 53 * 59 * 61 * 67 * 71 * 73 * 79 * 83 * 89 * 97 * 101 * 103 * 107 * 109 * 113 * 127 * 131 * 137 * 139 * 157 * 163 * 181 * 191 * 241 * 251 * 263)-th term of this sequence is an abundant number. - Jon E. Schoenfield, Nov 19 2019
The only perfect power (i.e., a perfect square, a perfect cube, and so forth) in the present sequence is a(0) by Mihăilescu's theorem (since 10^n is a perfect power not equal to 2^3 - see the links in A001597). - Marco Ripà, Feb 04 2025

			The continued fraction expansion of a(9)^(1/3) begins [1000; 3000000, 1000, 4500000, 800, 5357142, 1, 6, 14, 6, 1, 5999999, 6, 1, 12, 7, 1, ...] with 5 large partial quotients immediately following the integer part of the number. - _Peter Bala_, Sep 25 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..100
John Rafael M. Antalan, A Recreational Application of Two Integer Sequences and the Generalized Repetitious Number Puzzle, arXiv:1908.06014 [math.HO], 2019.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A002283, A066138, A083318, A138144, A138145, A152756, A168624.

[10^n + 1 - 0^n: n in [0..30]]; // Vincenzo Librandi, Jul 15 2011

Join[{1},LinearRecurrence[{11,-10},{11,101},20]] (* Harvey P. Dale, May 01 2014 *)

a(n)=if(n,10^n+1,1) \\ Charles R Greathouse IV, Oct 28 2014

a(n) = 10^n + 1 - 0^n. - Reinhard Zumkeller, Jun 10 2003
From Paul Barry, Feb 05 2005: (Start)
G.f.: (1-10*x^2)/((1-x)*(1-10*x));
a(n) = Sum_{k=0..n} binomial(n, k)*0^(k(n-k))*10^k. (End)
a(n) = A178500(n) + 1. - Reinhard Zumkeller, May 28 2010
E.g.f.: exp(x) + exp(10*x) - 1. - Ilya Gutkovskiy, Jun 03 2016

A008591 Multiples of 9: a(n) = 9*n.

Original entry on oeis.org

0, 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 135, 144, 153, 162, 171, 180, 189, 198, 207, 216, 225, 234, 243, 252, 261, 270, 279, 288, 297, 306, 315, 324, 333, 342, 351, 360, 369, 378, 387, 396, 405, 414, 423, 432, 441, 450, 459, 468, 477

Offset: 0

N. J. A. Sloane

An Iraqi tablet dating from the Middle Babylonian period (1400-1100 BC) gives a(1)-a(20), a(30), a(40), and a(50). See CDLI link for images and more information. - Charles R Greathouse IV, Jan 21 2017
Apart from 0, numbers whose digital root is 9. - Halfdan Skjerning, Mar 15 2018
Also numbers such that when the leftmost digit is moved to the unit's place the result is divisible by 9. - Stefano Spezia, Jul 08 2025

Ivan Panchenko, Table of n, a(n) for n = 0..200
CDLI, MKT 1, 039, HS 0217a
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 321
Tanya Khovanova, Recursive Sequences
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.
Leo Tavares, Illustration: Triangular Frames
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A008588, A008589, A008590.
Cf. A168182, A168183.
Cf. A002283, A007953.
Cf. A060544.

Range[0, 1000, 9] (* Vladimir Joseph Stephan Orlovsky, May 27 2011 *)

makelist(9*n,n,0,20); /* Martin Ettl, Dec 17 2012 */

a(n)=9*n \\ Charles R Greathouse IV, Jan 30 2012

Complement of A168183; A168182(a(n)) = 0. - Reinhard Zumkeller, Nov 30 2009
a(n) = A007953(A002283(n)). - Reinhard Zumkeller, Aug 06 2010
From Vincenzo Librandi, Dec 24 2010: (Start)
a(n) = 9*n = 2*a(n-1) - a(n-2).
G.f.: 9x/(x-1)^2. (End)
a(n) = A060544(n+1) - A060544(n). - Leo Tavares, Jul 17 2022
E.g.f.: 9*x*exp(x). - Stefano Spezia, Oct 08 2022

A051885 Smallest number whose sum of digits is n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 29, 39, 49, 59, 69, 79, 89, 99, 199, 299, 399, 499, 599, 699, 799, 899, 999, 1999, 2999, 3999, 4999, 5999, 6999, 7999, 8999, 9999, 19999, 29999, 39999, 49999, 59999, 69999, 79999, 89999, 99999, 199999, 299999, 399999, 499999

Offset: 0

Felice Russo, Dec 15 1999

This is also the list of lunar triangular numbers: A087052 with duplicates removed. - N. J. A. Sloane, Jan 25 2011
Numbers n such that A061486(n) = n. - Amarnath Murthy, May 06 2001
The product of digits incremented by 1 is the same as the number incremented by 1. If a(n) = abcd...(a,b,c,d, etc. are digits of a(n)) {a(n) + 1} = (a+1)*(b+1)(c+1)*(d+1)*..., e.g., 299 + 1 = (2+1)*(9+1)*(9+1) = 300. - Amarnath Murthy, Jul 29 2003
A138471(a(n)) = 0. - Reinhard Zumkeller, Mar 19 2008
a(n+1) = A108971(A179988(n)). - Reinhard Zumkeller, Aug 09 2010, Jul 10 2011
Positions of records in A003132: A080151(n) = A003132(a(n)). - Reinhard Zumkeller, Jul 10 2011
a(n) = A242614(n,1). - Reinhard Zumkeller, Jul 16 2014
A254524(a(n)) = 1. - Reinhard Zumkeller, Oct 09 2015
The slowest strictly increasing sequence of nonnegative integers such that, for any two terms, calculating the difference of their decimal representations requires no borrowing. - Rick L. Shepherd, Aug 11 2017

Iain Fox, Table of n, a(n) for n = 0..9000 (first 101 terms from Reinhard Zumkeller)
D. Applegate, M. LeBrun and N. J. A. Sloane, Dismal Arithmetic, arXiv:1107.1130 [math.NT], 2011. [Note: we have now changed the name from "dismal arithmetic" to "lunar arithmetic" - the old name was too depressing]
A. Murthy, Exploring some new ideas on Smarandache type sets, functions and sequences, Smarandache Notions Journal Vol. 11 N. 1-2-3 Spring 2000.
Index entries for sequences related to dismal (or lunar) arithmetic
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,10,-10).

Cf. A061104, A061105, A061486, A007953, A067043, A087052.
Numbers of form i*b^j-1 (i=1..b-1, j >= 0) for bases b = 2 through 9: A000225, A062318, A180516, A181287, A181288, A181303, A165804, A140576. - N. J. A. Sloane, Jan 25 2011
Cf. A002283.
Cf. A254524.

a051885 n = (m + 1) * 10^n' - 1 where (n',m) = divMod n 9
-- Reinhard Zumkeller, Jul 10 2011

Magma
```
[i*10^j-1: i in [1..9], j in [0..5]];
```

b:=10; t1:=[]; for j from 0 to 15 do for i from 1 to b-1 do t1:=[op(t1), i*b^j-1]; od: od: t1; # N. J. A. Sloane, Jan 25 2011

a[n_] := (Mod[n, 9] + 1)*10^Floor[n/9] - 1; Table[a[n], {n, 0, 49}](* Jean-François Alcover, Dec 01 2011, after Henry Bottomley *)

A051885(n) = (n%9+1)*10^(n\9)-1  \\ M. F. Hasler, Jun 17 2012

first(n) = Vec(x*(x^2 + x + 1)*(x^6 + x^3 + 1)/((x - 1)*(10*x^9 - 1)) + O(x^n), -n) \\ Iain Fox, Dec 30 2017

def A051885(n): return ((n % 9)+1)*10**(n//9)-1 # Chai Wah Wu, Apr 04 2021

These are the numbers i*10^j-1 (i=1..9, j >= 0). - N. J. A. Sloane, Jan 25 2011
a(n) = ((n mod 9) + 1)*10^floor(n/9) - 1 = a(n-1) + 10^floor((n-1)/9). - Henry Bottomley, Apr 24 2001
a(n) = A037124(n+1) - 1. - Reinhard Zumkeller, Jan 03 2008, Jul 10 2011
G.f.: x*(x^2+x+1)*(x^6+x^3+1) / ((x-1)*(10*x^9-1)). - Colin Barker, Feb 01 2013

More terms from James Sellers, Dec 16 1999

Offset fixed by Reinhard Zumkeller, Jul 10 2011

A002277 a(n) = 3*(10^n - 1)/9.

Original entry on oeis.org

0, 3, 33, 333, 3333, 33333, 333333, 3333333, 33333333, 333333333, 3333333333, 33333333333, 333333333333, 3333333333333, 33333333333333, 333333333333333, 3333333333333333, 33333333333333333, 333333333333333333, 3333333333333333333, 33333333333333333333, 333333333333333333333

Offset: 0

N. J. A. Sloane

From Wolfdieter Lang, Feb 08 2017: (Start)
This sequence (for n >= 1) appears in n-families satisfying so-called curious cubic identities based on the Armstrong numbers 153, 370 and 371, A005188(10) - A005188(12).
153 also involves A246057(n-1) and A093143(n). See a comment in A246057 with the van Poorten et al. reference, and A281857.
370 and 371 also involve A067275(n+1). See the comment there, and A281858 and A281860. (End)

			From _Wolfdieter Lang_, Feb 08 2017: (Start)
Curious cubic identities (see a comment above):
1^3 + 5^3 + 3^3 = 153, 16^3 + 50^3 + 33^3 = 165033, 166^3 + 500^3 + 333^3 = 166500333, ...
3^3 + 7^3 + 0^3 = 370; 336700 = 33^3 + 67^3 + (00)^3 = 336700,  333^3 + 667^3 + (000)^3 = 333667000, ...
3^3 + 7^3 + 1^3 = 371, 33^3 + 67^3 + (01)^3 = 336701, 333^3 + 667^3 + (001)^3 = 333667001, ... (End)

Ivan Panchenko, Table of n, a(n) for n = 0..200
Eric Weisstein's World of Mathematics, Repdigit.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A002275, A002276, A002278, A002279, A002280, A002281, A002282, A002283, A010785, A017197.

Cf. A005188, A067275, A075412, A093143, A135702, A178631, A178633, A246057, A281857, A281858, A281860.

[(10^n - 1)/3 : n in [0..30]]; // Wesley Ivan Hurt, Apr 01 2016

A002277:=n->(10^n-1)/3: seq(A002277(n), n=0..30); # Wesley Ivan Hurt, Apr 01 2016

LinearRecurrence[{11, -10}, {0, 3}, 20] (* Robert G. Wilson v, Jul 06 2013 *)
(10^Range[0, 30] - 1)/3 (* Wesley Ivan Hurt, Apr 01 2016 *)

A002277(n):=(10^n - 1)/3$
makelist(A002277(n),n,0,20); /* Martin Ettl, Nov 12 2012 */

a(n)=(10^n-1)/3 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 3*A002275(n).
a(n) = A178631(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
From Vincenzo Librandi, Jul 22 2010: (Start)
a(n) = a(n-1) + 3*10^(n-1) with a(0)=0;
a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=3. (End)
G.f.: 3*x/((1 - x)*(1 - 10*x)). - Ilya Gutkovskiy, Feb 24 2017
Sum_{n>=1} 1/a(n) = A135702. - Amiram Eldar, Nov 13 2020
E.g.f.: exp(x)*(exp(9*x) - 1)/3. - Stefano Spezia, Sep 13 2023
From Elmo R. Oliveira, Jul 20 2025: (Start)
a(n) = (A246057(n) - 1)/5.
a(n) = A010785(A017197(n-1)) for n >= 1. (End)

A002276 a(n) = 2*(10^n - 1)/9.

Original entry on oeis.org

0, 2, 22, 222, 2222, 22222, 222222, 2222222, 22222222, 222222222, 2222222222, 22222222222, 222222222222, 2222222222222, 22222222222222, 222222222222222, 2222222222222222, 22222222222222222, 222222222222222222, 2222222222222222222, 22222222222222222222, 222222222222222222222

Offset: 0

N. J. A. Sloane

a(n) is also the total number of holes in a variation of a box fractal as in illustration. - Kival Ngaokrajang, May 23 2014 [As observed by Hans Havermann, this seems to be incorrect: e.g., for n = 2 the illustration shows 28 small holes plus two larger holes. - M. F. Hasler, Oct 05 2020]

Ivan Panchenko, Table of n, a(n) for n = 0..200
Kival Ngaokrajang, Illustration for n = 1..4.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A002275, A002277, A002278, A002279, A002280, A002281, A002282, A002283, A178630, A178634.

Cf. A010785, A017185.

LinearRecurrence[{11, -10}, {0, 2}, 50] (* Jinyuan Wang, Feb 27 2020 *)

A002276(n):=2*(10^n - 1)/9$
makelist(A002276(n),n,0,20); /* Martin Ettl, Nov 12 2012 */

a(n)=10^n\9*2 \\ M. F. Hasler, Mar 27 2015

a(n) = A178630(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
From Vincenzo Librandi, Jul 22 2010: (Start)
a(n) = a(n-1) + 2*10^(n-1) with a(0) = 0.
a(n) = 11*a(n-1) - 10*a(n-2) with a(0) = 0, a(1) = 2. (End)
G.f.: 2*x/((1 - x)*(1 - 10*x)). - Ilya Gutkovskiy, Feb 24 2017
E.g.f.: 2*exp(x)*(exp(9*x) - 1)/9. - Stefano Spezia, Sep 13 2023
From Elmo R. Oliveira, Jul 19 2025: (Start)
a(n) = 2*A002275(n).
a(n) = A010785(A017185(n-1)) for n >= 1. (End)

A002282 a(n) = 8*(10^n - 1)/9.

Original entry on oeis.org

0, 8, 88, 888, 8888, 88888, 888888, 8888888, 88888888, 888888888, 8888888888, 88888888888, 888888888888, 8888888888888, 88888888888888, 888888888888888, 8888888888888888, 88888888888888888, 888888888888888888, 8888888888888888888, 88888888888888888888, 888888888888888888888

Offset: 0

N. J. A. Sloane

If the initial term is omitted, might be called eightful (or hateful) numbers!

			Curious multiplications:
9*9 + 7 = 88;
98*9 + 6 = 888;
987*9 + 5 = 8888;
9876*9 + 4 = 88888;
98765*9 + 3 = 888888;
987654*9 + 2 = 8888888;
9876543*9 + 1 = 88888888;
98765432*9 + 0 = 888888888;
987654321*9 - 1 = 8888888888;
9876543210*9 - 2 = 88888888888. - _Philippe Deléham_, Mar 09 2014

Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 32.

Harry J. Smith, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A002275, A002276, A002277, A002278, A002279, A002280, A002281, A002283, A059988, A075412, A178635.

Cf. A010785, A017257, A051003, A059482, A246058.

A002282:=n->8*(10^n - 1)/9; seq(A002282(n), n=0..20); # Wesley Ivan Hurt, Mar 10 2014

LinearRecurrence[{11,-10}, {0,8}, 20] (* Harvey P. Dale, May 30 2013 *)

{ a=-4/5; for (n = 0, 200, a+=8*10^(n - 1); write("b002282.txt", n, " ", a); ) } \\ Harry J. Smith, Jun 27 2009

def a(n): return 8*(10**n - 1)//9 # Martin Gergov, Oct 19 2022

From Jaume Oliver Lafont, Feb 03 2009: (Start)
a(n) = 11*a(n-1) - 10*a(n-2), with a(0)=0, a(1)=8.
G.f.: 8*x/((1-x)*(1-10*x)). (End)
a(n) = A178635(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 8*10^(n-1), with a(0)=0. - Vincenzo Librandi, Jul 22 2010
a(n) = 8*A002275(n) = A002283(n) - A002275(n). - Carauleanu Marc, Sep 03 2016
From Ilya Gutkovskiy, Sep 03 2016: (Start)
E.g.f.: 8*(exp(9*x) - 1)*exp(x)/9.
a(n) = floor(8*10^n/9). (End)
From Elmo R. Oliveira, Jul 20 2025: (Start)
a(n) = (A246058(n) - 1)/2.
a(n) = A010785(A017257(n-1)) for n >= 1. (End)

A062397 a(n) = 10^n + 1.

Original entry on oeis.org

2, 11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001, 100000000001, 1000000000001, 10000000000001, 100000000000001, 1000000000000001, 10000000000000001, 100000000000000001

Offset: 0

Henry Bottomley, Jun 22 2001

The first three terms (indices 0, 1 and 2) are the only known primes. Moreover, the terms not of the form a(2^k) are all composite, except for a(0). Indeed, for all n >= 0, a(2n+1) is divisible by 11, a(4n+2) is divisible by 101, a(8n+4) is divisible by 73, a(16n+8) is divisible by 17, a(32n+16) is divisible by 353, a(64n+32) is divisible by 19841, etc. - M. F. Hasler, Nov 03 2018 [Edited based on the comment by Jeppe Stig Nielsen, Oct 17 2019]
This sequence also results when each term is generated by converting the previous term into a Roman numeral, then replacing each letter with its corresponding decimal value, provided that the vinculum is used and numerals are written in a specific way for integers greater than 3999, e.g., IV with a vinculum over the I and V for 4000. - Jamie Robert Creasey, Apr 14 2021
By Mihăilescu's theorem, a(n) can never be a perfect power (see "Catalan's conjecture" in Links). - Marco Ripà, Mar 10 2025

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Wikipedia, Catalan's conjecture.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Except for the initial term, essentially the same as A000533. Cf. A054977, A007395, A000051, A034472, A052539, A034474, A062394, A034491, A062395, A062396, A007689, A063376, A063481, A074600-A074624, A034524, A178248, A228081 for numbers one more than powers, i.e., this sequence translated from base n (> 2) to base 10.
Cf. A002283, A011557.
Cf. A038371 (smallest prime factor), A185121.

[10^n + 1: n in [0..35]]; // Vincenzo Librandi, Apr 30 2011

LinearRecurrence[{11, -10},{2, 11},18] (* Ray Chandler, Aug 26 2015 *)
10^Range[0,20]+1 (* Harvey P. Dale, Jan 21 2020 *)

a(n)=10^n+1 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 10*a(n-1) - 9 = A011557(n) + 1 = A002283(n) + 2.
From Mohammad K. Azarian, Jan 02 2009: (Start)
G.f.: 1/(1-x) + 1/(1-10*x).
E.g.f.: exp(x) + exp(10*x). (End)

cp's OEIS Frontend

A099676 Partial sums of repdigits of A002283.

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A093851 a(n) = A002283(n-1) + floor(A052268(n)/(1+n)).

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A002275 Repunits: (10^n - 1)/9. Often denoted by R_n.

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A000533 a(0)=1; a(n) = 10^n + 1, n >= 1.

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A008591 Multiples of 9: a(n) = 9*n.

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A051885 Smallest number whose sum of digits is n.

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