A178635 -id:A178635 - cp's OEIS frontend

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

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)

A002283 a(n) = 10^n - 1.

Original entry on oeis.org

0, 9, 99, 999, 9999, 99999, 999999, 9999999, 99999999, 999999999, 9999999999, 99999999999, 999999999999, 9999999999999, 99999999999999, 999999999999999, 9999999999999999, 99999999999999999, 999999999999999999, 9999999999999999999, 99999999999999999999, 999999999999999999999, 9999999999999999999999

Offset: 0

N. J. A. Sloane

A friend from Germany remarks that the sequence 9, 99, 999, 9999, 99999, 999999, ... might be called the grumpy German sequence: nein!, nein! nein!, nein! nein! nein!, ...
The Regan link shows that integers of the form 10^n -1 have binary representations with exactly n trailing 1 bits. Also those integers have quinary expressions with exactly n trailing 4's. For example, 10^4 -1 = (304444)5. The first digits in quinary correspond to the number 2^n -1, in our example (30)5 = 2^4 -1. A similar pattern occurs in the binary case. Consider 9 = (1001)2. - Washington Bomfim Dec 23 2010
a(n) is the number of positive integers with less than n+1 digits. - Bui Quang Tuan, Mar 09 2015
From Peter Bala, Sep 27 2015: (Start)
For n >= 1, the simple continued fraction expansion of sqrt(a(2*n)) = [10^n - 1; 1, 2*(10^n - 1), 1, 2*(10^n - 1), ...] has period 2. The simple continued fraction expansion of sqrt(a(2*n))/a(n) = [1; 10^n - 1, 2,  10^n - 1, 2, ...] also has period 2. Note the occurrence of large partial quotients in both expansions.
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 the presence of unexpectedly large partial quotients early in the continued fraction expansions of the m-th roots of the numbers a(m*n) for m >= 3. Some typical examples are given below. (End)
For n > 0, numbers whose smallest decimal digit is 9. - Stefano Spezia, Nov 16 2023

			From _Peter Bala_, Sep 27 2015: (Start)
Continued fraction expansions showing large partial quotients:
a(12)^(1/3) = [9999; 1, 299999998, 1, 9998, 1, 449999998, 1, 7998, 1, 535714284, 1, 2, 2, 142, 2, 2, 1, 599999999, 3, 1, 1,...].
Compare with a(30)^(1/3) = [9999999999; 1, 299999999999999999998, 1, 9999999998, 1, 449999999999999999998, 1, 7999999998, 1, 535714285714285714284, 1, 2, 2, 142857142, 2, 2, 1, 599999999999999999999, 3, 1, 1,...].
a(24)^(1/4) = [999999; 1, 3999999999999999998, 1, 666665, 1, 1, 1, 799999999999999999, 3, 476190, 7, 190476190476190476, 21, 43289, 1, 229, 1, 1864801864801863, 1, 4, 6,...].
Compare with a(48)^(1/4) = [999999999999; 1, 3999999999999999999999999999999999998, 1, 666666666665, 1, 1, 1, 799999999999999999999999999999999999, 3, 476190476190, 7, 190476190476190476190476190476190476, 21, 43290043289, 1, 229, 1, 1864801864801864801864801864801863, 1, 4, 6,...].
a(25)^(1/5) = [99999, 1, 499999999999999999998, 1, 49998, 1, 999999999999999999998, 1, 33332, 3, 151515151515151515151, 5, 1, 1, 1947, 1, 1, 38, 3787878787878787878, 1, 3, 5,...].
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Peter Bala, A002283 and some continued fraction expansions.
Rick Regan, Nines in quinary, September 8th, 2009.
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.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A000533, A003020, A007138, A007953, A008591, A010888, A048379, A066138, A073668, A168624, A276352.
Cf. A002275, A002276, A002277, A002278, A002279, A002280, A002281, A002282.
Cf. A075412, A075415, A178630, A178631, A178632, A178633, A178634, A178635.
Cf. A010785.

a002283 = subtract 1 . (10 ^)  -- Reinhard Zumkeller, Feb 21 2014

[(10^n-1): n in [0..20]]; // Vincenzo Librandi, Apr 26 2011

Table[10^n - 1, {n, 0, 22}] (* Michael De Vlieger, Sep 27 2015 *)

A002283(n):=10^n-1$
makelist(A002283(n),n,0,20); /* Martin Ettl, Nov 08 2012 */

a(n)=10^n-1; \\ Charles R Greathouse IV, Jan 30 2012

def a(n): return 10**n-1 # Michael S. Branicky, Feb 25 2023

From Mohammad K. Azarian, Jan 14 2009: (Start)
G.f.: 1/(1-10*x)-1/(1-x).
E.g.f.: e^(10*x)-e^x. (End)
a(n) = A075412(n)/A002275(n) = A178630(n)/A002276(n) = A178631(n)/A002277(n) = A075415(n)/A002278(n) = A178632(n)/A002279(n) = A178633(n)/A002280(n) = A178634(n)/A002281(n) = A178635(n)/A002282(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 9*10^(n-1) with a(0)=0; Also: a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=9. - Vincenzo Librandi, Jul 22 2010
For n>0, A007953(a(n)) = A008591(n) and A010888(a(n)) = 9. - Reinhard Zumkeller, Aug 06 2010
A048379(a(n)) = 0. - Reinhard Zumkeller, Feb 21 2014
a(n) = Sum_{k=1..n} 9*10^k. - Carauleanu Marc, Sep 03 2016
Sum_{n>=1} 1/a(n) = A073668. - Amiram Eldar, Nov 13 2020
From Elmo R. Oliveira, Jul 19 2025: (Start)
a(n) = 9*A002275(n).
a(n) = A010785(A008591(n)). (End)

More terms from Michael De Vlieger, Sep 27 2015

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)

A002477 Wonderful Demlo numbers: a(n) = ((10^n - 1)/9)^2.

Original entry on oeis.org

1, 121, 12321, 1234321, 123454321, 12345654321, 1234567654321, 123456787654321, 12345678987654321, 1234567900987654321, 123456790120987654321, 12345679012320987654321, 1234567901234320987654321

Offset: 1

N. J. A. Sloane

Only the first nine terms of this sequence are palindromes. - Bui Quang Tuan, Mar 30 2015

Not all of the terms are Demlo numbers as defined by Kaprekar, i.e., concat(L,M,R) with M and L+R repdigits using the same digit. For example, a(10), a(19), a(28) are not, but a(k) for k = 11, 12, ..., 18 are. - M. F. Hasler, Nov 18 2017

			From _José de Jesús Camacho Medina_, Apr 01 2016: (Start)
n=1: ....................... 1 = 9 / 9;
n=2: ..................... 121 = 1089 / 9;
n=3: ................... 12321 = 110889 / 9;
n=4: ................. 1234321 = 11108889 / 9;
n=5: ............... 123454321 = 1111088889 / 9;
n=6: ............. 12345654321 = 111110888889 / 9;
n=7: ........... 1234567654321 = 11111108888889 / 9;
n=8: ......... 123456787654321 = 1111111088888889 / 9;
n=9: ....... 12345678987654321 = 111111110888888889 / 9.        (End)
a(11) = concat(L = 1234567901, R = 20987654321), with L + R = 22222222222 = 2*(10^11-1)/9, of same length as R. - _M. F. Hasler_, Nov 23 2017

D. R. Kaprekar, On Wonderful Demlo numbers, Math. Stud., 6 (1938), 68.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 29.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..300
Lubomira Dvorakova, Stanislav Kruml, and David Ryzak, Antipalindromic numbers, arXiv:2008.06864 [math.CO], 2020. [Mentions this sequence.]
K. R. Gunjikar and D. R. Kaprekar, Theory of Demlo numbers, J. Univ. Bombay, Vol. VIII, Part 3, Nov. 1939, pp. 3-9. [Annotated scanned copy]
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Demlo Number
Eric Weisstein's World of Mathematics, Repunit
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A002275, A080151.

[((10^n - 1)/9)^2: n in [1..20]]; // Vincenzo Librandi, Jul 26 2011

A002477 := proc(n)
    (10^n-1)^2/81 ;
end proc:
seq(A002477(n),n=1..12) ; # R. J. Mathar, Aug 06 2019

Table[FromDigits[PadRight[{},n,1]]^2,{n,15}] (* Harvey P. Dale, Oct 16 2012 *)
(10^Range[20] - 1)^2/81 (* Paolo Xausa, Aug 03 2025 *)

A002477(n):=((10^n - 1)/9)^2$
makelist(A002477(n),n,1,10); /* Martin Ettl, Nov 12 2012 */

a(n) = (10^n\9)^2 \\ Charles R Greathouse IV, Jul 25 2011

G.f.: x*(1+10*x) / ((1-x)*(1-10*x)*(1-100*x)). - Simon Plouffe in his 1992 dissertation
a(n+1) = 100*a(n) + 20*A000042(n) + 1; a(1) = 1. - Reinhard Zumkeller, May 31 2010
a(n) = A000042(n)^2.
a(n) = A075412(n)/9 = A178630(n)/18 = A178631(n)/27 = A075415(n)/36 = A178632(n)/45 = A178633(n)/54 = A178634(n)/63 = A178635(n)/72 = A059988(n)/81. - Reinhard Zumkeller, May 31 2010
a(n+2) = -1000*a(n)+110*a(n+1)+11. - Alexander R. Povolotsky, Jun 06 2014
E.g.f.: exp(x)*(1 - 2*exp(9*x) + exp(99*x))/81. - Stefano Spezia, May 23 2025

Minor edits from N. J. A. Sloane, Aug 18 2009

Further edits from Reinhard Zumkeller, May 12 2010

A075415 Squares of A002280 or numbers (666...6)^2.

Original entry on oeis.org

0, 36, 4356, 443556, 44435556, 4444355556, 444443555556, 44444435555556, 4444444355555556, 444444443555555556, 44444444435555555556, 4444444444355555555556, 444444444443555555555556, 44444444444435555555555556, 4444444444444355555555555556, 444444444444443555555555555556

Offset: 0

Michael Taylor (michael.taylor(AT)vf.vodafone.co.uk), Sep 14 2002

A transformation of the Wonderful Demlo numbers (A002477).

			a(2) = 66^2 = 4356.
From _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ..................... 36 = 9 * 4;
n=2: ................... 4356 = 99 * 44;
n=3: ................. 443556 = 999 * 444;
n=4: ............... 44435556 = 9999 * 4444;
n=5: ............. 4444355556 = 99999 * 44444;
n=6: ........... 444443555556 = 999999 * 444444;
n=7: ......... 44444435555556 = 9999999 * 4444444;
n=8: ....... 4444444355555556 = 99999999 * 44444444;
n=9: ..... 444444443555555556 = 999999999 * 444444444. (End)

Seiichi Manyama, Table of n, a(n) for n = 0..500
Gérard Villemin, Variations sur les carrés.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A075411, A075412, A075413, A075414, A075415, A075416, A075417.
Cf. A059988, A178630, A178631, A178632, A178633, A178634, A178635. [From Reinhard Zumkeller, May 31 2010]
Cf. A002275, A002278, A002279,  A002280, A002283, A002477, A052041, A102807, A309827.

Table[FromDigits[PadRight[{},n,6]]^2,{n,0,20}] (* or *) LinearRecurrence[ {111,-1110,1000},{0,36,4356},20] (* Harvey P. Dale, May 20 2021 *)

a(n) = A002280(n)^2 = (6*A002275(n))^2 = 36*A002275(n)^2.
a(n) = (6*(10^n-1)/9)^2 = (4/9)*(10^(2*n) - 2*10^n + 1), which is n-1 4's, followed by a 3, n-1 5's and a 6. - Ignacio Larrosa Cañestro, Feb 26 2005
From Reinhard Zumkeller, May 31 2010: (Start)
a(n) = ((A002278(n-1)*10 + 3)*10^(n-1) + A002279(n-1))*10 + 6 for n>0.
a(n) = A002283(n)*A002278(n). (End)
G.f.: 36*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Arkadiusz Wesolowski, Dec 26 2011
From Elmo R. Oliveira, Jul 27 2025: (Start)
E.g.f.: 4*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/9.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
a(n) = 36*A002477(n). (End)

Edited by Alois P. Heinz, Aug 21 2019 (merged with A102794, submitted by Richard C. Schroeppel, Feb 26 2005)

A059988 a(n) = (10^n - 1)^2.

Original entry on oeis.org

0, 81, 9801, 998001, 99980001, 9999800001, 999998000001, 99999980000001, 9999999800000001, 999999998000000001, 99999999980000000001, 9999999999800000000001, 999999999998000000000001, 99999999999980000000000001, 9999999999999800000000000001, 999999999999998000000000000001

Offset: 0

Henry Bottomley, Mar 07 2001

From James D. Klein, Feb 05 2012: (Start)
The periods of the reciprocals of a(n) are the consecutive integers from 0 to 10^n-1, omitting the one integer 10^n-2, right-justified in field widths of size n.
E.g.:
  1/81 = 0.012345679...
  1/9801 = 0.000102030405060708091011...9799000102...
  1/998001 = 0.000001002003004005...997999000001002... (End)
Sum of first 10^n - 1 odd numbers. - Arkadiusz Wesolowski, Jun 12 2013

			From _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ..................... 81 = 9^2;
n=2: ................... 9801 = 99^2;
n=3: ................. 998001 = 999^2;
n=4: ............... 99980001 = 9999^2;
n=5: ............. 9999800001 = 99999^2;
n=6: ........... 999998000001 = 999999^2;
n=7: ......... 99999980000001 = 9999999^2;
n=8: ....... 9999999800000001 = 99999999^2;
n=9: ..... 999999998000000001 = 999999999^2. (End)

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 32 at p. 61.
Walther Lietzmann, Lustiges und Merkwuerdiges von Zahlen und Formen, (F. Hirt, Breslau 1921-43), p. 149.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 34.

Harry J. Smith, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A075411, A075412, A075413, A075414, A075415, A075416, A075417, A178630, A178631, A178632, A178633, A178634, A178635, A272066, A272067, A272068.
Subsequence of A238237.
Cf. A002275, A002283, A002477.

A059988:=n->(10^n-1)^2; seq(A059988(n), n=0..20); # Wesley Ivan Hurt, Nov 21 2013

Table[(10^n-1)^2, {n,0,20}] (* Wesley Ivan Hurt, Nov 21 2013 *)

a(n) = { (10^n - 1)^2 } \\ Harry J. Smith, Jul 01 2009

def a(n): return (10**n - 1)**2  # Martin Gergov, Sep 10 2022

a(n) = 81*A002477(n) = A002283(n)^2 = (9*A002275(n))^2.
a(n) = {999... (n times)}^2 = {999... (n times), 000... (n times)} - {999... (n times)}. For example, 999^2 = 999000 - 999 = 998001. - Kyle D. Balliet, Mar 07 2009
a(n) = (A002283(n-1)*10 + 8) * 10^(n-1) + 1, for n>0. - Reinhard Zumkeller, May 31 2010
From Ilya Gutkovskiy, Apr 19 2016: (Start)
O.g.f.: 81*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).
E.g.f.: (1 - 2*exp(9*x) + exp(99*x))*exp(x). (End)
Sum_{n>=1} 1/a(n) = (log(10)*(QPolyGamma(0, 1, 1/10) - log(10/9)) + QPolyGamma(1, 1, 1/10))/log(10)^2 = 0.012448721523422795191... . - Stefano Spezia, Jul 31 2024
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3). - Elmo R. Oliveira, Aug 02 2025

A075412 Squares of A002277.

Original entry on oeis.org

0, 9, 1089, 110889, 11108889, 1111088889, 111110888889, 11111108888889, 1111111088888889, 111111110888888889, 11111111108888888889, 1111111111088888888889, 111111111110888888888889, 11111111111108888888888889, 1111111111111088888888888889, 111111111111110888888888888889

Offset: 0

Michael Taylor (michael.taylor(AT)vf.vodafone.co.uk), Sep 14 2002

A transformation of the Wonderful Demlo numbers (A002477).

			a(2) = 33^2 = 1089.
Contribution from _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ...................... 9 = 9 * 1;
n=2: ................... 1089 = 99 * 11;
n=3: ................. 110889 = 999 * 111;
n=4: ............... 11108889 = 9999 * 1111;
n=5: ............. 1111088889 = 99999 * 11111;
n=6: ........... 111110888889 = 999999 * 111111;
n=7: ......... 11111108888889 = 9999999 * 1111111;
n=8: ....... 1111111088888889 = 99999999 * 11111111;
n=9: ..... 111111110888888889 = 999999999 * 111111111. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Gérard Villemin, Variations sur les carrés.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A075411, A075412, A075413, A075414, A075415, A075416, A075417, A002283, A178630, A178631, A178632, A178633, A178634, A178635.

Cf. A000042, A002277, A002275, A002282, A002283, A002477, A059988.

LinearRecurrence[{11, -10}, {0, 3}, 20]^2 (* Vincenzo Librandi, Mar 20 2014 *)
Table[FromDigits[PadRight[{},n,9]]FromDigits[PadRight[{},n,1]],{n,0,15}] (* Harvey P. Dale, Feb 12 2023 *)

a(n) = A002277(n)^2 = (3*A002275(n))^2 = 9*A002275(n)^2.
a(n) = {111111... (2n times)} - 2*{ 111... (n times)} a(n) = A000042(2*n) - 2*A000042(n). - Amarnath Murthy, Jul 21 2003
a(n) = {333... (n times)}^2 = {111...(n times)}{000... (n times)} - {111... (n times)}. For example, 333^2 = 111000 - 111 = 110889. - Kyle D. Balliet, Mar 07 2009
From Reinhard Zumkeller, May 31 2010: (Start)
a(n) = A002283(n)*A002275(n).
For n>0, a(n) = (A002275(n-1)*10^n + A002282(n-1))*10 + 9. (End)
a(n) = (10^(n+1)-10)^2/900. - José de Jesús Camacho Medina, Apr 01 2016
From Elmo R. Oliveira, Jul 27 2025: (Start)
G.f.: 9*x*(1+10*x)/((1-x)*(1-10*x)*(1-100*x)).
E.g.f.: exp(x)*(1 - 2*exp(9*x) + exp(99*x))/9.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
a(n) = 9*A002477(n). (End)

A178630 a(n) = 18*((10^n - 1)/9)^2.

Original entry on oeis.org

18, 2178, 221778, 22217778, 2222177778, 222221777778, 22222217777778, 2222222177777778, 222222221777777778, 22222222217777777778, 2222222222177777777778, 222222222221777777777778, 22222222222217777777777778, 2222222222222177777777777778, 222222222222221777777777777778

Offset: 1

Reinhard Zumkeller, May 31 2010

			n=1: ..................... 18 = 9 * 2;
n=2: ................... 2178 = 99 * 22;
n=3: ................. 221778 = 999 * 222;
n=4: ............... 22217778 = 9999 * 2222;
n=5: ............. 2222177778 = 99999 * 22222;
n=6: ........... 222221777778 = 999999 * 222222;
n=7: ......... 22222217777778 = 9999999 * 2222222;
n=8: ....... 2222222177777778 = 99999999 * 22222222;
n=9: ..... 222222221777777778 = 999999999 * 222222222.

G. C. Greubel, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A059988, A075412, A075415, A178631, A178632, A178633, A178634, A178635.

Cf. A002276, A002281, A002283, A002477.

List([1..20], n -> 18*((10^n-1)/9)^2); # G. C. Greubel, Jan 28 2019

[18*((10^n - 1)/9)^2: n in [1..20]]; // Vincenzo Librandi, Dec 28 2010

Table[18*((10^n-1)/9)^2, {n, 1, 20}] (* G. C. Greubel, Jan 28 2019 *)

a(n)=18*(10^n\9)^2 \\ Charles R Greathouse IV, Aug 21 2011

[18*((10^n-1)/9)^2 for n in (1..20)] # G. C. Greubel, Jan 28 2019

a(n) = 18*A002477(n) = A002283(n)*A002276(n).
a(n)=((A002276(n-1)*10 + 1)*10^(n-1) + A002281(n-1))*10 + 8.
G.f.: 18*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Ilya Gutkovskiy, Feb 24 2017
From Elmo R. Oliveira, Jul 30 2025: (Start)
E.g.f.: 2*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/9.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 3. (End)

A178631 a(n) = 27*((10^n - 1)/9)^2.

Original entry on oeis.org

27, 3267, 332667, 33326667, 3333266667, 333332666667, 33333326666667, 3333333266666667, 333333332666666667, 33333333326666666667, 3333333333266666666667, 333333333332666666666667, 33333333333326666666666667, 3333333333333266666666666667, 333333333333332666666666666667

Offset: 1

Reinhard Zumkeller, May 31 2010

			n=1: ..................... 27 = 9 * 3;
n=2: ................... 3267 = 99 * 33;
n=3: ................. 332667 = 999 * 333;
n=4: ............... 33326667 = 9999 * 3333;
n=5: ............. 3333266667 = 99999 * 33333;
n=6: ........... 333332666667 = 999999 * 333333;
n=7: ......... 33333326666667 = 9999999 * 3333333;
n=8: ....... 3333333266666667 = 99999999 * 33333333;
n=9: ..... 333333332666666667 = 999999999 * 333333333.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A059988, A075412, A075415, A178630, A178632, A178633, A178634, A178635.

Cf. A002277, A002280, A002283, A002477.

[27*((10^n-1)/9)^2: n in [1..50]]; // Vincenzo Librandi, Dec 28 2010

27*(FromDigits/@Table[PadRight[{},n,1],{n,20}])^2 (* or *) LinearRecurrence[ {111,-1110,1000},{27,3267,332667},20] (* Harvey P. Dale, Oct 11 2012 *)

A178631(n):=27*((10^n-1)/9)^2$ makelist(A178631(n),n,1,10); /* Martin Ettl, Nov 12 2012 */

a(n)=27*(10^n\9)^2 \\ Charles R Greathouse IV, Jul 02 2013

a(n) = 27*A002477(n) = A002283(n)*A002277(n).
a(n) = ((A002277(n-1)*10 + 2)*10^(n-1) + A002280(n-1))*10 + 7.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>3, a(1)=27, a(2)=3267, a(3)=332667. - Harvey P. Dale, Oct 11 2012
G.f.: 27*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Ilya Gutkovskiy, Feb 24 2017
E.g.f.: exp(x)*(1 - 2*exp(9*x) + exp(99*x))/3. - Elmo R. Oliveira, Aug 01 2025

A178632 a(n) = 45*((10^n - 1)/9)^2.

Original entry on oeis.org

45, 5445, 554445, 55544445, 5555444445, 555554444445, 55555544444445, 5555555444444445, 555555554444444445, 55555555544444444445, 5555555555444444444445, 555555555554444444444445, 55555555555544444444444445, 5555555555555444444444444445, 555555555555554444444444444445

Offset: 1

Reinhard Zumkeller, May 31 2010

			n=1: ..................... 45 = 9 * 5;
n=2: ................... 5445 = 99 * 55;
n=3: ................. 554445 = 999 * 555;
n=4: ............... 55544445 = 9999 * 5555;
n=5: ............. 5555444445 = 99999 * 55555;
n=6: ........... 555554444445 = 999999 * 555555;
n=7: ......... 55555544444445 = 9999999 * 5555555;
n=8: ....... 5555555444444445 = 99999999 * 55555555;
n=9: ..... 555555554444444445 = 999999999 * 555555555.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A059988, A075412, A075415, A178630, A178631, A178633, A178634, A178635.

Cf. A002278, A002279, A002283, A002477.

[45*((10^n-1)/9)^2: n in [1..50]]; // Vincenzo Librandi, Dec 28 2010

45 (FromDigits/@Table[PadRight[{}, n, 1], {n, 20}])^2 (* Vincenzo Librandi, Mar 20 2014 *)
LinearRecurrence[{111,-1110,1000},{45,5445,554445},20] (* Harvey P. Dale, Jan 23 2019 *)

A178632(n):=45*((10^n-1)/9)^2$ makelist(A178632(n),n,1,12); /* Martin Ettl, Nov 08 2012 */

a(n)=45*(10^n\9)^2 \\ Charles R Greathouse IV, Jul 02 2013

a(n) = 45*A002477(n) = A002283(n)*A002279(n).
a(n) = (A002279(n-1)*10^n + A002278(n))*10 + 5.
G.f.: 45*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Ilya Gutkovskiy, Feb 24 2017
From Elmo R. Oliveira, Aug 01 2025: (Start)
E.g.f.: 5*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/9.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 3. (End)

cp's OEIS Frontend

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

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A002283 a(n) = 10^n - 1.

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

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A002477 Wonderful Demlo numbers: a(n) = ((10^n - 1)/9)^2.

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A075415 Squares of A002280 or numbers (666...6)^2.

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