cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 21 results. Next

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

Views

Author

Keywords

Comments

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

References

  • 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.

Crossrefs

Programs

  • Haskell
    a002275 = (`div` 9) . subtract 1 . (10 ^)
    a002275_list = iterate ((+ 1) . (* 10)) 0
    -- Reinhard Zumkeller, Jul 05 2013, Feb 05 2012
    
  • Magma
    [(10^n-1)/9: n in [0..25]]; // Vincenzo Librandi, Nov 06 2014
    
  • Maple
    seq((10^k - 1)/9, k=0..30); # Wesley Ivan Hurt, Sep 28 2013
  • Mathematica
    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 *)
  • Maxima
    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 */
    
  • PARI
    a(n)=(10^n-1)/9; \\ Michael B. Porter, Oct 26 2009
    
  • PARI
    my(x='x+O('x^30)); concat(0, Vec(x/((1-10*x)*(1-x)))) \\ Altug Alkan, Apr 10 2016
    
  • Python
    print([(10**n-1)//9 for n in range(100)]) # Michael S. Branicky, Apr 30 2022
  • Sage
    [lucas_number1(n, 11, 10) for n in range(21)]  # Zerinvary Lajos, Apr 27 2009
    

Formula

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

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Author

Keywords

Comments

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

Examples

			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)
		

Crossrefs

Programs

Formula

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)

Extensions

More terms from Michael De Vlieger, Sep 27 2015

A057951 Number of prime factors of 10^n - 1 (counted with multiplicity).

Original entry on oeis.org

2, 3, 4, 4, 4, 7, 4, 6, 6, 6, 4, 9, 5, 6, 8, 8, 4, 11, 3, 9, 9, 9, 3, 12, 7, 8, 9, 10, 7, 15, 5, 13, 8, 8, 9, 14, 5, 5, 8, 13, 6, 17, 6, 13, 12, 8, 4, 15, 6, 12, 10, 11, 6, 16, 10, 14, 8, 10, 4, 22, 9, 7, 16, 17, 9, 17, 5, 12, 8, 14, 4, 20, 5, 9, 14, 8, 10, 18
Offset: 1

Views

Author

Patrick De Geest, Nov 15 2000

Keywords

Crossrefs

bigomega(b^n-1): this sequence (b=10), A057952 (b=9), A057953 (b=8), A057954 (b=7), A057955 (b=6), A057956 (b=5), A057957 (b=4), A057958 (b=3), A046051 (b=2).

Programs

Formula

Mobius transform of A085035 - T. D. Noe, Jun 19 2003
a(n) = Omega(10^n -1) = Omega(R_n) + 2 = A046053(n) + 2 {where Omega(n) = A001222(n) and R_n = (10^n - 1)/9 = A002275(n)}. - Lekraj Beedassy, Jun 09 2006
a(n) = A001222(A002283(n)). - Ray Chandler, Apr 22 2017

Extensions

Erroneous b-file replaced by Ray Chandler, Apr 26 2017

A095370 Number of distinct prime factors of the repunit (-1 + 10^n)/9.

Original entry on oeis.org

0, 1, 2, 2, 2, 5, 2, 4, 3, 4, 2, 7, 3, 4, 6, 6, 2, 8, 1, 7, 7, 6, 1, 10, 5, 6, 5, 8, 5, 13, 3, 11, 6, 6, 7, 11, 3, 3, 6, 11, 4, 14, 4, 10, 9, 6, 2, 13, 4, 10, 8, 9, 4, 12, 8, 12, 6, 8, 2, 20, 7, 5, 13, 15, 7, 14, 3, 10, 6, 12, 2, 17, 3, 7, 12, 6, 8, 15, 6, 15, 10, 7, 3, 21, 7, 8, 10, 14, 5, 21, 12, 10
Offset: 1

Views

Author

Labos Elemer, Jun 04 2004; corrected Jun 09 2004

Keywords

Comments

Factoring certain repunits is especially difficult.

Examples

			a(62)=5 because
11111111111111111111111111111111111111111111111111111111111111 =
11 * 2791 * 6943319 * 57336415063790604359 * 909090909090909090909090909091.
a(97)=3 because (10^97 - 1)/9 = 12004721 * 846035731396919233767211537899097169 * 109399846855370537540339266842070119107662296580348039.
		

References

  • Yates, S. "Peculiar Properties of Repunits." J. Recr. Math. 2, 139-146,1969.
  • Yates, S. "Prime Divisors of Repunits." J. Recr. Math. 8, 33-38, 1975.

Crossrefs

Cf. A046053 (total number of prime factors).

Programs

Formula

a(n) = A001221(A002275(n)).
If 3|n, then a(n) = A102347(n); otherwise a(n) = A102347(n) - 1. - Max Alekseyev, Apr 25 2022

Extensions

Terms to a(322) in b-file from Ray Chandler, Apr 22 2017
a(323)-a(352) in b-file from Max Alekseyev, Apr 26 2022

A102380 Irregular triangle read by rows in which row n lists prime factors (with multiplicity) of the repunit (10^n - 1)/9 (A002275(n)).

Original entry on oeis.org

1, 11, 3, 37, 11, 101, 41, 271, 3, 7, 11, 13, 37, 239, 4649, 11, 73, 101, 137, 3, 3, 37, 333667, 11, 41, 271, 9091, 21649, 513239, 3, 7, 11, 13, 37, 101, 9901, 53, 79, 265371653, 11, 239, 4649, 909091, 3, 31, 37, 41, 271, 2906161, 11, 17, 73, 101
Offset: 1

Views

Author

N. J. A. Sloane, Nov 28 2006

Keywords

Comments

See A003020 for other links and references.

Examples

			First rows:
    1;
   11;
    3,   37;
   11,  101;
   41,  271;
    3,    7, 11, 13, 37;
  239, 4649;
  ...
		

Crossrefs

Programs

  • Maple
    [seq( ifactor((10^n-1)/9),n=1..20)];

Extensions

First 100 rows in b-file from T. D. Noe, Feb 27 2009
Rows n=101..322 in b-file from Ray Chandler, May 01 2017
Rows n=323..352 in b-file from Max Alekseyev, Apr 26 2022

A005422 Largest prime factor of 10^n - 1.

Original entry on oeis.org

3, 11, 37, 101, 271, 37, 4649, 137, 333667, 9091, 513239, 9901, 265371653, 909091, 2906161, 5882353, 5363222357, 333667, 1111111111111111111, 27961, 10838689, 513239, 11111111111111111111111, 99990001, 182521213001, 1058313049
Offset: 1

Views

Author

Keywords

References

  • J. Brillhart et al., Factorizations of b^n +- 1. Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 2nd edition, 1985; and later supplements.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Same as A003020 except for the additional a(1) = 3.
Cf. similar sequences listed in A274906.

Programs

Formula

For n > 1, a(n) = A003020(n). For 1 < n < 10, a(n) = A075024(n). - M. F. Hasler, Jul 30 2015
a(n) = A006530(A002283(n)). - Vincenzo Librandi, Jul 13 2016
a(A004023(n)) = A002275(A004023(n)). - Bernard Schott, May 24 2022

Extensions

Terms to a(100) in b-file from Yousuke Koide added by T. D. Noe, Dec 06 2006
Edited by M. F. Hasler, Jul 30 2015
a(101)-a(322) in b-file from Ray Chandler, Apr 22 2017
a(323)-a(352) in b-file from Max Alekseyev, Apr 26 2022

A067063 Smallest prime factor of repunit(n) = (10^n-1)/9 (A002275).

Original entry on oeis.org

11, 3, 11, 41, 3, 239, 11, 3, 11, 21649, 3, 53, 11, 3, 11, 2071723, 3, 1111111111111111111, 11, 3, 11, 11111111111111111111111, 3, 41, 11, 3, 11, 3191, 3, 2791, 11, 3, 11, 41, 3, 2028119, 11, 3, 11, 83, 3, 173, 11, 3, 11, 35121409, 3, 239
Offset: 2

Views

Author

Amarnath Murthy, Jan 03 2002

Keywords

Comments

a(n) = A003020(n) = R_(n) iff n is a term of A004023. - Bernard Schott, May 22 2022

References

  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers.

Crossrefs

Largest factor: A003020.

Programs

  • Maple
    'min(op(numtheory[factorset]((10^k-1)/9)))'$k=2..50; # M. F. Hasler, Nov 21 2006
  • Mathematica
    a = {}; Do[a = Append[a, FactorInteger[(10^n - 1)/9][[1, 1]]], {n, 2, 111} ]; a
    Table[FactorInteger[FromDigits[PadRight[{},n,1]]][[1,1]],{n,2,50}] (* Harvey P. Dale, Dec 10 2013 *)

Formula

a(3n) = 3, a(6n-4) = a(6n-2) = 11, a(30n-25) = a(30n-5) = 41, ... - M. F. Hasler, Nov 21 2006
a(n) = A020639(A002275(n)). - Ray Chandler, Apr 22 2017

Extensions

More terms from Robert G. Wilson v, Jan 04 2002

A061075 Greatest prime number p(n) with decimal fraction period of length n.

Original entry on oeis.org

3, 11, 37, 101, 271, 13, 4649, 137, 333667, 9091, 513239, 9901, 265371653, 909091, 2906161, 5882353, 5363222357, 52579, 1111111111111111111, 27961, 10838689, 8779, 11111111111111111111111, 99990001, 182521213001, 1058313049, 440334654777631, 121499449, 77843839397
Offset: 1

Views

Author

Heiner Muller-Merbach (hmm(AT)sozwi.uni-kl.de), May 29 2001

Keywords

Examples

			1/271 = 0.0036900369, period of n=5 for p(5)=271.
		

Crossrefs

Last terms in rows of A046107.

Programs

  • Mathematica
    a[n_] := Cyclotomic[n, 10] // FactorInteger // Last // First; Table[a[n], {n, 1, 26}] (* Jean-François Alcover, Aug 05 2013, after Pari *)
  • PARI
    a(n) = my(p); if(n<1, 0, p=factor(polcyclo(n,10))[,1]; p[#p])

Formula

a(n) = A006530(A019328(n)). - Ray Chandler, May 10 2017

Extensions

Terms to a(322) in b-file from Ray Chandler, Apr 28 2017
a(323)-a(352) in b-file from Max Alekseyev, Apr 26 2022

A070529 Number of divisors of repunit 111...111 (with n digits).

Original entry on oeis.org

1, 2, 4, 4, 4, 32, 4, 16, 12, 16, 4, 128, 8, 16, 64, 64, 4, 384, 2, 128, 128, 96, 2, 1024, 32, 64, 64, 256, 32, 8192, 8, 2048, 64, 64, 128, 3072, 8, 8, 64, 2048, 16, 24576, 16, 1536, 768, 64, 4, 8192, 16, 1024, 256, 512, 16, 8192, 256, 4096
Offset: 1

Views

Author

Henry Bottomley, May 02 2002

Keywords

Examples

			a(9) = 12 since the divisors of 111111111 are 1, 3, 9, 37, 111, 333, 333667, 1001001, 3003003, 12345679, 37037037, 111111111.
		

Crossrefs

Programs

Formula

a(n) = A000005(A002275(n)).
a(n) = A070528(n)*A051064(n)/(A051064(n)+2).
a(A004023(n)) = 2. - Michel Marcus, Sep 09 2015
a(A046413(n)) = 4. - Bruno Berselli, Sep 09 2015

Extensions

Terms to a(280) in b-file from Hans Havermann, Aug 20 2011
a(281)-a(322) in b-file from Ray Chandler, Apr 22 2017
a(323)-a(352) ib b-file from Max Alekseyev, May 04 2022

A102146 a(n) = sigma(10^n - 1), where sigma(n) is the sum of positive divisors of n.

Original entry on oeis.org

13, 156, 1520, 15912, 148512, 2042880, 14508000, 162493344, 1534205464, 16203253248, 144451398000, 2063316971520, 14903272088640, 158269280832000, 1614847741624320, 17205180696931968, 144444514193267496
Offset: 1

Views

Author

Jun Mizuki (suzuki32(AT)sanken.osaka-u.ac.jp), Feb 14 2005

Keywords

Crossrefs

Programs

  • Mathematica
    DivisorSigma[1,10^Range[20]-1] (* Harvey P. Dale, Jan 05 2012 *)
  • PARI
    a(n) = sigma(10^n-1); \\ Michel Marcus, Apr 22 2017

Formula

a(n) = A000203(A002283(n)). - Ray Chandler, Apr 22 2017
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