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 1-9 of 9 results.

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

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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)

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

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Author

Keywords

Comments

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

References

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

Crossrefs

Programs

  • Haskell
    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
    
  • Magma
    [(n-9*Floor((n-1)/9))*(10^Floor((n+8)/9)-1)/9: n in [0..50]]; // Vincenzo Librandi, Nov 10 2014
    
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    a(n)=10^((n+8)\9)\9*((n-1)%9+1) \\ Charles R Greathouse IV, Jun 15 2011
    
  • PARI
    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
    
  • PARI
    is(n)={1==#Set(digits(n))}
    inv(n) = 9*#Str(n) + n%10 - 9 \\ David A. Corneth, Jun 24 2016
    
  • Python
    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
    
  • Python
    print([0]+[int(d*r) for r in range(1, 7) for d in "123456789"]) # Michael S. Branicky, Dec 29 2021
    
  • Python
    # 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

Formula

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

Extensions

Name clarified by Jon E. Schoenfield, Nov 10 2023

A000042 Unary representation of natural numbers.

Original entry on oeis.org

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

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Author

Keywords

Comments

Or, numbers written in base 1.
If p is a prime > 5 then d_{a(p)} == 1 (mod p) where d_{a(p)} is a divisor of a(p). This also gives an alternate elementary proof of the infinitude of prime numbers by the fact that for every prime p there exists at least one prime of the form k*p + 1. - Amarnath Murthy, Oct 05 2002
11 = 1*9 + 2; 111 = 12*9 + 3; 1111 = 123*9 + 4; 11111 = 1234*9 + 5; 111111 = 12345*9 + 6; 1111111 = 123456*9 + 7; 11111111 = 1234567*9 + 8; 111111111 = 12345678*9 + 9. - Vincenzo Librandi, Jul 18 2010

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See pp. 57-58.
  • K. G. Kroeber, Mathematik der Palindrome; p. 348; 2003; ISBN 3 499 615762; Rowohlt Verlag; Germany.
  • D. Olivastro, Ancient Puzzles. Bantam Books, NY, 1993, p. 276.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 32.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    A000042 n = (10^n-1) `div` 9 -- James Spahlinger, Oct 08 2012
    (Common Lisp) (defun a000042 (n) (truncate (expt 10 n) 9)) ; James Spahlinger, Oct 12 2012
    
  • Magma
    [(10^n - 1)/9: n in [1..20]]; // G. C. Greubel, Nov 04 2018
    
  • Maple
    a:= n-> parse(cat(1$n)):
    seq(a(n), n=1..25);  # Alois P. Heinz, Mar 23 2018
  • Mathematica
    Table[(10^n - 1)/9, {n, 1, 18}]
    FromDigits/@Table[PadLeft[{},n,1],{n,20}] (* Harvey P. Dale, Aug 21 2011 *)
  • PARI
    a(n)=if(n<0,0,(10^n-1)/9)
    
  • Python
    def a(n): return int("1"*n) # Michael S. Branicky, Jan 01 2021
  • Sage
    [gaussian_binomial(n, 1, 10) for n in range(1, 19)]  # Zerinvary Lajos, May 28 2009
    

Formula

a(n) = (10^n - 1)/9.
G.f.: 1/((1-x)*(1-10*x)).
Binomial transform of A003952. - Paul Barry, Jan 29 2004
From Paul Barry, Aug 24 2004: (Start)
a(n) = 10*a(n-1) + 1, n > 1, a(1)=1. [Offset 1.]
a(n) = Sum_{k=0..n} binomial(n+1, k+1)*9^k. [Offset 0.] (End)
a(2n) - 2*a(n) = (3*a(n))^2. - Amarnath Murthy, Jul 21 2003
a(n) is the binary representation of the n-th Mersenne number (A000225). - Ross La Haye, Sep 13 2003
The Hankel transform of this sequence is [1,-10,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007
E.g.f.: (exp(10*x) - exp(x))/9. - G. C. Greubel, Nov 04 2018
a(n) = 11*a(n-1) - 10*a(n-2). - Wesley Ivan Hurt, May 28 2021
a(n+m-2) = a(m)*a(n-1) - (a(m)-1)*a(n-2), n>1, m>0. - Matej Veselovac, Jun 07 2021
Sum_{n>=1} 1/a(n) = A065444. - Stefano Spezia, Jul 30 2024

Extensions

More terms from Paul Barry, Jan 29 2004

A073668 Decimal expansion of Sum_{k>=1} 1/(10^k - 1).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 5, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8, 2, 6, 4, 4, 4, 9, 2, 4, 4, 8, 2, 8, 2, 6, 6, 4, 3, 0, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 3, 2, 2, 4, 6, 7, 4, 8, 2, 6, 4, 8, 3, 2, 2, 4, 6, 6, 4, 8, 3, 0, 5, 4, 3, 2, 4, 4, 4, 8, 3, 2, 4, 6, 4, 4, 5, 2, 2, 6, 6, 9, 2, 8, 2, 8, 8
Offset: 0

Views

Author

Robert G. Wilson v, Aug 29 2002

Keywords

Comments

Parallels A000005 up to a(46).
Sum_{k>=1} x^k/(1-x^k) = Sum_{k>=1} tau(k)*x^k. Choosing x = 1/10 gives the result. - Amarnath Murthy, Oct 21 2002

Examples

			0.122324243426244526264428344628264449244... = A065444/9.
		

References

  • Amarnath Murthy, Some interesting results on d(N), the number of divisors of a natural number, page 463, Octogon Mathematical Magazine, Vol. 8 No. 2, October 2000.

Crossrefs

Programs

  • Maple
    evalf(Sum(1/(10^k - 1), k = 1..infinity), 200) # Vaclav Kotesovec, Jul 16 2019
    # second program with faster converging series after Joerg Arndt
    evalf( add( (1/10)^(n^2)*(1 + 2/(10^n - 1)), n = 1..8), 105); # Peter Bala, Jan 30 2022
  • Mathematica
    RealDigits[ N[ Sum[1/(10^k - 1), {k, 1, Infinity}], 120]] [[1]]
  • PARI
    suminf(k=1,1/(10^k-1)) \\ Charles R Greathouse IV, Oct 05 2014

Formula

From Eric Desbiaux, Mar 11 2009: (Start)
Equals Sum_{k >= 1} 1/((2^k*5^k)-1).
Equals Sum_{k >= 1} (1/2^k)*(1/5^k)/(1-((1/2^k)*(1/5^k))).
Sum_{k >= 1} 1/(5^k) = 1/4.
Sum_{k >= 1} 1/(2^k) = 1.
Sum_{k >= 1} (1/5^k)/(1-((1/2^k)*(1/5^k))) = 0.2726344339156...
Sum_{k >= 1} (1/2^k)/(1-((1/2^k)*(1/5^k))) = 1.0582125127815...
Sum_{k >= 1} 1/(1-((1/2^k)*(1/5^k))) - 1 = A073668.
(End)
Fast computation via Lambert series: 0.122324243426... = Sum_{n>=1} x^(n^2)*(1+x^n)/(1-x^n) where x=1/10. - Joerg Arndt, Oct 18 2020

A083230 Number of repunit divisors of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Reinhard Zumkeller, Jun 01 2003

Keywords

Comments

Differs from A043284 (maximal run length in decimal expansion) from a(100) on. - M. F. Hasler, Oct 18 2019

Examples

			n = 110, divisors are {1, 2, 5, 10, 11, 22, 55, 110} with two repunits: 1 and 11, therefore a(110) = 2.
n = 111, divisors are {1, 3, 37, 111} with two repunits: 1 and 111, therefore a(111) = 2.
n = 111111, divisors are {1, 3, 7, 11, 13, 21, 33, 37, 39, 77, 91, 111, 143, 231, 259, 273, 407, 429, 481, 777, 1001, 1221, 1443, 2849, 3003, 3367, 5291, 8547, 10101, 15873, 37037, 111111} with four repunits: 1, 11, 111 and 111111, therefore a(111111) = 4.
		

Crossrefs

Programs

  • Mathematica
    A083230[n_]:=Count[IntegerDigits[Divisors[n]],{1..}];Array[A083230,100] (* Paolo Xausa, Sep 27 2023 *)
  • PARI
    a(n)={my(s=0, k=1); while(k<=n, if(n%k==0, s++); k=10*k+1); s} \\ Andrew Howroyd, Aug 07 2018

Formula

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A065444. - Amiram Eldar, Apr 17 2025

A338712 Numbers with all digits equal and from the set {1, 3, 7, 9}.

Original entry on oeis.org

1, 3, 7, 9, 11, 33, 77, 99, 111, 333, 777, 999, 1111, 3333, 7777, 9999, 11111, 33333, 77777, 99999, 111111, 333333, 777777, 999999, 1111111, 3333333, 7777777, 9999999, 11111111, 33333333, 77777777, 99999999, 111111111, 333333333, 777777777, 999999999, 1111111111, 3333333333, 7777777777, 9999999999
Offset: 1

Views

Author

N. J. A. Sloane, Nov 08 2020, following a suggestion from Hugo Pfoertner

Keywords

Comments

Candidates for prefixes and suffixes in A090287.

Crossrefs

Programs

  • Maple
    a:= n-> [1, 3, 7, 9][1+irem(n-1, 4)]*(10^iquo(n+3, 4)-1)/9:
    seq(a(n), n=1..50);  # Alois P. Heinz, Nov 09 2020
  • Mathematica
    A338712={}; Do[AppendTo[A338712, FromDigits[ConstantArray[#,i]] & /@{ 1,3,7,9}], {i,10}]; A338712//Flatten (* Robert Price, Sep 21 2023 *)

Formula

From Colin Barker, Nov 09 2020: (Start)
G.f.: x*(1 + 3*x + 7*x^2 + 9*x^3) / ((1 - x)*(1 + x)*(1 + x^2)*(1 - 10*x^4)).
a(n) = 11*a(n-4) - 10*a(n-8) for n>8. (End)
Sum_{n>=1} 1/a(n) = (100/63) * A065444. - Amiram Eldar, Aug 31 2025

A135702 Decimal expansion of 3*Sum_{k=1..inf} 1/(10^k-1).

Original entry on oeis.org

3, 6, 6, 9, 7, 2, 7, 3, 0, 2, 7, 8, 7, 3, 3, 5, 7, 8, 7, 9, 3, 2, 8, 5, 0, 3, 3, 8, 8, 4, 7, 9, 3, 3, 4, 7, 7, 3, 4, 4, 8, 4, 7, 9, 9, 2, 9, 1, 0, 9, 3, 8, 8, 5, 4, 5, 3, 2, 9, 6, 7, 4, 0, 2, 4, 4, 7, 9, 4, 4, 9, 6, 7, 3, 9, 9, 4, 4, 9, 1, 6, 2, 9, 7, 3, 3, 4, 4, 9, 7, 3, 9, 3, 3, 5, 6, 8, 0, 0, 7, 8, 4, 8, 6, 5
Offset: 0

Views

Author

Eric Desbiaux, Mar 03 2008

Keywords

Comments

Equals: 3*A073668, also A065444/3.

Examples

			0.36697273...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[9*N[ (1/3) Sum[1/(10^k - 1), {k, 1, Infinity}], 120]] [[1]]
  • PARI
    3*suminf(k=1, 1/(10^k-1)) \\ Michel Marcus, Oct 30 2016

A355698 a(n) is the number of repdigits divisors of n (A010785).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 4, 3, 3, 2, 5, 1, 3, 3, 4, 1, 5, 1, 4, 3, 4, 1, 6, 2, 2, 3, 4, 1, 5, 1, 4, 4, 2, 3, 6, 1, 2, 2, 5, 1, 5, 1, 6, 4, 2, 1, 6, 2, 3, 2, 3, 1, 5, 4, 5, 2, 2, 1, 6, 1, 2, 4, 4, 2, 8, 1, 3, 2, 4, 1, 7, 1, 2, 3, 3, 4, 4, 1, 5, 3, 2, 1, 6, 2, 2, 2, 8, 1, 6, 2, 3, 2, 2, 2, 6, 1, 3, 6, 4, 1, 4, 1, 4, 4
Offset: 1

Views

Author

Bernard Schott, Jul 14 2022

Keywords

Comments

More than the usual number of terms are displayed in order to show the difference from A087990.
The first 100 terms are the same first 100 terms of A087990, then a(101) = 1 while A087990(101) = 2, because 101 is the smallest palindrome that is not repdigit; the next difference is 121.
Inequalities: 1 <= a(n) <= A087990(n).

Examples

			66 has 8 divisors: {1, 2, 3, 6, 11, 22, 33, 66} that are all repdigits, hence a(66) = 8.
121 has 3 divisors: {1, 11, 121} of which 2 are repdigits: {1, 11}, hence a(121) = 2.
		

Crossrefs

Programs

  • Maple
    isrepdig:= proc(n) nops(convert(convert(n,base,10),set))=1 end proc:
    f:= proc(n) nops(select(isrepdig, numtheory:-divisors(n))) end proc:
    map(f, [$1..200]); # Robert Israel, Aug 07 2024
  • Mathematica
    a[n_] := DivisorSum[n, 1 &, Length[Union[IntegerDigits[#]]] == 1 &]; Array[a, 100] (* Amiram Eldar, Jul 14 2022 *)
  • PARI
    a(n) = my(ret=0,u=1); while(u<=n, ret+=sum(d=1,9, n%(u*d)==0); u=10*u+1); ret; \\ Kevin Ryde, Jul 14 2022
    
  • PARI
    isrep(n) = {1==#Set(digits(n))}; \\ A010785
    a(n) = sumdiv(n, d, isrep(d)); \\ Michel Marcus, Jul 15 2022
  • Python
    from sympy import divisors
    def c(n): return len(set(str(n))) == 1
    def a(n): return sum(1 for d in divisors(n, generator=True) if c(d))
    print([a(n) for n in range(1, 105)]) # Michael S. Branicky, Jul 14 2022
    

Formula

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = (7129/2520) * A065444 = 3.11446261209177581335... . - Amiram Eldar, Apr 17 2025

A067617 Continued fraction for 9 * sum( 1/(10^k -1), k=1..0/0 ).

Original entry on oeis.org

1, 9, 1, 9, 1, 109, 1, 8, 5, 1, 1, 8, 1, 1, 1, 2, 1, 91, 10, 5, 1, 1, 1, 3, 6, 2, 1, 3, 2, 1, 1, 1, 24, 3, 1, 5, 1, 21, 3, 1, 1, 1, 1, 1, 2, 7, 1, 2, 1, 1, 1, 1, 15, 7, 1, 11, 2, 2, 3, 3, 1, 1, 60, 3, 1, 2, 1, 3, 18, 1, 2, 7, 1, 2, 3, 1, 3, 4, 1, 2, 1, 36, 1, 1, 9, 1, 1, 1, 2, 1, 3, 1, 58, 2, 1, 62
Offset: 0

Views

Author

Frank Ellermann, Feb 23 2002

Keywords

Examples

			1.10091819083620073637985510165438004320345439787328165635989...
		

Crossrefs

Cf. A065444 (decimal expansion).

Extensions

Offset changed by Andrew Howroyd, Jul 04 2024
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