A010785 Repdigit numbers, or numbers whose digits are all equal.
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
References
- Albert H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, p. 83.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
- Eric F. Bravo, Carlos A. Gómez and Florian Luca, Product of Consecutive Tribonacci Numbers With Only One Distinct Digit, J. Int. Seq., Vol. 22 (2019), Article 19.6.3.
- Eric Fernando Bravo, On concatenations of Padovan and Perrin numbers, Math. Commun. (2023) Vol 28, 105-119.
- Mahadi Ddamulira, Repdigits as sums of three balancing numbers, Mathematica Slovaca, (2019), hal-02405969.
- Mahadi Ddamulira, Padovan numbers that are concatenations of two distinct repdigits, arXiv:2003.10705 [math.NT], 2020.
- Mahadi Ddamulira, Tribonacci numbers that are concatenations of two repdigits, hal-02547159, Mathematics [math] / Number Theory [math.NT], 2020.
- Mahadi Ddamulira, Padovan numbers that are concatenations of two distinct repdigits, Mathematica Slovaca, Vol. 71, No. 2 (2021), pp. 275-284.
- Bart Goddard and Jeremy Rouse, Sum of two repdigits a square, arXiv:1607.06681 [math.NT], 2016. Mentions this sequence.
- Bir Kafle, Florian Luca and Alain Togbé, Triangular Repblocks, Fibonacci Quart., Vol. 56, No. 4 (2018), pp. 325-328.
- Bir Kafle, Florian Luca and Alain Togbé, Pentagonal and heptagonal repdigits, Annales Mathematicae et Informaticae, Vol. 52 (2020), pp. 137-145.
- Benedict Vasco Normenyo, Bir Kafle, and Alain Togbé, Repdigits as Sums of Two Fibonacci Numbers and Two Lucas Numbers, Integers, Vol. 19 (2019), Article A55.
- Salah Eddine Rihane and Alain Togbé, Repdigits as products of consecutive Padovan or Perrin numbers, Arab. J. Math., Vol. 10 (2021), pp. 469-480.
- Charles W. Trigg, Infinite sequences of palindromic triangular numbers, The Fibonacci Quarterly, Vol. 12, No. 2 (1974), pp. 209-212.
- Eric Weisstein's World of Mathematics, Repdigit.
- Wikipedia, Repdigit.
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,0,-10).
Crossrefs
Programs
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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
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Magma
[(n-9*Floor((n-1)/9))*(10^Floor((n+8)/9)-1)/9: n in [0..50]]; // Vincenzo Librandi, Nov 10 2014
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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
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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 *)
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PARI
a(n)=10^((n+8)\9)\9*((n-1)%9+1) \\ Charles R Greathouse IV, Jun 15 2011
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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
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PARI
is(n)={1==#Set(digits(n))} inv(n) = 9*#Str(n) + n%10 - 9 \\ David A. Corneth, Jun 24 2016
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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
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Python
print([0]+[int(d*r) for r in range(1, 7) for d in "123456789"]) # Michael S. Branicky, Dec 29 2021
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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 > 10: a(n) mod 10 = floor(a(n)/10) mod 10.
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
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
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
Comments