A178500 -id:A178500 - cp's OEIS frontend

A011557 Powers of 10: a(n) = 10^n.

1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000, 10000000000, 100000000000, 1000000000000, 10000000000000, 100000000000000, 1000000000000000, 10000000000000000, 100000000000000000, 1000000000000000000

Offset: 0

N. J. A. Sloane

Same as Pisot sequences E(1, 10), L(1, 10), P(1, 10), T(1, 10). Essentially same as Pisot sequences E(10, 100), L(10, 100), P(10, 100), T(10, 100). See A008776 for definitions of Pisot sequences.
Same as k^n in base k. - Dominick Cancilla, Aug 02 2010 [Corrected by Jianing Song, Sep 17 2022]
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 10-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Smallest n+1 digit number greater than 0 (with offset 0). - Wesley Ivan Hurt, Jan 17 2014
Numbers with digit sum = 1, or, A007953(a(n)) = 1. - Reinhard Zumkeller, Jul 17 2014
Does not satisfy Benford's law. - N. J. A. Sloane, Feb 14 2017

Philip Morrison et al., Powers of Ten, Scientific American Press, 1982 and later editions.
S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

T. D. Noe, Table of n, a(n) for n = 0..100
Kees Boeke, Cosmic View: The Universe in 40 Jumps (1957) [The original "powers of ten" book]
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
Charles and Ray Eames, Powers of Ten
Tanya Khovanova, Recursive Sequences
Robert Price, Comments on A011557 concerning Elementary Cellular Automata, Feb 21 2016
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Science, Optics and You, Secret Worlds: The Universe Within [Powers of Ten]
Eric Weisstein's World of Mathematics, 10
Eric Weisstein's World of Mathematics, Digitaddition
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
Wikipedia, Powers of Ten
S. Wolfram, A New Kind of Science
Index entries for linear recurrences with constant coefficients, signature (10).
Index to Elementary Cellular Automata
Index entries for sequences related to cellular automata
Index entries for sequences related to Benford's law

Cf. A000005, A000012, A000290, A000400, A007953, A008776, A159991, A178500, A242614.
Cf. A178501: this sequence with 0 prefixed.
Row 5 of A329332.

a011557 = (10 ^)
a011557_list = iterate (* 10) 1
-- Reinhard Zumkeller, Jul 05 2013, Feb 05 2012

A011557:=n->10^n; seq(A011557(n), n=0..40); # Wesley Ivan Hurt, Jan 17 2014

Table[10^n,{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 15 2011 *)
10^Range[0,20] (* Harvey P. Dale, Sep 17 2023 *)

A011557(n):=10^n$ makelist(A011557(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

a(n)=10^n \\ Charles R Greathouse IV, Jun 15 2011

print([10**n for n in range(19)]) # Michael S. Branicky, Jan 10 2021

a(n) = 10^n.
a(n) = 10*a(n-1).
G.f.: 1/(1-10*x).
E.g.f.: exp(10*x).
A000005(a(n)) = A000290(n+1). - Reinhard Zumkeller, Mar 04 2007
a(n) = 60^n/6^n = A159991(n)/A000400(n). - Reinhard Zumkeller, May 02 2009
a(n) = A178501(n+1); for n > 0: a(n) = A178500(n). - Reinhard Zumkeller, May 28 2010
Sum_{n>0} 1/a(n) = 1/9 = A000012. - Stefano Spezia, Apr 28 2024

Links to "Powers of Ten" books and videos added by N. J. A. Sloane, Nov 07 2009

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

A061601 9's complement of n: a(n) = 10^d - 1 - n where d is the number of digits in n. If a is a digit in n replace it with 9 - a.

Original entry on oeis.org

9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 89, 88, 87, 86, 85, 84, 83, 82, 81, 80, 79, 78, 77, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28

Offset: 0

Amarnath Murthy, May 19 2001

A109002 and A178500 give record values and where they occur: A109002(n+1)=a(A178500(n)) and a(m)<A109002(n+1) for m<A178500(n). - Reinhard Zumkeller, May 28 2010
If n is divisible by 3, so is a(n). The same goes for 9. - Alonso del Arte, Dec 01 2011
For n > 0, a(n-1) consists of the A055642(n) least significant digits of the 10-adic integer -n. - Stefano Spezia, Jan 21 2021

			a(7) = 2 = 10 - 1 -7. a(123) = 1000 -1 -123 = 876.

Kjartan Poskitt, Murderous Maths: Numbers, The Key to the Universe, Scholastic Ltd, 2002. See p 159.

Indranil Ghosh, Table of n, a(n) for n = 0..25000 (terms 0..1000 from Harry J. Smith)

Cf. A035327, A171960.
Cf. A055120.
See A267193 for complement obverse of n.

a061601 n = if n <= 9 then 9 - n else 10 * ad n' + 9 - d
            where (n',d) = divMod n 10
-- Reinhard Zumkeller, Feb 21 2014, Oct 04 2011

A061601 := proc(n)
        10^A055642(n)-1-n ;
end proc: # R. J. Mathar, Nov 30 2011

nineComplement[n_] := FromDigits[Table[9, {Length[IntegerDigits[n]]}] - IntegerDigits[n]]; Table[nineComplement[n], {n, 0, 71}] (* Alonso del Arte, Nov 30 2011 *)

A061601(n)=my(e=length(Str(n)));10^e-1 - n; \\ Joerg Arndt, Aug 28 2013

def A061601(n):
    return 10**len(str(n))-1-n # Indranil Ghosh, Jan 30 2017

a(n) = if n<10 then 9 - n else 10*a([n/10]) + 9 - n mod 10. - Reinhard Zumkeller, Jan 20 2010

a(n) <= 9n - 1. - Charles R Greathouse IV, Nov 15 2022

Corrected and extended by Matthew Conroy, Jan 19 2002

A105279 a(0)=0; a(n) = 10*a(n-1) + 10.

Original entry on oeis.org

0, 10, 110, 1110, 11110, 111110, 1111110, 11111110, 111111110, 1111111110, 11111111110, 111111111110, 1111111111110, 11111111111110, 111111111111110, 1111111111111110, 11111111111111110, 111111111111111110, 1111111111111111110, 11111111111111111110, 111111111111111111110

Offset: 0

Alexandre Wajnberg, Apr 25 2005

a(n) is the smallest even number with digits in {0,1} having digit sum n; in other words, the base 10 reading of the binary string of A000918(n). Cf. A069532. - Jason Kimberley, Nov 02 2011

Also, except for a(0), the binary representation of the diagonal from the corner to the origin of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 645", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Jul 19 2017

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Robert Price, Diagrams of first 20 stages of the Cellular Automata.
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton.
S. Wolfram, A New Kind of Science.
Wolfram Research, Wolfram Atlas of Simple Programs.
Index entries for sequences related to cellular automata.
Index to 2D 5-Neighbor Cellular Automata.
Index to Elementary Cellular Automata.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A000918, A002275, A029858, A052379, A052386, A069532, A080674, A124166, A161770.
Row n=10 of A228275.
Partial sums of A178500.

a105279 n = a105279_list !! n
a105279_list = iterate ((* 10) . (+ 1)) 0
-- Reinhard Zumkeller, Feb 05 2012

[-10/9+(10/9)*10^n: n in [0..20]]; // Vincenzo Librandi, Jul 04 2011

NestList[10*(# + 1) &, 0, 25] (* Paolo Xausa, Jul 17 2024 *)

a(n) = (10/9)*(10^n - 1), with n>=0.
a(n) = Sum_{k=1..n} 10^k.
Repunits times 10: a(n) = 10 * A002275(n). - Reinhard Zumkeller, Feb 05 2012
From Stefano Spezia, Sep 15 2023: (Start)
O.g.f.: 10*x/((1 - x)*(1 - 10*x)).
E.g.f.: 10*exp(x)*(exp(9*x) - 1)/9. (End)
From Elmo R. Oliveira, Jun 18 2025: (Start)
a(n) = 11*a(n-1) - 10*a(n-2).
a(n) = A124166(n)/10.
a(n) = A161770(n)/100 for n >= 1. (End)

A178501 Zero followed by powers of ten.

Original entry on oeis.org

0, 1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000, 10000000000, 100000000000, 1000000000000, 10000000000000, 100000000000000, 1000000000000000, 10000000000000000, 100000000000000000, 1000000000000000000, 10000000000000000000, 100000000000000000000

Offset: 0

Reinhard Zumkeller, May 28 2010

The sequence S consisting of the nonnegative numbers arranged in lexicographic order according to their decimal expansion begins 0, 1, 10, 100, 1000, ..., 2, 20, 200, 2000, ..., 3, 30, ... does not have an OEIS entry, since there are uncountably many terms before 2 appears (or even before 100000010000 appears). However, S does begin with the present sequence. - N. J. A. Sloane, Dec 09 2024

a(n)^k + reverse(a(n))^k is a palindrome for any positive integer k. - Bui Quang Tuan, Mar 31 2015

Index entries for linear recurrences with constant coefficients, signature (10).

Cf. A093136, A131577, A140429, A178500; subsequence of A029793.

The powers of 10, A011557, is a subsequence.

Join[{0}, 10^Range[0, 19]] (* Alonso del Arte, Mar 31 2015 *)

a(n)=10^n\10 \\ Charles R Greathouse IV, Oct 07 2015

concat(0, Vec(-x/(10*x-1) + O(x^22))) \\ Elmo R. Oliveira, Jul 21 2025

a(n+1) = A011557(n).
a(n) = A178500(n)/10.
From Paul Barry, Jul 09 2003: (Start)
a(n) = (10^n - 0^n)/10.
E.g.f.: exp(5*x)*sinh(5*x)/5.
Binomial transform of A015577. (End)
G.f.: x/(1 - 10*x). - Chai Wah Wu, Jun 17 2020
From Elmo R. Oliveira, Jul 21 2025: (Start)
a(n) = 10*a(n-1) for n > 1.
a(n) = A093136(n)/2 for n >= 1. (End)

More terms from Elmo R. Oliveira, Jul 21 2025

A109002 Maximal difference between two n-digit numbers.

Original entry on oeis.org

9, 89, 899, 8999, 89999, 899999, 8999999, 89999999, 899999999, 8999999999, 89999999999, 899999999999, 8999999999999, 89999999999999, 899999999999999, 8999999999999999, 89999999999999999, 899999999999999999, 8999999999999999999, 89999999999999999999, 899999999999999999999

Offset: 1

Amarnath Murthy, Aug 14 2005

			a(1) = 9 - 0 = 9, a(4) = 9999 - 1000 = 8999.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A061601, A178500.

[9] cat [9*10^n-1: n in [1..30]]; // Vincenzo Librandi, Oct 29 2011

Join[{9},Table[FromDigits[PadRight[{8},n,9]],{n,2,20}]] (* or *) LinearRecurrence[{11,-10},{9,89,899},20] (* Harvey P. Dale, May 09 2021 *)

a(n)=if(n>1,(10^n-1)-10^(n-1),9) \\ Charles R Greathouse IV, Oct 29 2011

a(n) = (10^n - 1) - 10^(n-1), n > 1.
From Reinhard Zumkeller, May 28 2010: (Start)
a(n) = A061601(A178500(n-1)).
a(n+1) = 10*a(n) + 9. (End)
G.f.: 9*x - x^2*(-89+80*x)/((10*x-1)*(x-1)). - R. J. Mathar, Oct 29 2011
From Elmo R. Oliveira, Jun 12 2025: (Start)
E.g.f.: (1 + 10*x - 10*exp(x) - exp(10*x) + 10*exp(10*x))/10.
a(n) = 11*a(n-1) - 10*a(n-2) for n >= 4. (End)

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A011557 Powers of 10: a(n) = 10^n.

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

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A061601 9's complement of n: a(n) = 10^d - 1 - n where d is the number of digits in n. If a is a digit in n replace it with 9 - a.

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A105279 a(0)=0; a(n) = 10*a(n-1) + 10.

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A178501 Zero followed by powers of ten.

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A109002 Maximal difference between two n-digit numbers.

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