A168624 -id:A168624 - cp's OEIS frontend

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

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

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

A066138 a(n) = 10^(2*n) + 10^n + 1.

Original entry on oeis.org

3, 111, 10101, 1001001, 100010001, 10000100001, 1000001000001, 100000010000001, 10000000100000001, 1000000001000000001, 100000000010000000001, 10000000000100000000001, 1000000000001000000000001, 100000000000010000000000001, 10000000000000100000000000001, 1000000000000001000000000000001

Offset: 0

Henry Bottomley, Dec 07 2001

Palindromes whose digit sum is 3.
Essentially the same as A135577. - R. J. Mathar Apr 29 2008
From Peter Bala, Sep 25 2015: (Start)
For n >= 1, the simple continued fraction expansion of sqrt(a(n)) = [10^n; 1, 1, 2/3*(10^n - 1), 1, 1, 2*10^n, ...] has period 6. As n increases, the expansion has the large partial quotients 2/3*(10^n - 1) and 2*10^n.
For n >= 1, the continued fraction expansion of sqrt(a(2*n))/a(n) = [0; 1, 10^n - 1, 1, 1, 1/3*(10^n - 4), 1, 4, 1, 1/3*(10^n - 4), 1, 1, 10^n - 1, 2, 10^n - 1, ...] has pre-period of length 3 and period 12 beginning 1, 1, 1/3*(10^n - 4), ....  As n increases, the expansion has the large partial quotients 10^n - 1 and 1/3*(10^n - 4).
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), for n >= 2, begins [10^(2*n); 3*10^n - 1, 1, 0.5*10^(2*n) - 1, 1.44*10^n - 1, 1, ...]. Cf. A000533, A002283 and A168624. (End)

			From _Peter Bala_, Sep 25 2015: (Start)
Simple continued fraction expansions showing large partial quotients:
a(9)^(1/3) =[1000000; 2999, 1, 499999, 1439, 1, 2582643, 1, 1, 1, 2, 3, 3, ...].
a(20)^(1/4) = [10000000000; 39999999999, 1, 3999999999, 16949152542, 2, 1, 2, 6, 1, 4872106, 3, 9, 2, 3, ...].
a(25)^(1/5) = [10000000000; 4999999999999999, 1, 3333333332, 2, 1, 217391304347825, 2, 2, 1, 1, 1, 2, 1, 23980814, 1, 1, 1, 1, 1, 7, ...]. (End)

Harry J. Smith, Table of n, a(n) for n=0..100
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A000533, A002283, A033934, A062397, A074992, A135577, A168624.

[10^(2*n) + 10^n + 1: n in [0..20]]; // Vincenzo Librandi, Sep 27 2015

Table[10^(2 n) + 10^n + 1, {n, 0, 15}] (* Michael De Vlieger, Sep 27 2015 *)
CoefficientList[Series[(3 - 222 x + 1110 x^2)/((1 - 100 x) (1 - 10 x) (1 - x)), {x, 0, 33}], x] (* Vincenzo Librandi, Sep 27 2015 *)

a(n) = { 10^(2*n) + 10^n + 1 } \\ Harry J. Smith, Feb 02 2010

Vec(-3*(370*x^2-74*x+1)/((x-1)*(10*x-1)*(100*x-1)) + O(x^20)) \\ Colin Barker, Sep 27 2015

A168624(n) = a(2*n)/a(n). - Peter Bala, Sep 24 2015
G.f.: (3 - 222*x + 1110*x^2)/((1 - 100*x)*(1 - 10*x)*(1 - x)). - Vincenzo Librandi, Sep 27 2015
From Colin Barker, Sep 27 2015: (Start)
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.
G.f.: -3*(370*x^2-74*x+1)/((x-1)*(10*x-1)*(100*x-1)). (End)
From Elmo R. Oliveira, Aug 27 2024: (Start)
E.g.f.: exp(x)*(exp(99*x) + exp(9*x) + 1).
a(n) = 3*A074992(n). (End)

Offset changed from 1 to 0 by Harry J. Smith, Feb 02 2010

More terms from Michael De Vlieger, Sep 27 2015

A054215 Squares that are concatenations of two consecutive decreasing numbers.

Original entry on oeis.org

8281, 82428241, 98029801, 538277538276, 998002998001, 7783702677837025, 9998000299980001, 79225472657922547264, 86432513458643251344, 92237976109223797609, 99998000029999800001, 106710893290106710893289

Offset: 1

Patrick De Geest, Feb 15 2000

Infinitely many terms of this sequence are provided by A168624(k)^2 for k>0. - Bruno Berselli, Mar 13 2018

			E.g. '8242' + '8242-1' gives 82428241 which is 9079^2.

Luca, Florian, and Pantelimon Stănică. "Perfect Squares as Concatenation of Consecutive Integers." The American Mathematical Monthly 126.8 (2019): 728-734.

Cf. A054214, A054216, A030465, A030466, A030467, A020339, A020340.

a(n) = concatenation of A054214(n) and A054214(n)-1. - Max Alekseyev, May 14 2007

More terms from Max Alekseyev, May 14 2007

82848241 corrected to 82428241 by Dominick Cancilla, Jul 21 2010

A243162 Numbers n such that n^2 divides n.n.n where dot "." means concatenation.

Original entry on oeis.org

1, 3, 13, 21, 37, 39, 91, 1443, 3367, 9901, 157737, 333667, 999001, 3075403, 9226209, 14287143, 33336667, 99990001, 1171182883, 1224848037, 1286294191, 1397863441, 1428557143, 1469179621, 1535254357, 1568996211, 1753536967, 1792076241, 1839599913, 1891910811

Offset: 1

Hans Havermann, May 31 2014

Number of d-digit solutions for d = 1..100: 2, 5, 0, 3, 0, 3, 2, 3, 0, 39, 0, 2, 0, 106, 0, 3, 3, 2, 0, 441, 4, 14, 0, 5, 0, 15, 2, 283, 0, 23, 0, 61, 0, 24, 21, 4, 0, 22, 0, 240, 0, 34, 0, 96, 3, 30, 0, 6, 16, 281, 0, 216, 0, 22, 5, 3894, 2, 10, 0, 149, 2, 11, 0, 407, 0, 25, 0, 2136, 0, 53983, 0, 12, 1, 29, 11, 1872, 99, 20, 0, 6984, 0, 45, 0, 279, 32, 10, 5, 15928, 0, 213, 24, 791, 0, 20, 14, 44, 0, 713, 12, 89804.

Numbers n such that n divides 100^d+10^d+1, where 10^(d-1)<=n<10^d. - Robert Israel, Jan 11 2017

			21^2 divides 212121; 91^2 divides 919191; so both 21 and 91 are in the sequence.

Hans Havermann, Table of n, a(n) for n = 1..1366

Cf. A147553 (n^2 divides n.n), A147554 (primes in this sequence).

Contains A074992 and A168624.

Res:= {}:
for d from 1 to 15 do
  Res:= Res union select(t -> t >= 10^(d-1) and t < 10^d,
   numtheory:-divisors(100^d+10^d+1))
od:
sort(convert(Res,list)); # Robert Israel, Jan 11 2017

Do[d=Divisors[100^i+10^i+1];s=Select[d,Length[IntegerDigits[#]]==i&];If[Length[s]>0,Do[Print[s[[j]]],{j,Length[s]}]],{i,42}]

A276352 a(n) = 100^n - 10^n.

Original entry on oeis.org

0, 90, 9900, 999000, 99990000, 9999900000, 999999000000, 99999990000000, 9999999900000000, 999999999000000000, 99999999990000000000, 9999999999900000000000, 999999999999000000000000, 99999999999990000000000000, 9999999999999900000000000000, 999999999999999000000000000000

Offset: 0

Carauleanu Marc, Aug 31 2016

Index entries for linear recurrences with constant coefficients, signature (110,-1000).

Cf. A002283, A011557, A138147, A168624.

[100^n-10^n : n in [0..20]]; // Wesley Ivan Hurt, Sep 07 2016

A276352:=n->100^n-10^n: seq(A276352(n), n=0..20); # Wesley Ivan Hurt, Sep 07 2016

Table[100^n - 10^n, {n, 0, 10}] (* Alonso del Arte, Aug 31 2016 *)

a(n)=100^n-10^n \\ Charles R Greathouse IV, Sep 01 2016

a(n) = floor((1000^n)/(10^n + 1)).
a(n) = A002283(n)*A011557(n).
a(n) = 9*A138147(n), for n>0.
a(n) = A168624(n) - 1.
a(n) = Sum_{k=1..n} 9*10^(2n-k).
a(n) = ((10^n)*A002283(2n))/(10^n + 1).
From Chai Wah Wu, Sep 01 2016: (Start)
a(n) = 110*a(n-1) - 1000*a(n-2) for n > 1.
G.f.: 90*x/((10*x - 1)*(100*x - 1)). (End)
E.g.f.: exp(10*x)*(exp(90*x) - 1). - Elmo R. Oliveira, Aug 14 2024

cp's OEIS Frontend

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

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

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A066138 a(n) = 10^(2*n) + 10^n + 1.

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A054215 Squares that are concatenations of two consecutive decreasing numbers.

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A243162 Numbers n such that n^2 divides n.n.n where dot "." means concatenation.

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A276352 a(n) = 100^n - 10^n.

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A187868 Primes of the form 100^k - 10^k + 1.