A066138 -id:A066138 - 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

A074992 a(n) = (10^(2*n) + 10^n + 1)/3.

Original entry on oeis.org

1, 37, 3367, 333667, 33336667, 3333366667, 333333666667, 33333336666667, 3333333366666667, 333333333666666667, 33333333336666666667, 3333333333366666666667, 333333333333666666666667, 33333333333336666666666667, 3333333333333366666666666667, 333333333333333666666666666667

Offset: 0

Amarnath Murthy, Aug 31 2002

Apart from the initial 1, common difference of the arithmetic progression pertaining to the sequence A074991.
This is also a root sequence pertaining to the patterned perfect square sequence 1369, 11336689,111333666889,... i.e., k ones, k threes and k sixes followed by (k-1) 8's and a 9. (37^2 = 1369).
This is a self-complementing sequence: each term has even number of digits (the first one has to be read 01, the leading zero is important). If you add the first half to the second half of any term, you get the sequence A011557, powers of 10. Furthermore, the reciprocals of the sequence terms, except the first one, give a sequence of periodic terms with period sequence as in A008585, a(n) = 3*n, and value given by A086574, a(n)=k where R(k+3)=3. - Rodolfo A. Fiorini, Jul 14 2016

Rodolfo A. Fiorini, Computerized tomography noise reduction by CICT optimized exponential cyclic sequences (OECS) co-domain, Fundamenta Informaticae, 141(2015), pp. 115-134.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A008585, A011557, A066138, A074991, A086574, A274766.

A074992 := proc(n)
    (10^(2*n)+10^n+1)/3 ;
end proc:
seq(A074992(n),n=0..15) ; # R. J. Mathar, May 06 2017

{01}~Join~Table[FromDigits@ Flatten@ Map[IntegerDigits, {#, 10^n - #}] &@ Floor[10^n/3], {n, 12}] (* Michael De Vlieger, Jul 22 2016 *)

a(n) = (10^(2*n) + 10^n + 1)/3; \\ Michel Marcus, Sep 14 2013

Vec(-x*(1000*x^2-740*x+37)/((x-1)*(10*x-1)*(100*x-1)) + O(x^100)) \\ Colin Barker, Sep 23 2013

a(n)=my(x=10^n); (x^2+x+1)/3 \\ Charles R Greathouse IV, Jul 22 2016

a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3), for n > 2, a(0)=1, a(1)=37, a(2)=3367.
G.f.: (1 - 74*x + 370*x^2)/((1-x)*(1-10*x)*(1-100*x)). - Colin Barker, Sep 23 2013 and Robert Israel, Jul 22 2016
From Elmo R. Oliveira, Sep 12 2024: (Start)
E.g.f.: exp(x)*(exp(99*x) + exp(9*x) + 1)/3.
a(n) = A066138(n)/3. (End)

Entry revised (new definition, new offset, new initial term, etc.) by N. J. A. Sloane, Jul 27 2016 (Some of the old programs may need slight modifications.)

A135577 Numbers that have only the digit "1" as first, central and final digit. For numbers with 5 or more digits the rest of digits are "0".

Original entry on oeis.org

1, 111, 10101, 1001001, 100010001, 10000100001, 1000001000001, 100000010000001, 10000000100000001, 1000000001000000001, 100000000010000000001, 10000000000100000000001, 1000000000001000000000001, 100000000000010000000000001, 10000000000000100000000000001

Offset: 1

Omar E. Pol, Feb 24 2008

Also, equal to A135576(n), written in base 2.
Essentially the same as A066138. - R. J. Mathar Apr 29 2008
a(n) has 2n-1 digits.

			----------------------------
n ............ a(n)
----------------------------
1 ............. 1
2 ............ 111
3 ........... 10101
4 .......... 1001001
5 ......... 100010001
6 ........ 10000100001
7 ....... 1000001000001
8 ...... 100000010000001
9 ..... 10000000100000001
10 ... 1000000001000000001

G. C. Greubel, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A001576, A066138, A135576, A138118, A138119, A138120, A138147, A138826.

Join[{1}, LinearRecurrence[{111, -1110, 1000}, {111, 10101, 1001001}, 25]] (* G. C. Greubel, Oct 19 2016 *)
Join[{1},Table[FromDigits[Join[{1},PadRight[{},n,0],{1},PadRight[{},n,0],{1}]],{n,0,10}]] (* Harvey P. Dale, Aug 15 2022 *)

Vec(-x*(2000*x^3-1110*x^2+1)/((x-1)*(10*x-1)*(100*x-1))  + O(x^100)) \\ Colin Barker, Sep 16 2013

a(n) = A135576(n), written in base 2.
Also, a(1)=1, for n>1; a(n)=(concatenation of 1, n-2 digits 0, 1, n-2 digits 0 and 1).
From Colin Barker, Sep 16 2013: (Start)
a(n) = 1 + 10^(n-1) + 100^(n-1) for n>1.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>4.
G.f.: x*(2000*x^3 - 1110*x^2 + 1)/((1-x)*(10*x-1)*(100*x-1)). (End)
E.g.f.: (-111 - 200*x + 100*exp(x) + 10*exp(10*x) + exp(100*x))/100. - Elmo R. Oliveira, Jun 13 2025

A033934 a(n) = (10^n + 1)^2.

Original entry on oeis.org

4, 121, 10201, 1002001, 100020001, 10000200001, 1000002000001, 100000020000001, 10000000200000001, 1000000002000000001, 100000000020000000001, 10000000000200000000001, 1000000000002000000000001, 100000000000020000000000001, 10000000000000200000000000001

Offset: 0

Jeff Burch

The members of this sequence are both perfect squares and palindromes. Therefore A002779 is an infinite sequence. - Ant King, Jun 26 2011

Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A002779 (palindromic squares), A000290 (squares), A002113 (palindromes).

Cf. A011557, A062397, A066138.

(10^Range[0,20]+1)^2 (* or *) LinearRecurrence[{111,-1110,1000},{4,121,10201},20] (* Harvey P. Dale, Feb 16 2016 *)

my(x='x+O('x^15)); Vec((1210*x^2-323*x+4)/(-1000*x^3+1110*x^2-111*x+1)) \\ Elmo R. Oliveira, Jul 04 2025

a(n) = A062397(n)^2 = A066138(n) + A011557(n).
From Elmo R. Oliveira, Jul 04 2025: (Start)
G.f.: (4 - 323*x + 1210*x^2)/((1-x)*(1-10*x)*(1-100*x)).
E.g.f.: exp(x)*(1 + 2*exp(9*x) + exp(99*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3). (End)

Better description from Henry Bottomley, Dec 07 2001

More terms from Harvey P. Dale, Feb 16 2016

A168624 a(n) = 1 - 10^n + 100^n.

Original entry on oeis.org

1, 91, 9901, 999001, 99990001, 9999900001, 999999000001, 99999990000001, 9999999900000001, 999999999000000001, 99999999990000000001, 9999999999900000000001, 999999999999000000000001, 99999999999990000000000001, 9999999999999900000000000001, 999999999999999000000000000001

Offset: 0

Jason Earls, Dec 01 2009

Prime values for n = 2,4,6,8, with no others up to n = 3400. Beiler mentions this pattern in the reference.
From Peter Bala, Sep 27 2015: (Start)
Calculation suggests the continued fraction expansion of sqrt(a(n)), for n >= 1, begins [10^n - 1, 1, 1, 1/3*(2*10^n - 5), 1, 5, 1/9*(2*10^n - 11), 1, 17, (2*10^n - 20 - 9*(1 - MOD(n, 3)))/27, ...]. Note the large partial quotients early in the expansion.
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 = 2, 3, 4,... as n increases. Some examples are given below. Cf. A000533, A002283, A066138. (End)

			Simple continued fraction expansions showing large partial quotients:
sqrt(a(10)) = [9999999999; 1, 1, 6666666665, 1, 5, 2222222221, 1, 17, 740740740, 1, 1, 1, 5, 2, 1, 246913579, 1, 1, 4, 1, 1, 3, 1, 1, ...].
a(18)^(1/3) = [999999999999; 1, 2999999, 499999999999, 1, 1439999, 2582644628099, 5, 1, 3, 4, 1, 58, 1, 1, 1, 8, ...].
a(30)^(1/5) = [999999999999; 1, 4999999999999999999, 333333333333, 3, 217391304347826086, 1, 1, 1, 1, 1, 8, 2398081534, 1, 1, 1, 9, 1, 98, 1, 125052522059263, 1, 9, 7, 1, ...]. - _Peter Bala_, Sep 27 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 85.

Colin Barker, Table of n, a(n) for n = 0..499
P. Bala, A168624 and some empirical continued fraction expansions.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A000533, A002283, A066138, A187868.

Table[1-10^n+100^n,{n,0,20}] (* Harvey P. Dale, Dec 01 2013 *)

Vec(-(910*x^2-20*x+1)/((x-1)*(10*x-1)*(100*x-1)) + O(x^20)) \\ Colin Barker, 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.: -(910*x^2-20*x+1)/((x-1)*(10*x-1)*(100*x-1)). (End)
E.g.f.: exp(x)*(exp(99*x) - exp(9*x) + 1). - Elmo R. Oliveira, Sep 12 2024

A190839 a(n) is the greatest prime divisor of 10^(2*n)+10^n+1.

Original entry on oeis.org

37, 37, 333667, 9901, 2906161, 333667, 10838689, 99990001, 440334654777631, 2906161, 1344628210313298373, 999999000001, 900900900900990990990991, 10838689, 4185502830133110721, 9999999900000001, 13168164561429877, 440334654777631, 3931123022305129377976519, 39526741

Offset: 1

Vladimir Shevelev, May 21 2011

(10^(2*n)+10^n+1)^3 is palindrome.

Amiram Eldar, Table of n, a(n) for n = 1..158
factordb, 10^(2*n)+10^n+1.

Cf. A002780, A006530, A066138.

f[n_]:=Max[Transpose[FactorInteger[10^(2n)+10^n+1]][[1]]]; Array[f,20] (* Harvey P. Dale, Jun 10 2011 *)

u(n)=10^(2*n)+10^n+1; for(n=1,30,f=factor(u(n));print1(f[matsize(f)[1],1],", ")) \\ Joerg Arndt, May 27 2011

a(n) = A006530(A066138(n)). - R. J. Mathar, Oct 30 2015

A225810 a(n) = (10^n)^2 + 4*(10^n) + 1.

Original entry on oeis.org

6, 141, 10401, 1004001, 100040001, 10000400001, 1000004000001, 100000040000001, 10000000400000001, 1000000004000000001, 100000000040000000001, 10000000000400000000001, 1000000000004000000000001, 100000000000040000000000001, 10000000000000400000000000001

Offset: 0

Lance J. Weingartz, Jul 29 2013

This is an instance of (10^n)^2 + x(10^n) + 1 which umbrellas A066138, A033934, A171375, A171410, A171461, A171513 and A171553 which all produce palindromes of the form 1...n...1 when n <> 0.

Colin Barker, Table of n, a(n) for n = 0..499
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A033934, A066138, A171375, A171410, A171461, A171513, A171553.

Table[(10^n)^2 + 4*(10^n) + 1, {n, 0, 20}] (* T. D. Noe, Aug 12 2013 *)
LinearRecurrence[{111,-1110,1000},{6,141,10401},20] (* Harvey P. Dale, Oct 28 2017 *)

Vec(-3*(470*x^2-175*x+2)/((x-1)*(10*x-1)*(100*x-1)) + O(x^100)) \\ Colin Barker, Apr 27 2015

From Colin Barker, Apr 27 2015: (Start)
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
G.f.: -3*(470*x^2-175*x+2)/((x-1)*(10*x-1)*(100*x-1)). (End)
E.g.f.: exp(x)*(1 + 4*exp(9*x) + exp(99*x)). - Elmo R. Oliveira, Jul 04 2025

A225813 a(n) = (10^n)^2 + 7*(10^n) + 1.

Original entry on oeis.org

9, 171, 10701, 1007001, 100070001, 10000700001, 1000007000001, 100000070000001, 10000000700000001, 1000000007000000001, 100000000070000000001, 10000000000700000000001, 1000000000007000000000001, 100000000000070000000000001, 10000000000000700000000000001

Offset: 0

Lance J. Weingartz, Jul 29 2013

This is an instance of (10^n)^2 + x(10^n) + 1 which umbrellas A066138, A033934, A171375, A171410, A171461, A171513 and A171553 which all produce palindromes of the form 1...x...1 when n <> 0.

Colin Barker, Table of n, a(n) for n = 0..499
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A033934, A066138, A171375, A171410, A171461, A171513, A171553.

A225813:=n->(10^n)^2 + 7*(10^n) + 1: seq(A225813(n), n=0..20); # Wesley Ivan Hurt, Apr 08 2017

Table[(10^n)^2 + 7*(10^n) + 1, {n, 0, 20}] (* T. D. Noe, Aug 12 2013 *)
LinearRecurrence[{111,-1110,1000},{9,171,10701},20] (* Harvey P. Dale, Apr 12 2020 *)

Vec(-9*(190*x^2-92*x+1)/((x-1)*(10*x-1)*(100*x-1)) + O(x^100)) \\ Colin Barker, Apr 27 2015

From Colin Barker, Apr 27 2015: (Start)
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
G.f.: -9*(190*x^2-92*x+1)/((x-1)*(10*x-1)*(100*x-1)). (End)
E.g.f.: exp(x)*(1 + 7*exp(9*x) + exp(99*x)). - Elmo R. Oliveira, Jul 04 2025

A344423 a(n) = 10^(2n+2) + 11110^n + 1.

Original entry on oeis.org

212, 11111, 1011101, 100111001, 10001110001, 1000011100001, 100000111000001, 10000001110000001, 1000000011100000001, 100000000111000000001, 10000000001110000000001, 1000000000011100000000001, 100000000000111000000000001, 10000000000001110000000000001

Offset: 0

Felix Fröhlich, May 18 2021

For n > 1, palindromic numbers of the form 10..01110..01.
This is the earliest sequence of the form 10^(2*n+t) + A002275(t+1)*10^n + 1 that contains primes of the form mentioned in the previous comment. For example, the terms of the sequence for t = 0 are all divisible by 3 (see A066138, where 3 is the only prime), while each term b(i) of the sequence with t = 1 (A319667) is divisible by 10^i+1.
For the values of n such that a(n) is prime, see A344424.

Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A002275, A066138, A319667, A344424.

PARI
```
a(n) = 10^(2*n+2) + 111*10^n + 1
```

G.f.: -(13100*x^2 - 12421*x + 212)/(1000*x^3 - 1110*x^2 + 111*x - 1). - Jinyuan Wang, May 22 2021
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3). - Wesley Ivan Hurt, May 22 2021
E.g.f.: exp(x)*(1 + 111*exp(9*x) + 100*exp(99*x)). - Stefano Spezia, May 22 2021

cp's OEIS Frontend

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

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Magma

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

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A074992 a(n) = (10^(2*n) + 10^n + 1)/3.

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A135577 Numbers that have only the digit "1" as first, central and final digit. For numbers with 5 or more digits the rest of digits are "0".

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A033934 a(n) = (10^n + 1)^2.

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A168624 a(n) = 1 - 10^n + 100^n.

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A344423 a(n) = 10^(2n+2) + 11110^n + 1.