A152756 -id:A152756 - cp's OEIS frontend

A153500 First 3 terms coincide with A152756. For n>3, a(n) is the palindromic number formed from concatenation of 1, 0, A147759(n-3), 0, A147759(n-3), 0 and 1.

1, 101, 10001, 1010101, 101101101, 10101010101, 1010010100101, 101010101010101, 10101101010110101, 1010101010101010101, 101010010101010010101, 10101010101010101010101, 1010101101010101011010101, 101010101010101010101010101, 10101010010101010101001010101

Offset: 1

Omar E. Pol, Dec 27 2008, Feb 18 2009

a(n) is also A153499(n) written in base 2.

			n ............ a(n)
1 ............. 1
2 ............ 101
3 ........... 10001
4 .......... 1010101
5 ......... 101101101
6 ........ 10101010101
7 ....... 1010010100101
8 ...... 101010101010101
9 ..... 10101101010110101
10 ... 1010101010101010101
======================================
Another visualization of the structure
======================================
1 ............. *
2 ............ /.\
3 ........... /...\
4 .......... /.*.*.\
5 ......... /./|.|\.\
6 ........ /./.|.|.\.\
7 ....... /./..|.|..\.\
8 ...... /./.*.|.|.*.\.\
9 ..... /././|.|.|.|\.\.\
10 ... /././.|.|.|.|.\.\.\

Index entries for linear recurrences with constant coefficients, signature (101,-1110,102010,-111000,1010000,-1000000).

Cf. A135577, A138146, A152576, A152764, A153497, A153498, A153499, A144564.

a(n) = 101*a(n-1)-1110*a(n-2)+102010*a(n-3)-111000*a(n-4)+1010000*a(n-5)-1000000*a(n-6), n>7. [R. J. Mathar, Feb 20 2009]

G.f.: -x*(1000000*x^6-1010000*x^5+10000*x^4-10100*x^3-910*x^2-1) / ((x-1)*(100*x-1)*(10*x^2+1)*(1000*x^2+1)). [Colin Barker, Sep 17 2013]

More terms from R. J. Mathar, Feb 20 2009
Keyword:base added by Charles R Greathouse IV, Apr 26 2010
More terms from Colin Barker, Sep 17 2013

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

A144564 Bisection of A147757.

Original entry on oeis.org

1, 101, 10101, 1011101, 101111101, 10111111101, 1011111111101, 101111111111101, 10111111111111101, 1011111111111111101, 101111111111111111101, 10111111111111111111101, 1011111111111111111111101, 101111111111111111111111101, 10111111111111111111111111101

Offset: 1

Omar E. Pol, Dec 14 2008

			n ...... a(n)
1 ....... 1
2 ...... 101
3 ..... 10101
4 .... 1011101
5 ... 101111101

Index entries for linear recurrences with constant coefficients, signature (101,-100).

Cf. A135577, A138120, A138146, A147757, A152756.

Rest[CoefficientList[Series[x(1+10x)(100x^2-10x+1)/((100x-1)(x-1)),{x,0,20}],x]] (* or *) Join[{1,101},Table[FromDigits[Join[{1,0},PadRight[ {},2n+1,1],{0,1}]],{n,0,20}]] (* Harvey P. Dale, Dec 26 2014 *)

G.f.: x*(1+10*x)*(100*x^2-10*x+1)/((100*x-1)*(x-1)). - R. J. Mathar, Aug 24 2011

A152764 Bisection of A138144.

Original entry on oeis.org

1, 111, 11011, 1100011, 110000011, 11000000011, 1100000000011, 110000000000011, 11000000000000011, 1100000000000000011, 110000000000000000011, 11000000000000000000011

Offset: 1

Omar E. Pol, Dec 14 2008

			n ...... a(n)
1 ....... 1
2 ...... 111
3 ..... 11011
4 .... 1100011
5 ... 110000011

Index entries for linear recurrences with constant coefficients, signature (101,-100).

Cf. A135577, A138120, A138146, A152756, A152758, A058764.

LinearRecurrence[{101,-100},{1,111,11011,1100011},20] (* Harvey P. Dale, Nov 26 2019 *)

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

From Colin Barker, Sep 16 2013: (Start)
a(n) = 11+11*10^(2*n-3) for n>2.
a(n) = 101*a(n-1)-100*a(n-2) for n>4.
G.f.: -x*(10*x-1)*(10*x+1)^2 / ((x-1)*(100*x-1)). (End)

A271527 a(n) = 1000^n + 1.

Original entry on oeis.org

2, 1001, 1000001, 1000000001, 1000000000001, 1000000000000001, 1000000000000000001, 1000000000000000000001, 1000000000000000000000001, 1000000000000000000000000001, 1000000000000000000000000000001, 1000000000000000000000000000000001

Offset: 0

Ilya Gutkovskiy, Apr 09 2016

All terms in this sequence are palindromes (A002113).
Also, A062395 written in base 2 (see example).
a(n) minus one gives the number of nodes at n-th level of a 1000-ary tree.
More generally, the ordinary generating function for sequences of the form k^n + m, is (1 + m - (1 + k*m)*x)/((1 - x)*(1 - k*x)), and the exponential generating function is exp(k*x) + m*exp(x).

			a(n), n>0, is the binary representation of A062395(n)
n  ------------------------------------------
0  2........................................2
1  1001.....................................9
2  1000001.................................65
3  1000000001.............................513
4  1000000000001.........................4097
5  1000000000000001.....................32769
6  1000000000000000001.................262145
7  1000000000000000000001.............2097153
8  1000000000000000000000001.........16777217
9  1000000000000000000000000001.....134217729

Ilya Gutkovskiy, Examples of the ordinary generating function for the sequences of the form k^n + m
Index entries for linear recurrences with constant coefficients, signature (1001,-1000)

Cf. A000035, A000533, A007088, A007395, A007953, A057427, A060365, A062395, A152756.

Table[1000^n + 1, {n, 0, 11}]
LinearRecurrence[{1001, -1000}, {2, 1001}, 12]

x='x+O('x^99); Vec((2-1001*x)/((1-x)*(1-1000*x))) \\ Altug Alkan, Apr 09 2016

for n in range(0,10**4):print(1000**n+1)
# Soumil Mandal, Apr 10 2016

G.f.: (2 - 1001*x)/((1 - x)*(1 - 1000*x)).
E.g.f.: exp(1000*x) + exp(x).
a(n) = 1001*a(n-1) - 1000*a(n-2).
a(n) = A060365(n) + 1.
a(n) = A000533(3n), n>0.
a(n) = A007088(A062395(n)).
A007953(a(n)) = A007395(n).
A000035(a(n)) = A057427(n).
Sum_{n>=0} 1/a(n) = 0.501000001999002...
Lim_{n->infinity} a(n + 1)/a(n) = 1000.

cp's OEIS Frontend

A153500 First 3 terms coincide with A152756. For n>3, a(n) is the palindromic number formed from concatenation of 1, 0, A147759(n-3), 0, A147759(n-3), 0 and 1.

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

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A144564 Bisection of A147757.

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A152764 Bisection of A138144.

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A271527 a(n) = 1000^n + 1.

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