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

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

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

Examples

			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
		

Links

• 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).

Crossrefs

Cf. A002283, A066138, A083318, A138144, A138145, A152756, A168624.

Programs

Magma

[10^n + 1 - 0^n: n in [0..30]]; // Vincenzo Librandi, Jul 15 2011

Mathematica

Join[{1},LinearRecurrence[{11,-10},{11,101},20]] (* Harvey P. Dale, May 01 2014 *)

PARI

a(n)=if(n,10^n+1,1) \\ Charles R Greathouse IV, Oct 28 2014

Formula

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