

A000533


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


270



1, 11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001, 100000000001, 1000000000001, 10000000000001, 100000000000001, 1000000000000001, 10000000000000001
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OFFSET

0,2


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
a(n) = A178500(n) + 1.  Reinhard Zumkeller, May 28 2010
It seems that the sequence gives 'all' positive integers m with m^4 is 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 mth 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


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)


FORMULA

a(n) = 10^n + 1  0^n.  Reinhard Zumkeller, Jun 10 2003
From Paul Barry, Feb 05 2005: (Start)
G.f.: (110*x^2)/((1x)*(110*x));
a(n) = Sum_{k=0..n} binomial(n, k)*0^(k(nk))*10^k. (End)
E.g.f.: exp(x) + exp(10*x)  1.  Ilya Gutkovskiy, Jun 03 2016


EXAMPLE

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


MATHEMATICA

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


PROG

(MAGMA) [10^n + 1  0^n: n in [0..30]]; // Vincenzo Librandi, Jul 15 2011
(PARI) a(n)=if(n, 10^n+1, 1) \\ Charles R Greathouse IV, Oct 28 2014


CROSSREFS

Cf. A083318, A138144, A138145, A152756, A002283, A066138, A168624.
Sequence in context: A056810 A116098 A116129 * A147759 A147757 A089183
Adjacent sequences: A000530 A000531 A000532 * A000534 A000535 A000536


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane


STATUS

approved



