%I #113 Sep 08 2022 08:44:28

%S 1,11,101,1001,10001,100001,1000001,10000001,100000001,1000000001,

%T 10000000001,100000000001,1000000000001,10000000000001,

%U 100000000000001,1000000000000001,10000000000000001

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

%C Also, b^n+1 written in base b, for any base b >= 2.

%C Also, A083318 written in base 2. - _Omar E. Pol_, Feb 24 2008, Dec 30 2008

%C Also, palindromes formed from the reflected decimal expansion of the concatenation of 1 and infinite 0's. - _Omar E. Pol_, Dec 14 2008

%C a(n) = A178500(n) + 1. - _Reinhard Zumkeller_, May 28 2010

%C 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

%C a(n)^2 starts with a(n)+1 for n >= 1. - _Dhilan Lahoti_, Aug 31 2015

%C From _Peter Bala_, Sep 25 2015: (Start)

%C The simple continued fraction expansion of sqrt(a(2*n)) = [10^n; 2*10^n, 2*10^n, ...] has period 1.

%C 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.

%C 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.

%C As n increases, these expansions have large partial quotients.

%C A theorem of Kuzmin in the measure theory of continued fractions says that large partial quotients are the exception in continued fraction expansions.

%C 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.

%C (End)

%C a(1) and a(2) are the only prime terms up to n=100000. - _Daniel Arribas_, Jun 04 2016

%C 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

%C 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

%H Vincenzo Librandi, <a href="/A000533/b000533.txt">Table of n, a(n) for n = 0..100</a>

%H John Rafael M. Antalan, <a href="https://arxiv.org/abs/1908.06014">A Recreational Application of Two Integer Sequences and the Generalized Repetitious Number Puzzle</a>, arXiv:1908.06014 [math.HO], 2019.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-10)

%F a(n) = 10^n + 1 - 0^n. - _Reinhard Zumkeller_, Jun 10 2003

%F From _Paul Barry_, Feb 05 2005: (Start)

%F G.f.: (1-10*x^2)/((1-x)*(1-10*x));

%F a(n) = Sum_{k=0..n} binomial(n, k)*0^(k(n-k))*10^k. (End)

%F E.g.f.: exp(x) + exp(10*x) - 1. - _Ilya Gutkovskiy_, Jun 03 2016

%e 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

%t Join[{1},LinearRecurrence[{11,-10},{11,101},20]] (* _Harvey P. Dale_, May 01 2014 *)

%o (Magma) [10^n + 1 - 0^n: n in [0..30]]; // _Vincenzo Librandi_, Jul 15 2011

%o (PARI) a(n)=if(n,10^n+1,1) \\ _Charles R Greathouse IV_, Oct 28 2014

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

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_