A066138 a(n) = 10^(2*n) + 10^n + 1.

3, 111, 10101, 1001001, 100010001, 10000100001, 1000001000001, 100000010000001, 10000000100000001, 1000000001000000001, 100000000010000000001, 10000000000100000000001, 1000000000001000000000001, 100000000000010000000000001, 10000000000000100000000000001, 1000000000000001000000000000001

Offset: 0

Author

Henry Bottomley, Dec 07 2001

Keywords

• nonn
• base
• easy

Comments

Palindromes whose digit sum is 3.

Essentially the same as A135577. - R. J. Mathar Apr 29 2008

From Peter Bala, Sep 25 2015: (Start)

For n >= 1, the simple continued fraction expansion of sqrt(a(n)) = [10^n; 1, 1, 2/3*(10^n - 1), 1, 1, 2*10^n, ...] has period 6. As n increases, the expansion has the large partial quotients 2/3*(10^n - 1) and 2*10^n.

For n >= 1, the continued fraction expansion of sqrt(a(2*n))/a(n) = [0; 1, 10^n - 1, 1, 1, 1/3*(10^n - 4), 1, 4, 1, 1/3*(10^n - 4), 1, 1, 10^n - 1, 2, 10^n - 1, ...] has pre-period of length 3 and period 12 beginning 1, 1, 1/3*(10^n - 4), .... As n increases, the expansion has the large partial quotients 10^n - 1 and 1/3*(10^n - 4).

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), for n >= 2, begins [10^(2*n); 3*10^n - 1, 1, 0.5*10^(2*n) - 1, 1.44*10^n - 1, 1, ...]. Cf. A000533, A002283 and A168624. (End)

Examples

			From _Peter Bala_, Sep 25 2015: (Start)
Simple continued fraction expansions showing large partial quotients:
a(9)^(1/3) =[1000000; 2999, 1, 499999, 1439, 1, 2582643, 1, 1, 1, 2, 3, 3, ...].
a(20)^(1/4) = [10000000000; 39999999999, 1, 3999999999, 16949152542, 2, 1, 2, 6, 1, 4872106, 3, 9, 2, 3, ...].
a(25)^(1/5) = [10000000000; 4999999999999999, 1, 3333333332, 2, 1, 217391304347825, 2, 2, 1, 1, 1, 2, 1, 23980814, 1, 1, 1, 1, 1, 7, ...]. (End)
		

Links

• Harry J. Smith, Table of n, a(n) for n=0..100
• Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Crossrefs

Cf. A000533, A002283, A033934, A062397, A074992, A135577, A168624.

Programs

Magma

[10^(2*n) + 10^n + 1: n in [0..20]]; // Vincenzo Librandi, Sep 27 2015

Mathematica

Table[10^(2 n) + 10^n + 1, {n, 0, 15}] (* Michael De Vlieger, Sep 27 2015 *)
CoefficientList[Series[(3 - 222 x + 1110 x^2)/((1 - 100 x) (1 - 10 x) (1 - x)), {x, 0, 33}], x] (* Vincenzo Librandi, Sep 27 2015 *)

PARI

a(n) = { 10^(2*n) + 10^n + 1 } \\ Harry J. Smith, Feb 02 2010

PARI

Vec(-3*(370*x^2-74*x+1)/((x-1)*(10*x-1)*(100*x-1)) + O(x^20)) \\ Colin Barker, Sep 27 2015

Formula

A168624(n) = a(2*n)/a(n). - Peter Bala, Sep 24 2015

G.f.: (3 - 222*x + 1110*x^2)/((1 - 100*x)*(1 - 10*x)*(1 - x)). - Vincenzo Librandi, 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.: -3*(370*x^2-74*x+1)/((x-1)*(10*x-1)*(100*x-1)). (End)

From Elmo R. Oliveira, Aug 27 2024: (Start)

E.g.f.: exp(x)*(exp(99*x) + exp(9*x) + 1).

a(n) = 3*A074992(n). (End)

Extensions

Offset changed from 1 to 0 by Harry J. Smith, Feb 02 2010

More terms from Michael De Vlieger, Sep 27 2015