A066138 - OEIS

%I #31 Dec 08 2024 20:41:26

%S 3,111,10101,1001001,100010001,10000100001,1000001000001,

%T 100000010000001,10000000100000001,1000000001000000001,

%U 100000000010000000001,10000000000100000000001,1000000000001000000000001,100000000000010000000000001,10000000000000100000000000001,1000000000000001000000000000001

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

%C Palindromes whose digit sum is 3.

%C Essentially the same as A135577. - _R. J. Mathar_ Apr 29 2008

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

%C 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.

%C 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).

%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), 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)

%H Harry J. Smith, <a href="/A066138/b066138.txt">Table of n, a(n) for n=0..100</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (111,-1110,1000).

%F A168624(n) = a(2*n)/a(n). - _Peter Bala_, Sep 24 2015

%F G.f.: (3 - 222*x + 1110*x^2)/((1 - 100*x)*(1 - 10*x)*(1 - x)). - _Vincenzo Librandi_, Sep 27 2015

%F From _Colin Barker_, Sep 27 2015: (Start)

%F a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.

%F G.f.: -3*(370*x^2-74*x+1)/((x-1)*(10*x-1)*(100*x-1)). (End)

%F From _Elmo R. Oliveira_, Aug 27 2024: (Start)

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

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

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

%e Simple continued fraction expansions showing large partial quotients:

%e a(9)^(1/3) =[1000000; 2999, 1, 499999, 1439, 1, 2582643, 1, 1, 1, 2, 3, 3, ...].

%e a(20)^(1/4) = [10000000000; 39999999999, 1, 3999999999, 16949152542, 2, 1, 2, 6, 1, 4872106, 3, 9, 2, 3, ...].

%e 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)

%t Table[10^(2 n) + 10^n + 1, {n, 0, 15}] (* _Michael De Vlieger_, Sep 27 2015 *)

%t 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 *)

%o (PARI) a(n) = { 10^(2*n) + 10^n + 1 } \\ _Harry J. Smith_, Feb 02 2010

%o (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

%o (Magma) [10^(2*n) + 10^n + 1: n in [0..20]]; // _Vincenzo Librandi_, Sep 27 2015

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

%K nonn,base,easy

%O 0,1

%A _Henry Bottomley_, Dec 07 2001

%E Offset changed from 1 to 0 by _Harry J. Smith_, Feb 02 2010

%E More terms from _Michael De Vlieger_, Sep 27 2015