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 A066138 a(n) = 10^(2n) + 10^n + 1. 11
 3, 111, 10101, 1001001, 100010001, 10000100001, 1000001000001, 100000010000001, 10000000100000001, 1000000001000000001, 100000000010000000001, 10000000000100000000001, 1000000000001000000000001, 100000000000010000000000001, 10000000000000100000000000001, 1000000000000001000000000000001 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 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) 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). 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) EXAMPLE 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) 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 *) PROG (PARI) { for (n=0, 100, write("b066138.txt", n, " ", 10^(2*n) + 10^n + 1) ) } \\ Harry J. Smith, Feb 02 2010 (Magma) [10^(2*n) + 10^n + 1: n in [0..20]]; // Vincenzo Librandi, Sep 27 2015 (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 CROSSREFS Cf. A033934, A062397, A000533, A002283, A168624. Sequence in context: A301612 A180765 A232769 * A114207 A128684 A292336 Adjacent sequences: A066135 A066136 A066137 * A066139 A066140 A066141 KEYWORD nonn,base,easy AUTHOR Henry Bottomley, Dec 07 2001 EXTENSIONS Offset changed from 1 to 0 by Harry J. Smith, Feb 02 2010 More terms from Michael De Vlieger, Sep 27 2015 STATUS approved

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Last modified February 26 12:26 EST 2024. Contains 370352 sequences. (Running on oeis4.)