%I #70 Aug 15 2024 07:04:15
%S 0,1,10,101,1010,10101,101010,1010101,10101010,101010101,1010101010,
%T 10101010101,101010101010,1010101010101,10101010101010,
%U 101010101010101,1010101010101010,10101010101010101,101010101010101010
%N Alternate digits 1 and 0.
%C Fibonacci bit-representations of numbers for which there is only one possible representation and for which the maximal and minimal bit-representations (A104326 and A014417) are equal. The numbers represented are equal to the numbers in A000071 (subtract the first term of that sequence). For example, 10101 = 12 because 8+5+1. - _Casey Mongoven_, Mar 19 2006
%C Sequence A000975 written in base 2. - _Jaroslav Krizek_, Aug 05 2009
%C The absolute value of alternating sum of the first n repunits: a(n) = abs(Sum_{k=0..n} (-1)^k*A002275(n)). - _Ilya Gutkovskiy_, Dec 02 2015
%C Binary representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - _Robert Price_, Dec 05 2016
%H Vincenzo Librandi, <a href="/A056830/b056830.txt">Table of n, a(n) for n = 0..800</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (10,1,-10).
%H <a href="/index/Ar#2-automatic">Index entries for 2-automatic sequences</a>.
%F a(n) = +10*a(n-1) + a(n-2) - 10*a(n-3).
%F a(n) = floor(10^(n+1)/(10^2-1)) = a(n-2)+10^(n-1) = 10*a(n-1) + (1 - (-1)^n)/2.
%F From _Paul Barry_, Nov 12 2003: (Start)
%F a(n+1) = Sum_{k=0..floor(n/2)} 10^(n-2*k).
%F a(n+1) = Sum_{k=0..n} Sum_{j=0..k} (-1)^(j+k)*10^j.
%F G.f.: x/((1-x)*(1+x)*(1-10*x)).
%F a(n) = 9*a(n-1) + 10*a(n-2) + 1.
%F a(n) = 10^(n+1)/99 - (-1)^n/22 - 1/18. (End)
%F a(n) = A007088(A107909(A104161(n))) = A007088(A000975(n)). - _Reinhard Zumkeller_, May 28 2005
%F a(n) = round((20*10^n-11)/198) = floor((10*10^n-1)/99) = ceiling((10*10^n-10)/99) = round((10*10^n-10)/99). - _Mircea Merca_, Dec 27 2010
%F From _Daniel Forgues_, Sep 20 2018: (Start)
%F If a(n) is interpreted in base 2:
%F a(2n) = Sum_{k=1..n} 2^(2n-1), n >= 0; a(2n-1) = a(2n)/2, n >= 1.
%F a(2n) = A020988(n), n >= 0.
%F a(0) = 0; a(2n) = 4*a(2n-2) + 2, n >= 1. (End)
%e n a(n) A000975(n) (If a(n) is interpreted in base 2.)
%e ------------------------------
%e 0 0 ....................... 0
%e 1 1 ....................... 1
%e 2 10 ...................... 2 = 2^1
%e 3 101 ..................... 5
%e 4 1010 ................... 10 = 2^1 + 2^3
%e 5 10101 .................. 21
%e 6 101010 ................. 42 = 2^1 + 2^3 + 2^5
%e 7 1010101 ................ 85
%e 8 10101010 .............. 170 = 2^1 + 2^3 + 2^5 + 2^7
%e 9 101010101 ............. 341
%e 10 1010101010 ............ 682 = 2^1 + 2^3 + 2^5 + 2^7 + 2^9
%e 11 10101010101 .......... 1365
%e 12 101010101010 ......... 2730 = 2^1 + 2^3 + 2^5 + 2^7 + 2^9 + 2^11, etc.
%e - _Bruno Berselli_, Dec 02 2015
%p A056830 := proc(n) floor(10^(n+1)/99) ; end proc:
%t CoefficientList[Series[x/((1-x^2)*(1-10*x)), {x,0,30}], x] (* _G. C. Greubel_, Sep 26 2017 *)
%o (Magma) [Round((20*10^n-11)/198) : n in [0..30]]; // _Vincenzo Librandi_, Jun 25 2011
%o (PARI) Vec(x/((1-x)*(1+x)*(1-10*x))+O(x^30)) \\ _Charles R Greathouse IV_, Feb 13 2017
%o (Sage) [floor(10^(n+1)/99) for n in (0..30)] # _G. C. Greubel_, Jul 14 2019
%o (GAP) List([0..30], n-> Int(10^(n+1)/99) ); # _G. C. Greubel_, Jul 14 2019
%Y Partial sums of A015585.
%Y Cf. A000975, A033113, A033114, A033115, A033116, A033117, A033118, A033119, A059848, A062864.
%K nonn,base,easy
%O 0,3
%A _Henry Bottomley_, Aug 30 2000
%E More terms from _Casey Mongoven_, Mar 19 2006