A056830 Alternate digits 1 and 0.

0, 1, 10, 101, 1010, 10101, 101010, 1010101, 10101010, 101010101, 1010101010, 10101010101, 101010101010, 1010101010101, 10101010101010, 101010101010101, 1010101010101010, 10101010101010101, 101010101010101010

Offset: 0

Author

Henry Bottomley, Aug 30 2000

Keywords

• nonn
• base
• easy

Comments

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

Sequence A000975 written in base 2. - Jaroslav Krizek, Aug 05 2009

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

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

Examples

			n  a(n)             A000975(n)   (If a(n) is interpreted in base 2.)
------------------------------
0  0 ....................... 0
1  1 ....................... 1
2  10 ...................... 2 = 2^1
3  101 ..................... 5
4  1010 ................... 10 = 2^1 + 2^3
5  10101 .................. 21
6  101010 ................. 42 = 2^1 + 2^3 + 2^5
7  1010101 ................ 85
8  10101010 .............. 170 = 2^1 + 2^3 + 2^5 + 2^7
9  101010101 ............. 341
10 1010101010 ............ 682 = 2^1 + 2^3 + 2^5 + 2^7 + 2^9
11 10101010101 .......... 1365
12 101010101010 ......... 2730 = 2^1 + 2^3 + 2^5 + 2^7 + 2^9 + 2^11, etc.
- _Bruno Berselli_, Dec 02 2015
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..800
• Index entries for linear recurrences with constant coefficients, signature (10,1,-10).
• Index entries for 2-automatic sequences.

Crossrefs

Partial sums of A015585.

Cf. A000975, A033113, A033114, A033115, A033116, A033117, A033118, A033119, A059848, A062864.

Programs

GAP

List([0..30], n-> Int(10^(n+1)/99) ); # G. C. Greubel, Jul 14 2019

Magma

[Round((20*10^n-11)/198) : n in [0..30]]; // Vincenzo Librandi, Jun 25 2011

Maple

A056830 := proc(n) floor(10^(n+1)/99) ; end proc:

Mathematica

CoefficientList[Series[x/((1-x^2)*(1-10*x)), {x,0,30}], x] (* G. C. Greubel, Sep 26 2017 *)

PARI

Vec(x/((1-x)*(1+x)*(1-10*x))+O(x^30)) \\ Charles R Greathouse IV, Feb 13 2017

Sage

[floor(10^(n+1)/99) for n in (0..30)] # G. C. Greubel, Jul 14 2019

Formula

a(n) = +10*a(n-1) + a(n-2) - 10*a(n-3).

a(n) = floor(10^(n+1)/(10^2-1)) = a(n-2)+10^(n-1) = 10*a(n-1) + (1 - (-1)^n)/2.

From Paul Barry, Nov 12 2003: (Start)

a(n+1) = Sum_{k=0..floor(n/2)} 10^(n-2*k).

a(n+1) = Sum_{k=0..n} Sum_{j=0..k} (-1)^(j+k)*10^j.

G.f.: x/((1-x)*(1+x)*(1-10*x)).

a(n) = 9*a(n-1) + 10*a(n-2) + 1.

a(n) = 10^(n+1)/99 - (-1)^n/22 - 1/18. (End)

a(n) = A007088(A107909(A104161(n))) = A007088(A000975(n)). - Reinhard Zumkeller, May 28 2005

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

From Daniel Forgues, Sep 20 2018: (Start)

If a(n) is interpreted in base 2:

a(2n) = Sum_{k=1..n} 2^(2n-1), n >= 0; a(2n-1) = a(2n)/2, n >= 1.

a(2n) = A020988(n), n >= 0.

a(0) = 0; a(2n) = 4*a(2n-2) + 2, n >= 1. (End)

Extensions

More terms from Casey Mongoven, Mar 19 2006