

A268233


Excess of number of 1's over number of 0's in terms 0 through n of A047999.


2



1, 2, 3, 4, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 7, 8, 9, 8, 7, 8, 9, 10, 9, 10, 9, 10, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 18, 17, 16, 15, 14, 13, 12, 13, 14, 15, 14, 13, 12, 11, 10, 9, 10, 11, 12, 11, 12, 11, 10, 9, 8, 7, 8, 7, 8, 9, 10, 11, 12, 11, 10, 9, 8, 9, 10, 11, 12, 13, 12, 11, 10, 11, 10
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OFFSET

0,2


COMMENTS

If n and k are large with 0 <= k <= n, there is a high probability that there will be a position where the binary expansion of n is 0 and that of k is 1. By Lucas's theorem, this means that binomial(n,k) is even, and so A047999 is 0. This implies that A047999 is mostly zeros, and so we expect the present sequence to have slope 1, an observation which is supported by the graph.
In fact the above remark follows from the fact that the Hausdorff dimension of the 1's in the Sierpinski gasket (the limiting form of A047999) is 1.584...  N. J. A. Sloane, Feb 12 2016
The negative terms start at n = 178.  Georg Fischer, Feb 15 2019


LINKS

N. J. A. Sloane, Table of n, a(n) for n = 0..10584


MAPLE

# start with list of terms of A047999 in b1
ans:=[]; ct:=0; for n from 1 to nops(b1) do
if b1[n]=1 then ct:=ct+1 else ct:=ct1; fi;
ans:=[op(ans), ct]; od: ans;


CROSSREFS

Cf. A047999, A268231, A268232.
Sequence in context: A030323 A285872 A227181 * A309241 A065651 A322567
Adjacent sequences: A268230 A268231 A268232 * A268234 A268235 A268236


KEYWORD

sign,look


AUTHOR

N. J. A. Sloane, Feb 03 2016


STATUS

approved



