%I #19 May 24 2022 19:38:28
%S 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,
%T 14,15,16,17,18,19,18,17,16,15,14,13,12,13,14,15,14,13,12,11,10,9,10,
%U 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
%N Excess of number of 1's over number of 0's in terms 0 through n of A047999.
%C 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.
%C 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
%C The negative terms start at n = 178. - _Georg Fischer_, Feb 15 2019
%H N. J. A. Sloane, <a href="/A268233/b268233.txt">Table of n, a(n) for n = 0..10584</a>
%p # start with list of terms of A047999 in b1
%p ans:=[]; ct:=0; for n from 1 to nops(b1) do
%p if b1[n]=1 then ct:=ct+1 else ct:=ct-1; fi;
%p ans:=[op(ans),ct]; od: ans;
%t Accumulate[Mod[#,2]&/@Flatten[Table[Binomial[n,k],{n,0,20},{k,0,n}]]/.(0->-1)] (* _Harvey P. Dale_, May 24 2022 *)
%Y Cf. A047999, A268231, A268232.
%K sign,look
%O 0,2
%A _N. J. A. Sloane_, Feb 03 2016