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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
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
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:=ct-1; fi;
ans:=[op(ans), ct]; od: ans;
MATHEMATICA
Accumulate[Mod[#, 2]&/@Flatten[Table[Binomial[n, k], {n, 0, 20}, {k, 0, n}]]/.(0->-1)] (* Harvey P. Dale, May 24 2022 *)
CROSSREFS
KEYWORD
sign,look
AUTHOR
N. J. A. Sloane, Feb 03 2016
STATUS
approved