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A276798
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Partial sums of A276791.
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11
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1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15
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OFFSET
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0,5
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COMMENTS
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a(n+1) - 1 = z_C(n), where z_C(n) is the number of C numbers A276798 not exceeding n, for n >= 0, and z_C(-1) = 0. - Wolfdieter Lang, Dec 05 2018
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LINKS
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FORMULA
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a(n) = Sum_{k=0..n} A276791(k), for n >= 0.
a(n) = 2*n + 1 - B(n), where B(n) = A278039(n), n >= 0. For a proof see the comment on z_C and Proposition 7, eq. 43, of the W. Lang link given in A080843. - Wolfdieter Lang, Dec 05 2018
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MAPLE
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M:=12;
S[1]:=`0`; S[2]:=`01`; S[3]:=`0102`;
for n from 4 to M do S[n]:=cat(S[n-1], S[n-2], S[n-3]); od:
t0:=S[M]: # has 927 terms of tribonacci ternary word A080843
# get numbers of 0's, 1's, 2's
N0:=[]: N1:=[]: N2:=[]: c0:=0: c1:=0: c2:=0:
L:=length(t0);
for i from 1 to L do
js := substring(t0, i..i);
j:=convert(js, decimal, 10);
if j=0 then c0:=c0+1; elif j=1 then c1:=c1+1; else c2:=c2+1; fi;
N0:=[op(N0), c0]; N1:=[op(N1), c1]; N2:=[op(N2), c2];
od:
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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