A276797 Partial sums of A276794.

0, 0, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 10, 10, 10, 10, 11, 11, 12, 12, 12, 12, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 16, 16, 16, 16, 17, 17, 18, 18, 18, 18, 19, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 23, 23, 23, 23, 24, 24, 25

Offset: 0

Author

N. J. A. Sloane, Oct 28 2016

Keywords

• nonn
• easy

Comments

a(n+1) = z_A(n), the number of entries of A278040 (called A number in the W. Lang given there) not exceeding n, for n >= 1 and z_A(-1) := 0. - Wolfdieter Lang, Dec 06 2018

Conjecture: A140102(n) - n = a(n-1). - N. J. A. Sloane, Oct 26 2016 (added Mar 21 2019). This is true - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019

Links

• N. J. A. Sloane, Table of n, a(n) for n = 0..10000
• F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
• Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv preprint arXiv:1810.09787 [math.NT], 2018.
• Jeffrey Shallit, Some Tribonacci conjectures, arXiv:2210.03996 [math.CO], 2022.

Crossrefs

Cf. A003145, A140102, A276794, A278039, A278040.

A276793(n) + A276794(n) + A276791(n) = 1;

A276796(n) + A276797(n) + A276798(n) = n + 1.

Programs

Maple

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:
N0; N1; N2; # prints A276796, A276797, A276798 (except A276798 is off by 1 because it does not count the initial 0 in A003146). # N. J. A. Sloane, Jun 08 2018

Formula

a(n) = Sum_{k=0..n} A276794(k).

a(n) = n + 1 - (A276796(n) + A276798(n)).

a(n) = 2*B(n) - A(n) + 1, for n >= 0, where A(n) = A278040(n) and B(n) = A278039(n). For a proof see the W. Lang link in A278040, Proposition 7, eq. (41). - Wolfdieter Lang, Dec 06 2018