%I #19 Oct 11 2019 07:47:50
%S 3,7,11,15,18,22,26,30,34,38,42,45,49,53,57,61,65,68,72,76,80,84,88,
%T 92,95,99,103,107,110,114,118,122,126,130,134,137,141,145,149,153,157,
%U 160,164,168,172,176,180,184,187,191,195,199,203,207,211,214,218,222,226,230,234
%N Compound sequence with a(n) = A319198(A278041(n)), for n >= 0.
%C Old name was: Compound tribonacci sequence a(n) = A319198(A278041(n)), for n >= 0.
%C a(n) gives the sum of the entries of the tribonacci word sequence t = A080843 not exceeding t(C(n)), with C(n) = A278041(n).
%C The nine sequences A308199, A319967, A319968, A322410, A322409, A322411, A322413, A322412, A322414 are based on defining the tribonacci ternary word to start with index 0 (in contrast to the usual definition, in A080843 and A092782, which starts with index 1). As a result these nine sequences differ from the compound tribonacci sequences defined in A278040, A278041, and A319966-A319972. - _N. J. A. Sloane_, Apr 05 2019
%C The difference sequence (a(n+1)-a(n)) is equal to a change of alphabet of the tribonacci word t = A092782. The alphabet is {4,4,3}. This follows from the formula a(n) = A278039(n) + 2*n + 3. - _Michel Dekking_, Oct 05 2019
%F a(n) = z(C(n)) = Sum_{j=0..C(n)} t(j), n >= 0, with z = A319198, C = A278041 and t = A080843.
%F a(n) = B(n) + 2*n + 3, where B(n) = A278039(n). For a proof see the W. Lang link in A080843, Proposition 8, eq. (47).
%F a(n) = 3 + Sum_{k=1..n-1} d(k), where d is the tribonacci sequence on the alphabet (4,4,3}. - _Michel Dekking_, Oct 05 2019
%e n = 2: C(2) = 16, t = {0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, ...} which sums to 11 = a(2) = 4 + 7, because B(2) = 4.
%Y Cf. A080843, A278039, A278041, A319198, A321333, A322407.
%K nonn,easy
%O 0,1
%A _Wolfdieter Lang_, Jan 02 2019
%E Name changed by _Michel Dekking_, Oct 08 2019