A322408 Compound sequence with a(n) = A319198(A278041(n)), for n >= 0.

3, 7, 11, 15, 18, 22, 26, 30, 34, 38, 42, 45, 49, 53, 57, 61, 65, 68, 72, 76, 80, 84, 88, 92, 95, 99, 103, 107, 110, 114, 118, 122, 126, 130, 134, 137, 141, 145, 149, 153, 157, 160, 164, 168, 172, 176, 180, 184, 187, 191, 195, 199, 203, 207, 211, 214, 218, 222, 226, 230, 234

Offset: 0

Author

Wolfdieter Lang, Jan 02 2019

Keywords

• nonn
• easy

Comments

Old name was: Compound tribonacci sequence a(n) = A319198(A278041(n)), for n >= 0.

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).

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

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

Examples

			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.
		

Crossrefs

Cf. A080843, A278039, A278041, A319198, A321333, A322407.

Formula

a(n) = z(C(n)) = Sum_{j=0..C(n)} t(j), n >= 0, with z = A319198, C = A278041 and t = A080843.

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).

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

Extensions

Name changed by Michel Dekking, Oct 08 2019