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A322411
Compound tribonacci sequence with a(n) = A278040(A278041(n)), for n >= 0.
7
12, 36, 56, 80, 93, 117, 137, 161, 185, 205, 229, 242, 266, 286, 310, 330, 354, 367, 391, 411, 435, 459, 479, 503, 516, 540, 560, 584, 597, 621, 641, 665, 689, 709, 733, 746, 770, 790, 814, 834, 858, 871, 895, 915, 939, 963, 983, 1007, 1020, 1044, 1064, 1088, 1112, 1132, 1156, 1169, 1193, 1213, 1237, 1257, 1281
OFFSET
0,1
COMMENTS
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
LINKS
L. Carlitz, R. Scoville and V. E. Hoggatt, Jr., Fibonacci representations of higher order, Fib. Quart., 10 (1972), 43-69.
FORMULA
a(n) = A(C(n)) = A(C(n) + 1) - 2 = 4*A(n) + 3*B(n) + 2*n + 8, for n >= 0, with A = A278040 and C = A278041. For a proof see the W. Lang link in A278040, Proposition 9, eq. (50).
This formula already follows from Theorem 15 in the 1972 paper by Carlitz et al., which gives that b(c(n)) = a(n) + 2b(n) + 2c(n), where a, b and c are the classical positional sequences of the letters in the tribonacci word. The connection is made by using that c(n) = a(n) + b(n) + n, and by making the translation B(n) = a(n+1)-1, A(n) = b(n+1)-1, C(n) = c(n+1)-1. (Note the switching of A and B!). - Michel Dekking, Apr 07 2019
a(n+1) = A319969(n)-1 = A003145(A003146(n))-1, the corresponding classical compound tribonacci sequence. - Michel Dekking, Apr 04 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jan 02 2019
STATUS
approved