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A322411
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Compound tribonacci sequence with a(n) = A278040(A278041(n)), for n >= 0.
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7
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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
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OFFSET
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0,1
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COMMENTS
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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
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LINKS
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FORMULA
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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
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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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