A278040 The tribonacci representation of a(n) is obtained by appending 0,1 to the tribonacci representation of n (cf. A278038).

1, 5, 8, 12, 14, 18, 21, 25, 29, 32, 36, 38, 42, 45, 49, 52, 56, 58, 62, 65, 69, 73, 76, 80, 82, 86, 89, 93, 95, 99, 102, 106, 110, 113, 117, 119, 123, 126, 130, 133, 137, 139, 143, 146, 150, 154, 157, 161, 163, 167, 170, 174, 178, 181, 185, 187, 191, 194, 198, 201, 205, 207, 211, 214, 218, 222, 225, 229, 231, 235

Offset: 0

Author

N. J. A. Sloane, Nov 18 2016

Keywords

• nonn
• base
• easy

Comments

This sequence gives the A(n) numbers of the W. Lang link. There the B(n) and C(n) numbers are A278039(n) and A278041(n), respectively. - Wolfdieter Lang, Dec 05 2018

Positions of letter b in the tribonacci word t generated by a->ab, b->ac, c->a, when given offset 0. - Michel Dekking, Apr 03 2019

This sequence gives the positions of the word ab in the tribonacci word t. This follows from the fact that the letter b is always preceded in t by the letter a, and the formula AA = B-1, where A := A003144, B := A003145, C := A003146. - Michel Dekking, Apr 09 2019

Examples

			The tribonacci representation of 7 is 1000 (see A278038), so a(7) has tribonacci representation 100001, which is 24+1 = 25, so a(7) = 25.
		

Links

• N. J. A. Sloane, Table of n, a(n) for n = 0..20000
• L. Carlitz, R. Scoville and V. E. Hoggatt, Jr., Fibonacci representations of higher order, Fib. Quart., 10 (1972), 43-69.
• Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv:1810.09787v1 [math.NT], 2018.

Crossrefs

Cf. A003145, A276789, A276796, A278038, A278039, A278041, A319198, A319968.

By analogy with the Wythoff compound sequences A003622 etc., the nine compounds of A003144, A003145, A003146 might be called the tribonacci compound sequences. They are A278040, A278041, and A319966-A319972.

Formula

a(n) = A003145(n+1) - 1.

a(n) = A003144(A003144(n)). - N. J. A. Sloane, Oct 05 2018

See Theorem 13 in the Carlitz, Scoville and Hoggatt paper. - Michel Dekking, Mar 20 2019

From Wolfdieter Lang, Dec 13 2018: (Start)

This sequence gives the indices k with A080843(k) = 1, ordered increasingly with offset 0.

a(n) = 1 + 4*n - A319198(n-1), n >= 0, with A319198(-1) = 0.

a(n) = A276796(C(n)) - 1, with C(n) = A278041(n).

For a proof see the W. Lang link, Proposition 5, and eq. (58).

a(n) - 1 = B1(n), where B1-numbers are B-numbers from A278039 followed by an A-number from A278040. See a comment and example in A319968.

a(n) - 1 = B(B(n)) = B(B(n) + 1) - 2, for n > = 0, where B = A278039.

(End)