A278041 The tribonacci representation of a(n) is obtained by appending 0,1,1 to the tribonacci representation of n (cf. A278038).

3, 10, 16, 23, 27, 34, 40, 47, 54, 60, 67, 71, 78, 84, 91, 97, 104, 108, 115, 121, 128, 135, 141, 148, 152, 159, 165, 172, 176, 183, 189, 196, 203, 209, 216, 220, 227, 233, 240, 246, 253, 257, 264, 270, 277, 284, 290, 297, 301, 308, 314, 321, 328, 334, 341, 345, 352, 358, 365, 371, 378, 382, 389, 395, 402, 409, 415

Offset: 0

Author

N. J. A. Sloane, Nov 18 2016

Keywords

• nonn
• base
• easy

Comments

This sequence gives the indices k for which A080843(k) = 2, sorted increasingly with offset 0. In the W. Lang link a(n) = C(n). - Wolfdieter Lang, Dec 06 2018

Positions of letter c 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 ac in the tribonacci word t. This follows from the fact that the letter c is always preceded in t by the letter a, and the formula AB = C-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 1000011, which is 44+2+1 = 47, so a(7) = 47.
		

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, Theorem 13.
• Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv preprint arXiv:1810.09787 [math.NT], 2018.

Crossrefs

Cf. A003145, A003146, A080843, A276797, A276798, A278038, A278039, A278040, A278041, 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) = A003146(n+1) - 1.

a(n) = A003144(A003145(n)). - N. J. A. Sloane, Oct 05 2018

From Wolfdieter Lang, Dec 06 2018: (Start)

a(n) = n + 2 + A(n) + B(n), where A(n) = A278040(n) and B = A278039(n).

a(n) = 7*n + 3 - (z_A(n-1) + 3*z_C(n-1)), where z_A(n) = A276797(n+1) and z_C(n) = A276798(n+1) - 1, n >= 0.

For proofs see the W. Lang link in A080843, eqs. 37 and 40.

a(n) - 1 = B2(n), where B2-numbers are B-numbers from A278039 followed by a C-number from A278041. See a comment and example in A319968.

(End)