A319431 The first differences (A319430) of the tribonacci representation numbers (A003726 or A278038) consists of runs of 1's separated by the terms of the present sequence.

2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 5, 2, 19, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 37, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 5, 2, 19, 2, 3, 2, 5, 2, 3, 74, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 5, 2, 19, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 37, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 5, 2, 147, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 5, 2, 19, 2, 3, 2, 5, 2

Offset: 1

Author

N. J. A. Sloane, Sep 30 2018

Keywords

• nonn

Comments

The runs of 1's in A319430 have lengths that apparently are given by A275925 (with a slight change at the start). The present sequence shows the terms greater than 1.

Let b = "2,3,2", c = "2,3,2,5,2,3", d = "2,3,2,5,2". The sequence appears to consist of a word over the alphabet {b,c,d} interspersed with the sequence 10, 19, 10, 37, 10, 19, 74, 10, 19, 10, 37, 10, 147, 10, 19, 10, 37, 10, ...:

c, 10, d, 19, c, 10, b, 37, c, 10, d, 19, c, 74, c, 10, d, 19, c, 10, b, 37, c, 10, d, 147, c, 10, d, 19, c, 10, b, 37, c, 10, d, 19, c, 74, ...

The interspersed sequence 10, 19, 10, 37, 10, 19, 74, ... appears to have the same kind of structure.

It would be nice to have a recurrence of some kind that produces this sequence.

Examples

			0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 16, 17, 18, 19, 20, 21, 22, 24, 25, ... = A003726 (trib. repres. numbers)
1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 1, ... = A319430 (differences)
2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 5, 2, 19, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 37, 2, ... (omit 1's, present sequence)
6, 1, 5, 1, 6, 1, 3, 1, 6, 1, 5, 1, 6, 1, 6, 1, 5, 1, 6, 1, 3, 1, 6, 1, 5, 1, ... = run lengths in differences
6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, ... = A275925 truncated (BISECTION of run lengths)
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..50000 (first 1606 terms from N. J. A. Sloane)

Crossrefs

Cf. A003726, A275925, A278038, A319430.

Programs

Mathematica

DeleteCases[Differences@ Select[Range[0, 1200], SequenceCount[IntegerDigits[#, 2], {1, 1, 1}] == 0 &] , 1] (* Michael De Vlieger, Dec 23 2019 *)