A349700 Difference between 4*A001590(n+2) and A075092(n).

-6, 4, 6, 0, 18, 24, 30, 92, 138, 236, 518, 856, 1570, 3072, 5374, 9972, 18714, 33684, 62406, 115440, 210482, 388776, 715934, 1312460, 2419114, 4449532, 8174406, 15049672, 27675714, 50884368, 93629694, 172187364, 316668474, 582540836, 1071371910, 1970517728

Offset: 0

Author

Wojciech Florek, Nov 25 2021

Keywords

• sign
• easy

Comments

Tribonacci numbers 4*A001590(n+2) (i.e., tribonacci numbers t(n) = t(n-1) + t(n-2) + t(n-3) with t(0) = 0, t(1) = 4, t(2) = 8) for n > 2 are numbers of ternary sequences (q(1),q(2),...,q(n)), q(i) = 0,1,2, of length n such that all triples (q(i),q(i+1),q(i+2)) contain digits 0 and 1 at least once. In the other words {0,1} is a subset of each {q(i),q(i+1),q(i+2)}. Similarly, A075092(n), for n > 2, presents a number of ternary cyclic sequences with this property. Hence, a(n), for n > 2, gives a number of (ordinary) sequences which do not lead to cyclic sequences with triples (q(n-1),q(n),q(1)) and (q(n),q(1),q(2)) satisfying the above formulated condition. The special cases, n = 1,2, can be included in a way proposed in A001644.

The recurrence formula is the same as this for A075092, with different initial conditions.

Examples

			If n = 4, then there are 24 finite sequences: 0010, 0011, 0012, 0100, 0101, 0102, 0110, 0120, 0210, nine analogous starting with 1, 2010, 2011, 2012, 2100, 2101, and 2102. Only six of them, namely  0011, 0101, 0110, 1001, 1010, 1100, yield cyclic sequences satisfying the restriction imposed. Therefore, a(4) = 24 - 6 = 18.
		

Links

• Wojciech Florek, A class of generalized Tribonacci sequences applied to counting problems, Appl. Math. Comput., 338 (2018), 809-821.
• W. O. J. Moser, Cyclic binary strings without long runs of like (alternating) bits, Fibonacci Quart. 31(1) (1993), 2-6.
• Index entries for linear recurrences with constant coefficients, signature (0,1,4,1,0,-1).

Crossrefs

Cf. A001590, A001644, A075092, A294627.

Programs

Mathematica

nterms=50;LinearRecurrence[{0,1,4,1,0,-1},{-6,4,6,0,18,24},nterms] (* Paolo Xausa, Nov 26 2021 *)

Formula

a(n) = a(n-2) + 4*a(n-3) + a(n-4) - a(n-6), a(0) = -6, a(1) = 4, a(2) = 6, a(3) = 0, a(4) = 18, a(5) = 24.

G.f.: (4*x^5-2*x^4-20*x^3-12*x^2-4*x+6)/(-x^6+x^4+4*x^3+x^2-1).