A321664 A sequence consisting of three disjoint copies of the Fibonacci sequence, one shifted, with the property that for any four consecutive terms the maximum term is the sum of the two minimum terms.

0, 1, 1, 1, 2, 1, 2, 3, 2, 4, 5, 3, 7, 8, 5, 12, 13, 8, 20, 21, 13, 33, 34, 21, 54, 55, 34, 88, 89, 55, 143, 144, 89, 232, 233, 144, 376, 377, 233, 609, 610, 377, 986, 987, 610, 1596, 1597, 987, 2583, 2584, 1597, 4180, 4181, 2584, 6764, 6765, 4181, 10945

Offset: 0

Author

David Nacin, Nov 23 2018

Keywords

• nonn

Comments

This sequence was constructed to show that there are many sequences, besides those merging with multiples of the Padovan sequence A000931, with the property that for any four consecutive terms the maximum term is the sum of the two minimum terms. This refutes a conjecture that was formerly in that entry.

Examples

			For n=13, as n is 1 (mod 3), we find a(3*4+1) is the 4+2=6th Fibonacci number which is 8.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000
• David Nacin, Van der Laan Sequences and a Conjecture on Padovan Numbers, J. Int. Seq., Vol. 26 (2023), Article 23.1.2.
• Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,0,0,0,-1).

Crossrefs

Exhibits a property shared with multiples of A000931.

Programs

Magma

m:=70; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!((x+x^2+x^3-x^5-x^7)/(1-2*x^3+x^9))); // Vincenzo Librandi, Nov 29 2018

Maple

seq(coeff(series(((x^4+x^3+x^2+x+1)/(1-x^3-x^6))-(1/(1-x^3)),x,n+1), x, n), n = 0 .. 60); # Muniru A Asiru, Nov 29 2018

Mathematica

CoefficientList[Series[(x+x^2+x^3-x^5-x^7)/(1-2x^3+x^9), {x, 0, 20}], x] (* or *)
LinearRecurrence[{0,0,2,0,0,0,0,0,-1}, {0,1,1,1,2,1,2,3,2}, 50] (* G. C. Greubel, Dec 04 2018 *)

PARI

my(x='x+O('x^70)); Vec((x+x^2+x^3-x^5-x^7)/(1-2*x^3+x^9)) \\ G. C. Greubel, Dec 04 2018

Python

def a(n):
    if n<6:
        return [0,1,1,1,2,1][n]
    return a(n-3)+a(n-6)+[1,0,0][n%3]

Racket

(define (f x) (cond [(< x 6) (list-ref (list 0 1 1 1 2 1) x)]
[else (+ (f (- x 3)) (f (- x 6)) (list-ref (list 1 0 0) (remainder x 3)))]))

Sage

s=((x+x^2+x^3-x^5-x^7)/(1-2*x^3+x^9)).series(x, 70); s.coefficients(x, sparse=False) # G. C. Greubel, Dec 04 2018

Formula

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

G.f.: (x + x^2 + x^3 - x^5 - x^7)/(1 - 2*x^3 + x^9).

a(3*n) = A000045(n+2)-1, a(3*n+1) = A000045(n+2), a(3*n+2) = A000045(n+1).

a(n) = 2*a(n-3) - a(n-9). - G. C. Greubel, Dec 04 2018