A057597 a(n) = -a(n-1) - a(n-2) + a(n-3), a(0)=0, a(1)=0, a(2)=1.

0, 0, 1, -1, 0, 2, -3, 1, 4, -8, 5, 7, -20, 18, 9, -47, 56, 0, -103, 159, -56, -206, 421, -271, -356, 1048, -963, -441, 2452, -2974, 81, 5345, -8400, 3136, 10609, -22145, 14672, 18082, -54899, 51489, 21492, -127880, 157877, -8505, -277252, 443634, -174887, -545999, 1164520, -793408, -917111

Offset: 0

Author

N. J. A. Sloane, Oct 06 2000

Keywords

• sign
• easy

Comments

Reflected (A074058) tribonacci numbers A000073: A000073(n) = a(1-n).

There is an alternative way to produce this sequence, from A000073, which is 0,0,1,1,2,4,7,13,24,44,... Call this {b(n)}. Taking x1 = (b(2))^2 - b(1)*b(3) = 0; x2 = (b(3))^2 - b(2)*b(4) = 1; x3 = (b(4))^2 - b(3)*b(5) = -1; x4 = 0, x5 = 2, we generate (0),0,1,-1,0,2,-3,1. - John McNamara, Jan 02 2004

Pisano period lengths: 1, 4, 13, 8, 31, 52, 48, 16, 39, 124, 110, 104, 168, 48, 403, 32, 96, 156, 360, 248, ... - R. J. Mathar, Aug 10 2012

The negative powers of the tribonacci constant t = A058265 are t^(-n) = a(n+1)*t^2 + b(n)*t + a(n+2)*1, for n >= 0, with b(n) = A319200(n) = -(a(n+1) - a(n)), for n >= 0. 1/t = t^2 - t - 1 = A192918. See the example in A319200 for the first powers. - Wolfdieter Lang, Oct 23 2018

References

• Petho Attila, Posting to Number Theory List (NMBRTHRY(AT)LISTSERV.NODAK.EDU), Oct 06 2000.

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• Milan Janjic, Recurrence Relations and Determinants, arXiv preprint arXiv:1112.2466 [math.CO], 2011.
• Milan Janjic, Determinants and Recurrence Sequences, Journal of Integer Sequences, 15 (2012), Article 12.3.5. - From _N. J. A. Sloane_, Sep 16 2012.
• Ronald C. King, Generating functions for some series of characters of classical Lie groups, arXiv:2303.00576 [math.CO], 2023, p. 9.
• A. G. Shannon, H. M. Srivastava, and József Sàndor, Towards a new generalized Simson's identity, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 479-490. See p. 482.
• Index entries for linear recurrences with constant coefficients, signature (-1,-1,1).

Crossrefs

Cf. A000073, A058265, A319200. First differences of A077908.

Programs

GAP

a:=[0,0,1];;  for n in [4..55] do a[n]:=-a[n-1]-a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Oct 23 2018

Haskell

a057597 n = a057597_list !! n
a057597_list = 0 : 0 : 1 : zipWith3 (\x y z -> - x - y + z)
               (drop 2 a057597_list) (tail a057597_list) a057597_list
-- Reinhard Zumkeller, Oct 07 2012

Maple

seq(coeff(series(x^2/(1+x+x^2-x^3),x,n+1), x, n), n = 0 .. 50); # Muniru A Asiru, Oct 23 2018

Mathematica

CoefficientList[Series[x^2/(1+x+x^2-x^3), {x, 0, 50}], x]

PARI

{a(n) = polcoeff( if( n<0, x / ( 1 - x - x^2 - x^3), x^2 / ( 1 + x + x^2 - x^3) ) + x*O(x^abs(n)), abs(n))} /* Michael Somos, Sep 03 2007 */

Formula

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

G.f.: Q(0)*x^2/2, where Q(k) = 1 + 1/(1 - x*(4*k+1 + x - x^2)/( x*(4*k+3 + x - x^2) - 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 09 2013

G.f. -x*T(1/x), where T is the g.f. of A000073. - Wolfdieter Lang, Oct 26 2018

Extensions

Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021