A319200 a(n) = -(A(n) - A(n-1)) where A(n) = A057597(n+1), for n >= 0.

0, -1, 2, -1, -2, 5, -4, -3, 12, -13, -2, 27, -38, 9, 56, -103, 56, 103, -262, 215, 150, -627, 692, 85, -1404, 2011, -522, -2893, 5426, -3055, -5264, 13745, -11536, -7473, 32754, -36817, -3410, 72981, -106388, 29997, 149372, -285757, 166382, 268747, -720886, 618521, 371112, -1710519, 1957928, 123703, -3792150

Offset: 0

Author

Wolfdieter Lang, Oct 23 2018

Keywords

• sign
• easy

Comments

This sequence appears in the reduction formula for negative powers of the tribonacci constant t = A058265: t^(-n) = A(n)*t^2 + a(n)*t + A(n+1)*1, with A(n) = A057597(n+1), for n >= 0. This follows from t^3 = t^2 + t + 1, or 1/t = t^2 - t - 1 = A192918, leading to the recurrence: A(n) = -A(n) - A(n-1) + A(n-2), with inputs A(-3) = 1, A(-2) = 1 and A(-1) = 0 and a(n) = -(A(n) - A(n-1)). See the example below.

Examples

			The coefficients of t^2, t, 1 for t^(-n) begin, for n >= -3:
n     t^2  t   1
-----------------
-3     1   1   1
-2     1   0   0
-1     0   1   0
----------------
+0     0   0   1
+1     1  -1  -1
+2    -1   2   0
+3     0  -1   2
+4     2  -2  -3
+5    -3   5   1
+6     1  -4   4
+7     4  -3  -8
+8    -8  12   5
+9     5 -13   7
10     7  -2 -20
...
		

Links

• Index entries for linear recurrences with constant coefficients, signature (-1,-1,1).

Crossrefs

Cf. A057597, A058265, A078016(n+1) (different signs), A192918.

Programs

Mathematica

LinearRecurrence[{-1,-1,1},{0,-1,2},60] (* Harvey P. Dale, Jul 20 2025 *)

PARI

a057597(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)) \\ after Michael Somos in A057597
a(n) = -(a057597(n+1)-a057597(n)) \\ Felix Fröhlich, Oct 23 2018

Formula

a(n) = -(A057597(n+1) - A057597(n)), for n >= 0.

Recurrence a(n) = -a(n-1) - a(n-2) + a(n-3), for n >=0, with a(-3) = 1, a(-2) = 0 and a(-1) = 1.

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