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A081172
Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3), with a(0) = 1, a(1) = 1, a(2) = 0.
37
1, 1, 0, 2, 3, 5, 10, 18, 33, 61, 112, 206, 379, 697, 1282, 2358, 4337, 7977, 14672, 26986, 49635, 91293, 167914, 308842, 568049, 1044805, 1921696, 3534550, 6501051, 11957297, 21992898, 40451246, 74401441, 136845585, 251698272, 462945298, 851489155
OFFSET
0,4
COMMENTS
The name "tribonacci number" is less well-defined than "Fibonacci number". The sequence A000073 (which begins 0, 0, 1) is probably the most important version, but the name has also been applied to A000213, A001590, and A081172. - N. J. A. Sloane, Jul 25 2024
Completes the set of tribonacci numbers starting with 0's and 1's in the first three terms:
0,0,0 A000004;
0,0,1 A000073;
0,1,0 A001590;
0,1,1 A000073 starting at a(1);
1,0,0 A000073 starting at a(-1);
1,0,1 A001590;
1,1,0 this sequence;
1,1,1 A000213.
LINKS
Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
N. G. Voll, Some identities for four term recurrence relations, Fib. Quart., 51 (2013), 268-273.
FORMULA
From R. J. Mathar, Mar 27 2009: (Start)
G.f.: (1-2*x^2)/(1 - x - x^2 - x^3).
a(n) = A000073(n+2) - 2*A000073(n). (End)
MAPLE
A081172 := proc(n)
option remember;
if n <= 2 then
op(n+1, [1, 1, 0]) ;
else
add(procname(n-i), i=1..3) ;
end if;
end proc: # R. J. Mathar, Aug 09 2012
MATHEMATICA
LinearRecurrence[{1, 1, 1}, {1, 1, 0}, 40] (* Vladimir Joseph Stephan Orlovsky, Jun 07 2011 *)
PROG
(PARI) { a1=1; a2=1; a3=0; write("b081172.txt", 0, " ", a1); write("b081172.txt", 1, " ", a2); write("b081172.txt", 2, " ", a3); for(n=3, 500, a=a1+a2+a3; a1=a2; a2=a3; a3=a; write("b081172.txt", n, " ", a) ) } \\ Harry J. Smith, Mar 20 2009
(PARI) my(x='x+O('x^40)); Vec((1-2*x^2)/(1-x-x^2-x^3)) \\ G. C. Greubel, Apr 23 2019
(Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1-2*x^2)/(1-x-x^2-x^3) )); // G. C. Greubel, Apr 23 2019
(Sage) ((1-2*x^2)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 23 2019
(GAP) a:=[1, 1, 0];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # G. C. Greubel, Apr 23 2019
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Harry J. Smith, Apr 19 2003
STATUS
approved