%I M1074 N0406 #673 Nov 23 2024 09:05:08
%S 0,0,1,1,2,4,7,13,24,44,81,149,274,504,927,1705,3136,5768,10609,19513,
%T 35890,66012,121415,223317,410744,755476,1389537,2555757,4700770,
%U 8646064,15902591,29249425,53798080,98950096,181997601,334745777,615693474,1132436852
%N Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) for n >= 3 with a(0) = a(1) = 0 and a(2) = 1.
%C 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
%C Also (for n > 1) number of ordered trees with n+1 edges and having all leaves at level three. Example: a(4)=2 because we have two ordered trees with 5 edges and having all leaves at level three: (i) one edge emanating from the root, at the end of which two paths of length two are hanging and (ii) one path of length two emanating from the root, at the end of which three edges are hanging. - _Emeric Deutsch_, Jan 03 2004
%C a(n) is the number of compositions of n-2 with no part greater than 3. Example: a(5)=4 because we have 1+1+1 = 1+2 = 2+1 = 3. - _Emeric Deutsch_, Mar 10 2004
%C Let A denote the 3 X 3 matrix [0,0,1;1,1,1;0,1,0]. a(n) corresponds to both the (1,2) and (3,1) entries in A^n. - _Paul Barry_, Oct 15 2004
%C Number of permutations satisfying -k <= p(i)-i <= r, i=1..n-2, with k=1, r=2. - _Vladimir Baltic_, Jan 17 2005
%C Number of binary sequences of length n-3 that have no three consecutive 0's. Example: a(7)=13 because among the 16 binary sequences of length 4 only 0000, 0001 and 1000 have 3 consecutive 0's. - _Emeric Deutsch_, Apr 27 2006
%C Therefore, the complementary sequence to A050231 (n coin tosses with a run of three heads). a(n) = 2^(n-3) - A050231(n-3) - _Toby Gottfried_, Nov 21 2010
%C Convolved with the Padovan sequence = row sums of triangle A153462. - _Gary W. Adamson_, Dec 27 2008
%C For n > 1: row sums of the triangle in A157897. - _Reinhard Zumkeller_, Jun 25 2009
%C a(n+2) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 1; 0, 0, 1; 1, 0, 0] or [1, 1, 0; 1, 0, 1; 1, 0, 0] or [1, 1, 1; 1, 0, 0; 0, 1, 0] or [1, 0, 1; 1, 0, 0; 1, 1, 0]. - _R. J. Mathar_, Feb 03 2014
%C a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 0, 1; 1, 1, 1; 0, 1, 0], [0, 1, 0; 0, 1, 1; 1, 1, 0], [0, 0, 1; 1, 0, 1; 0, 1, 1] or [0, 1, 0; 0, 0, 1; 1, 1, 1]. - _R. J. Mathar_, Feb 03 2014
%C Also row sums of A082601 and of A082870. - _Reinhard Zumkeller_, Apr 13 2014
%C Least significant bits are given in A021913 (a(n) mod 2 = A021913(n)). - _Andres Cicuttin_, Apr 04 2016
%C The nonnegative powers of the tribonacci constant t = A058265 are t^n = a(n)*t^2 + (a(n-1) + a(n-2))*t + a(n-1)*1, for n >= 0, with a(-1) = 1 and a(-2) = -1. This follows from the recurrences derived from t^3 = t^2 + t + 1. See the example in A058265 for the first nonnegative powers. For the negative powers see A319200. - _Wolfdieter Lang_, Oct 23 2018
%C The term "tribonacci number" was coined by Mark Feinberg (1963), a 14-year-old student in the 9th grade of the Susquehanna Township Junior High School in Pennsylvania. He died in 1967 in a motorcycle accident. - _Amiram Eldar_, Apr 16 2021
%C Andrews, Just, and Simay (2021, 2022) remark that it has been suggested that this sequence is mentioned in Charles Darwin's Origin of Species as bearing the same relation to elephant populations as the Fibonacci numbers do to rabbit populations. - _N. J. A. Sloane_, Jul 12 2022
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%F G.f.: x^2/(1 - x - x^2 - x^3).
%F G.f.: x^2 / (1 - x / (1 - x / (1 + x^2 / (1 + x)))). - _Michael Somos_, May 12 2012
%F G.f.: Sum_{n >= 0} x^(n+2) *[ Product_{k = 1..n} (k + k*x + x^2)/(1 + k*x + k*x^2) ] = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + ... may be proved by the method of telescoping sums. - _Peter Bala_, Jan 04 2015
%F a(n+1)/a(n) -> A058265. a(n-1)/a(n) -> A192918.
%F a(n) = central term in M^n * [1 0 0] where M = the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 1 1]. (M^n * [1 0 0] = [a(n-1) a(n) a(n+1)].) a(n)/a(n-1) tends to the tribonacci constant, 1.839286755... = A058265, an eigenvalue of M and a root of x^3 - x^2 - x - 1 = 0. - _Gary W. Adamson_, Dec 17 2004
%F a(n+2) = Sum_{k=0..n} T(n-k, k), where T(n, k) = trinomial coefficients (A027907). - _Paul Barry_, Feb 15 2005
%F A001590(n) = a(n+1) - a(n); A001590(n) = a(n-1) + a(n-2) for n > 1; a(n) = (A000213(n+1) - A000213(n))/2; A000213(n-1) = a(n+2) - a(n) for n > 0. - _Reinhard Zumkeller_, May 22 2006
%F Let C = the tribonacci constant, 1.83928675...; then C^n = a(n)*(1/C) + a(n+1)*(1/C + 1/C^2) + a(n+2)*(1/C + 1/C^2 + 1/C^3). Example: C^4 = 11.444...= 2*(1/C) + 4*(1/C + 1/C^2) + 7*(1/C + 1/C^2 + 1/C^3). - _Gary W. Adamson_, Nov 05 2006
%F a(n) = j*C^n + k*r1^n + L*r2^n where C is the tribonacci constant (C = 1.8392867552...), real root of x^3-x^2-x-1=0, and r1 and r2 are the two other roots (which are complex), r1 = m+p*i and r2 = m-p*i, where i = sqrt(-1), m = (1-C)/2 (m = -0.4196433776...) and p = ((3*C-5)*(C+1)/4)^(1/2) = 0.6062907292..., and where j = 1/((C-m)^2 + p^2) = 0.1828035330..., k = a+b*i, and L = a-b*i, where a = -j/2 = -0.0914017665... and b = (C-m)/(2*p*((C-m)^2 + p^2)) = 0.3405465308... . - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
%F a(n+1) = 3*c*((1/3)*(a+b+1))^n/(c^2-2*c+4) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3), c=(586+102*sqrt(33))^(1/3). Round to the nearest integer. - Al Hakanson (hawkuu(AT)gmail.com), Feb 02 2009
%F a(n) = round(3*((a+b+1)/3)^n/(a^2+b^2+4)) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3).. - _Anton Nikonov_
%F Another form of the g.f.: f(z) = (z^2-z^3)/(1-2*z+z^4). Then we obtain a(n) as a sum: a(n) = Sum_{i=0..floor((n-2)/4)} ((-1)^i*binomial(n-2-3*i,i)*2^(n-2-4*i)) - Sum_{i=0..floor((n-3)/4)} ((-1)^i*binomial(n-3-3*i,i)*2^(n-3-4*i)) with natural convention: Sum_{i=m..n} alpha(i) = 0 for m > n. - _Richard Choulet_, Feb 22 2010
%F a(n+2) = Sum_{k=0..n} Sum_{i=k..n, mod(4*k-i,3)=0} binomial(k,(4*k-i)/3)*(-1)^((i-k)/3)*binomial(n-i+k-1,k-1). - _Vladimir Kruchinin_, Aug 18 2010
%F a(n) = 2*a(n-2) + 2*a(n-3) + a(n-4). - _Gary Detlefs_, Sep 13 2010
%F Sum_{k=0..2*n} a(k+b)*A027907(n,k) = a(3*n+b), b >= 0 (see A099464, A074581).
%F a(n) = 2*a(n-1) - a(n-4), with a(0)=a(1)=0, a(2)=a(3)=1. - _Vincenzo Librandi_, Dec 20 2010
%F Starting (1, 2, 4, 7, ...) is the INVERT transform of (1, 1, 1, 0, 0, 0, ...). - _Gary W. Adamson_, May 13 2013
%F 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
%F a(n+2) = Sum_{j=0..floor(n/2)} Sum_{k=0..j} binomial(n-2*j,k)*binomial(j,k)*2^k. - _Tony Foster III_, Sep 08 2017
%F Sum_{k=0..n} (n-k)*a(k) = (a(n+2) + a(n+1) - n - 1)/2. See A062544. - _Yichen Wang_, Aug 20 2020
%F a(n) = A008937(n-1) - A008937(n-2) for n >= 2. - _Peter Luschny_, Aug 20 2020
%F From _Yichen Wang_, Aug 27 2020: (Start)
%F Sum_{k=0..n} a(k) = (a(n+2) + a(n) - 1)/2. See A008937.
%F Sum_{k=0..n} k*a(k) = ((n-1)*a(n+2) - a(n+1) + n*a(n) + 1)/2. See A337282. (End)
%F For n > 1, a(n) = b(n) where b(1) = 1 and then b(n) = Sum_{k=1..n-1} b(n-k)*A000931(k+2). - _J. Conrad_, Nov 24 2022
%F Conjecture: the congruence a(n*p^(k+1)) + a(n*p^k) + a(n*p^(k-1)) == 0 (mod p^k) holds for positive integers k and n and for all the primes p listed in A106282. - _Peter Bala_, Dec 28 2022
%F Sum_{k=0..n} k^2*a(k) = ((n^2-4*n+6)*a(n+1) - (2*n^2-2*n+5)*a(n) + (n^2-2*n+3)*a(n-1) - 3)/2. - _Prabha Sivaramannair_, Feb 10 2024
%F a(n) = Sum_{r root of x^3-x^2-x-1} r^n/(3*r^2-2*r-1). - _Fabian Pereyra_, Nov 23 2024
%e G.f. = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + 24*x^8 + 44*x^9 + 81*x^10 + ...
%p a:= n-> (<<0|1|0>, <0|0|1>, <1|1|1>>^n)[1,3]:
%p seq(a(n), n=0..40); # _Alois P. Heinz_, Dec 19 2016
%p # second Maple program:
%p A000073:=proc(n) option remember; if n <= 1 then 0 elif n=2 then 1 else procname(n-1)+procname(n-2)+procname(n-3); fi; end; # _N. J. A. Sloane_, Aug 06 2018
%t CoefficientList[Series[x^2/(1 - x - x^2 - x^3), {x, 0, 50}], x]
%t a[0] = a[1] = 0; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; Array[a, 36, 0] (* _Robert G. Wilson v_, Nov 07 2010 *)
%t LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 60] (* _Vladimir Joseph Stephan Orlovsky_, May 24 2011 *)
%t a[n_] := SeriesCoefficient[If[ n < 0, x/(1 + x + x^2 - x^3), x^2/(1 - x - x^2 - x^3)], {x, 0, Abs @ n}] (* _Michael Somos_, Jun 01 2013 *)
%t Table[-RootSum[-1 - # - #^2 + #^3 &, -#^n - 9 #^(n + 1) + 4 #^(n + 2) &]/22, {n, 0, 20}] (* _Eric W. Weisstein_, Nov 09 2017 *)
%o (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 */
%o (PARI) my(x='x+O('x^99)); concat([0, 0], Vec(x^2/(1-x-x^2-x^3))) \\ _Altug Alkan_, Apr 04 2016
%o (PARI) a(n)=([0,1,0;0,0,1;1,1,1]^n)[1,3] \\ _Charles R Greathouse IV_, Apr 18 2016, simplified by _M. F. Hasler_, Apr 18 2018
%o (Maxima) A000073[0]:0$
%o A000073[1]:0$
%o A000073[2]:1$
%o A000073[n]:=A000073[n-1]+A000073[n-2]+A000073[n-3]$
%o makelist(A000073[n], n, 0, 40); /* _Emanuele Munarini_, Mar 01 2011 */
%o (Haskell)
%o a000073 n = a000073_list !! n
%o a000073_list = 0 : 0 : 1 : zipWith (+) a000073_list (tail
%o (zipWith (+) a000073_list $ tail a000073_list))
%o -- _Reinhard Zumkeller_, Dec 12 2011
%o (Python)
%o def a(n, adict={0:0, 1:0, 2:1}):
%o if n in adict:
%o return adict[n]
%o adict[n]=a(n-1)+a(n-2)+a(n-3)
%o return adict[n] # _David Nacin_, Mar 07 2012
%o from functools import cache
%o @cache
%o def A000073(n: int) -> int:
%o if n <= 1: return 0
%o if n == 2: return 1
%o return A000073(n-1) + A000073(n-2) + A000073(n-3) # _Peter Luschny_, Nov 21 2022
%o (Magma) [n le 3 select Floor(n/3) else Self(n-1)+Self(n-2)+Self(n-3): n in [1..70]]; // _Vincenzo Librandi_, Jan 29 2016
%o (GAP) a:=[0,0,1];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # _Muniru A Asiru_, Oct 24 2018
%Y Cf. A000045, A000078, A000213, A000931, A001590 (first differences, also a(n)+a(n+1)), A001644, A008288 (tribonacci triangle), A008937 (partial sums), A021913, A027024, A027083, A027084, A046738 (Pisano periods), A050231, A054668, A062544, A063401, A077902, A081172, A089068, A118390, A145027, A153462, A230216.
%Y A057597 is this sequence run backwards: A057597(n) = a(1-n).
%Y Row 3 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).
%Y Partitions: A240844 and A117546.
%Y Cf. also A092836 (subsequence of primes), A299399 = A092835 + 1 (indices of primes).
%K nonn,easy,nice,changed
%O 0,5
%A _N. J. A. Sloane_
%E Minor edits by _M. F. Hasler_, Apr 18 2018
%E Deleted certain dangerous or potentially dangerous links. - _N. J. A. Sloane_, Jan 30 2021