This site is supported by donations to The OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A000073 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) with a(0)=a(1)=0, a(2)=1. (Formerly M1074 N0406) 218
 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537, 2555757, 4700770, 8646064, 15902591, 29249425, 53798080, 98950096, 181997601, 334745777, 615693474, 1132436852 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS Also (for n>2) number of ways writing 2^(n-2) as a product of decimal digits of some other number which has no digits equal to 1; e.g. n=8: 2^n=256, solutions = {488, ..., 8822, ..84222, .., 822222, ...4222222, 22222222}, their number is 81; so a(n+2)=A067374(2^n) - Labos E. (labos(AT)ana.sote.hu), Jan 28 2002. 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 a(n)=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 Let A=[0,0,1;1,1,1;0,1,0]. A000073(n) corresponds to both the (1,2) and (3,1) positions in A^n. - Paul Barry, Oct 15 2004 Number of permutations satisfying -k<=p(i)-i<=r, i=1..n-2, with k=1, r=2. - Vladimir Baltic, Jan 17 2005 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 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 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 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 the two others roots (complex) r1=m+pI r2=m-pI where m= (1-c)/2 (m=-0,4196433776) and p = ((3*c-5)*(c+1)/4)^(1/2) (p=0,6062907292) and where j= 1/((c-m)^2+p^2) (=0,1828035330) k = a+bI and l =a-bI where a= -j/2 (a=-0,0914017665) and b=(c-m)/(2*p*((c-m)^2+p^2)(b=0,3405465308) - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007 Convolved with the Padovan sequence = row sums of triangle A153462. [ Gary W. Adamson, Dec 27 2008] For n>1: row sums of the triangle in A157897. [Reinhard Zumkeller, Jun 25 2009] REFERENCES A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 47, ex. 4. Dairyko, Michael; Tyner, Samantha; Pudwell, Lara; Wynn, Casey. Non-contiguous pattern avoidance in binary trees. Electron. J. Combin. 19 (2012), no. 3, Paper 22, 21 pp. MR2967227. - From N. J. A. Sloane, Feb 01 2013 M. S. El Naschie, Statistical geometry of a Cantor discretum and semiconductors, Computers Math. Applic., 29 (No, 12, 1995), 103-110. Nathaniel D. Emerson, A Family of Meta-Fibonacci Sequences Defined by Variable-Order Recursions, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.8. M. Feinberg, Fibonacci-Tribonacci, Fib. Quart. 1(#3) (1963), 71-74. M. Feinberg, New slants, Fib. Quart., 2 (1964), 223-227. S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.2. M. D. Hirschhorn, Coupled third-order recurrences, Fib. Quart., 44 (2006), 26-31. F. T. Howard and Curtis Cooper, Some identities for r-Fibonacci numbers, http://www.math-cs.ucmo.edu/~curtisc/articles/howardcooper/genfib4.pdf. M. Janjic, Determinants and Recurrence Sequences, Journal of Integer Sequences, 2012, Article 12.3.5. - From N. J. A. Sloane, Sep 16 2012 O. Martin, A. M. Odlyzko and S. Wolfram, Algebraic properties of cellular automata, Comm. Math. Physics, 93 (1984), pp. 219-258, Reprinted in Theory and Applications of Cellular Automata, S. Wolfram, Ed., World Scientific, 1986, pp. 51-90 and in Cellular Automata and Complexity: Collected Papers of Stephen Wolfram, Addison-Wesley, 1994, pp. 71-113. See Eq. 5.5b. L. Pudwell, Pattern avoidance in trees (slides from a talk, mentions many sequences), http://faculty.valpo.edu/lpudwell/slides/notredame.pdf, 2012. - From N. J. A. Sloane, Jan 03 2013 J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, Princeton, NJ, 1978. N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). M. E. Waddill and L. Sacks, Another generalized Fibonacci sequence, Fib. Quart., 5 (1967), 209-222. LINKS T. D. Noe, Table of n, a(n) for n = 0..200 Joerg Arndt, Fxtbook, pp.307-309 Vladimir Baltic, On the number of certain types of strongly restricted permutations, Applicable Analysis and Discrete Mathematics Vol. 4, No 1 (April, 2010), 119-135 P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5. INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 10 Vladimir Kruchinin Composition of ordinary generating functions Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4 _Simon Plouffe_, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \{a} Montr\'{e}al, 1992. _Simon Plouffe_, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \{a} Montr\'{e}al, 1992. Eric Weisstein's World of Mathematics, Fibonacci n-Step Number Eric Weisstein's World of Mathematics, Tribonacci Number Wikipedia, Generalizations of Fibonacci numbers FORMULA G.f.: x^2/(1 - x - x^2 - x^3). G.f.: x^2 / (1 - x / (1 - x / (1 + x^2 / (1 + x)))). - Michael Somos, May 12 2012 a(n+1)/a(n) -> A058265. a(n) = center term in M^n * [1 0 0] where M = the 3X3 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...an eigenvalue of M and a root of x^3 - x^2 - x - 1 = 0. - Gary W. Adamson, Dec 17 2004 a(n+2) = sum{k=0..n, T(n-k, k)}, T(n, k) = trinomial coefficients (A027907); - Paul Barry, Feb 15 2005 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 a(n) = 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). The offset is 1. a(3)=2. Round off to the nearest integer. [Al Hakanson (hawkuu(AT)gmail.com), Feb 02 2009] Another form of the g.f.: f(z)=(z^2-z^3)/(1-2*z+z^4). Then we obtain a(n) as sum: a(n)=sum((-1)^i*binomial(n-2-3*i,i)*2^(n-2-4*i),i=0..floor((n-2)/4))-sum((-1)^i*binomial(n-3-3*i,i)*2^(n-3-4*i),i=0..floor((n-3)/4)) with natural convention: sum(alpha(i),i=m..n)=0 for m>n. [Richard Choulet (richard.choulet(AT)orange.fr), Feb 22 2010] a(n) = sum(k=1..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)) [Kruchinin Vladimir, Aug 18 2010] a(n) = 2*a(n-2)+2*a(n-3)+a(n-4) [Gary Detlefs, Sep 13 2010] sum_{k=0..2*n} A000073(k+b)*A027907(n,k) = A000073(3*n+b), b>=0 (see A099464, A074581). a(0)=a(1)=0, a(2)=a(3)=1, a(n)=2*a(n-1)-a(n-4). [Vincenzo Librandi, Dec 20 2010] For n>0 Lim_{n -> Inf} a(n-1)/a(n) = 1/((1+(19+3*sqrt(33))^(1/3)+(19-3*sqrt(33))^(1/3))/3)=0.54368901269... a root of 1-x-x^2-x^3. [Tim Monahan Aug 13 2011] G.f.: x^2*(1+x/(U(0) - x))/(1-x^2)  where U(k)=  1 - x^2/(1 - 1/(1 + 1/U(k+1)));(continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 16 2012 G.f.: x^2 + x^3*(G(0) - 1)/(x-1) where G(k) =  1 - (1+x+x^2)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 17 2013 G.f.: x^2 + x^3*(G(0)-1)/(x+1) where G(k) =  1 + (1+x+x^2)/(1-x/(x+1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 26 2013 Starting (1, 2, 4, 7,...) is the INVERT transform of (1, 1, 1, 0, 0, 0,...). [Gary W. Adamson, May 13 2013] MAPLE A000073:=-z/(-1+z+z**2+z**3); [Simon Plouffe in his 1992 dissertation.] a:=taylor((z^2-z^3)/(1-2*z+z^4), z=0, 31); for p from 0 to 30 do j(p):=coeff(a, z, p):od :seq(j(p), p=0..30); for n from 0 to 30 do k(n):=sum((-1)^i*binomial(n-2-3*i, i)*2^(n-2-4*i), i=0..floor((n-2)/4))-sum((-1)^i*binomial(n-3-3*i, i)*2^(n-3-4*i), i=0..floor((n-3)/4)):od:seq(k(n), n=0..30); [Richard Choulet (richard.choulet(AT)orange.fr), Feb 22 2010] MATHEMATICA CoefficientList[Series[x^2/(1 - x - x^2 - x^3), {x, 0, 50}], x] 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 *) LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, May 24 2011 *) PROG (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 */ (Maxima) a(n):=sum(sum(if mod(4*k-i, 3)>0 then 0 else binomial(k, (4*k-i)/3)*(-1)^((i-k)/3)*binomial(n-i+k-1, k-1), i, k, n), k, 1, n); [Kruchinin Vladimir, Aug 18 2010] (Maxima) A000073[0]:0$A000073[1]:0$ A000073[2]:1$A000073[n]:=A000073[n-1]+A000073[n-2]+A000073[n-3]$   makelist(A000073[n], n, 0, 40);  /* Emanuele Munarini, Mar 01 2011 */ (Haskell) a000073 n = a000073_list !! n a000073_list = 0 : 0 : 1 : zipWith (+) a000073_list (tail                           (zipWith (+) a000073_list \$ tail a000073_list)) -- Reinhard Zumkeller, Dec 12 2011 (Python) def a(n, adict={0:0, 1:0, 2:1}): .if n in adict: ..return adict[n] .adict[n]=a(n-1)+a(n-2)+a(n-3) .return adict[n] #- David Nacin, Mar 07 2012 CROSSREFS Cf. A000045, A000073, A000213, A001590, A081172, A145027, A001644, A063401, A008937, A089068, A027084, A062544, A077902, A054668, A027083, A027024, A118390, A050231, A046738 (Pisano periods). A057597 is this sequence run backwards: A057597(n) = a(1-n). Row 3 of arrays A048887 and A092921 (k-generalized Fibonacci numbers). Cf. A153462, A000931. Sequence in context: A107281 A006744 A054175 * A160254 A005318 A102111 Adjacent sequences:  A000070 A000071 A000072 * A000074 A000075 A000076 KEYWORD nonn,easy,nice,changed AUTHOR EXTENSIONS More terms from Larry Reeves (larryr(AT)acm.org), Jul 31 2000 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Puzzles | Hot | Classics
Recent Additions | More pages | Superseeker | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement .