0,5

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 (deutsch(AT)duke.poly.edu), 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 (deutsch(AT)duke.poly.edu), 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 (pbarry(AT)wit.ie), Oct 15 2004

Number of permutations satisfying -k<=p(i)-i<=r, i=1..n-2, with k=1, r=2. - Vladimir Baltic (baltic(AT)matf.bg.ac.yu), 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 (deutsch(AT)duke.poly.edu), Apr 27 2006

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 (qntmpkt(AT)yahoo.com), 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. [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 27 2008]

For n>1: row sums of the triangle in A157897. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jun 25 2009]

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 47, ex. 4.

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.

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.

Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.4.

J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, Princeton, NJ, 1978.

M. E. Waddill and L. Sacks, Another generalized Fibonacci sequence, Fib. Quart., 5 (1967), 209-222.

T. D. Noe, Table of n, a(n) for n = 0..200

Index entries for sequences related to linear recurrences with constant coefficients

Joerg Arndt, Fxtbook

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

S. Kak, The Golden Mean and the Physics of Aesthetics

T. Mansour, Permutations avoiding a set of patterns T \subseteq S_3 and a pattern \tau \in S_4

S. 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.

S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

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

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 (qntmpkt(AT)yahoo.com), Dec 17 2004

a(n+2)=sum{k=0..n, T(n-k, k)}, T(n, k) = trinomial coefficients (A027907); - Paul Barry (pbarry(AT)wit.ie), 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 (reinhard.zumkeller(AT)gmail.com), May 22 2006

a(n)=3*c*((1/3)*(a+b+1))^n/(c^2-2*c+4) where a=(19+3*sqrt33)^(1/3), b=(19-3*sqrt33)^(1/3), c=(586+102*sqrt33)^(1/3). The offset is 1. a(3)=2. Round off to the nearest integer.[From Al Hakanson (hawkuu(AT)gmail.com), Feb 02 2009]

A000073:=-z/(-1+z+z**2+z**3); [S. Plouffe in his 1992 dissertation.]

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 */

sage: from sage.combinat.sloane_functions import recur_gen3 sage: it = recur_gen3(0, 0, 1, 1, 1, 1) sage: [it.next() for i in range(38)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 24 2008

Cf. A000213, A001590, A081172, A145027, A001644

Cf. A063401, A008937, A089068, A027084.

Cf. A062544, A077902, A054668, A027083, A027024.

Cf. A118390.

A057597 is this sequence run backwards: A057597(n) = a(1-n).

Row 3 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).

A153462, A000931 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 27 2008]

Sequence in context: A107281 A006744 A054175 this_sequence A160254 A005318 A102111

Adjacent sequences: A000070 A000071 A000072 this_sequence A000074 A000075 A000076

nonn,easy,nice

N. J. A. Sloane (njas(AT)research.att.com).

More terms from Larry Reeves (larryr(AT)acm.org), Jul 31 2000