

A000071


Fibonacci numbers  1.
(Formerly M1056 N0397)


199



0, 0, 1, 2, 4, 7, 12, 20, 33, 54, 88, 143, 232, 376, 609, 986, 1596, 2583, 4180, 6764, 10945, 17710, 28656, 46367, 75024, 121392, 196417, 317810, 514228, 832039, 1346268, 2178308, 3524577, 5702886, 9227464, 14930351, 24157816, 39088168
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OFFSET

1,4


COMMENTS

a(n) is the number of allowable transition rules for passing from one change to the next (on n1 bells) in the English art of bellringing. This is also the number of involutions in the symmetric group S_{n1} which can be represented as a product of transpositions of consecutive numbers from {1,2,...,n1}. Thus for n=6 we have a(6) = from (12), (12)(34), (12)(45), (23), (23)(45), (34), (45), for instance. See my 1983 Math. Proc. Camb. Phil. Soc. paper.  Arthur T. White, letter to N. J. A. Sloane, Dec 18 1986
Number of permutations p of {1,2,...,n1} such that maxp(i)i=1. Example: a(4)=2 since only the permutations 132 and 213 of {1,2,3} satisfy the given condition.  Emeric Deutsch, Jun 04 2003
Number of 001avoiding binary words of length n3.
Also, sum of first n Fibonacci numbers.  Giorgi Dalakishvili (mcnamara_gio(AT)yahoo.com), Apr 02 2005
a(n)=number of partitions of {1,...,n1} into two blocks in which only 1 or 2strings of consecutive integers can appear in a block and there is at least one 2string. E.g. a(6) = 7 because the enumerated partitions of {1,2,3,4,5} are 124/35,134/25, 14/235,13/245,1245/3,145/23,125/34.  Augustine O. Munagi, Apr 11 2005
Numbers for which only one Fibonacci bitrepresentation is possible and for which the maximal and minimal Fibonacci bitrepresentations (A104326 and A014417) are equal. For example, a(12) = 10101 because 8+3+1 = 12.  Casey Mongoven, Mar 19 2006
Beginning with a(2), the 'Recamán transform' (see A005132) of the Fibonacci numbers (A000045).  Nick Hobson (nickh(AT)qbyte.org), Mar 01 2007
Starting with nonzero terms = row sums of triangle A158950.  Gary W. Adamson, Mar 31 2009
a(n+2) = minimum number of elements in an AVL tree of height n.  Lennert Buytenhek (buytenh(AT)wantstofly.org), May 31 2010
a(n) is the number of branch nodes in the Fibonacci tree of order n1. A Fibonacci tree of order n (n>=2) is a complete binary tree whose left subtree is the Fibonacci tree of order n1 and whose right subtree is the Fibonacci tree of order n2; each of the Fibonacci trees of order 0 and 1 is defined as a single node (see the Knuth reference, p. 417).  Emeric Deutsch, Jun 14 2010
a(n+3) is the number of distinct threestrand positive braids of length n (cf. Burckel).  Maxime Bourrigan, Apr 04 2011
A000119(a(n)) = 1.  Reinhard Zumkeller, Dec 28 2012
a(n+1) is the number of compositions of n with maximal part 2.  Joerg Arndt, May 21 2013
a(n) = A228074(n1,2) for n > 2.  Reinhard Zumkeller, Aug 15 2013
a(0) = 1. For a(5) = 4 we have 2143, 1324, 2134 and 1243.  Jon Perry, Sep 14 2013
a(n+2) is the number of leafs of greatgrandparent DAG (directed acyclic graph) of height n. A greatgrandparent DAG of height n is a single node for n=1; for n>1 each leaf of ggpDAG(n1) has two child nodes where pairs of adjacent new nodes are merged into single node iff they have disjoint grandparents and same greatgrandparent. Consequence: a(n) = 2*a(n1)  a(n3).  Hermann StammWilbrandt, Jul 06 2014
A083368(a(n+3)) = n.  Reinhard Zumkeller, Aug 10 2014
2 and 7 are the only prime numbers in this sequence.  Emmanuel Vantieghem, Oct 01 2014
From Russell Jay Hendel, Mar 15 2015: (Start)
We can settle Gerald McGarvey's conjecture mentioned below in the Formula section, however we require n>4. We need the following 4 prerequisites.
(1) a(n)=F(n)1, with {F(n)}_{n>=1} the Fibonacci numbers A000045. (2) (Binet Form) F(n) =(d^ne^n)/sqrt(5) with d=phi and e=1phi, de=1 and d+e=1. It follows that a(n)=(d(n)e(n))/sqrt(5)1. (3) To prove floor(x)=y is equivalent to proving that xy lies in the semiopen interval [0,1). (4) The series {s(n)=c1 x^n+c2}_{n>=1}, with 1<x<0, and c1 and c2 positive constants, converges by oscillation with s(1)<s(3)<s(5) < ... < s(6)<s(4)<s(2). If follows that for any odd n, the open interval (s(n),s(n+1)) contains the subsequence {s(t)}_{t>=n+2}. Using these prerequisites we can analyze the conjecture.
Using prerequisites (2) and (3) we see we must prove, for all n>4, that d(d^(n1)e^(n1))/sqrt(5)1)  (d^ne^n)/sqrt(5)+1+c lies in the interval [0,1). But de=1, implying de^(n1)=e^(n2). It follows that we must equivalently prove (for all n>4) that E(n,c)=(e^(n2)+e^n)/sqrt(5)+1d + c = e^(n2) (e^2+1)/sqrt(5)+e +c lies in [0,1). Clearly, for any particular n, E(n,c) has extrema (maxima, minima) when c=2(1d) and c=(1+d)(1d). Therefore, the proof is completed by using prerequistite (4). It suffices to verify E(5,2(1d))=0, E(6,2(1d))=0.236068, E(5,(1d)(1+d))=0.618034, E(6,(1d)(1+d))=0.854102, all lie in [0,1).
(End)


REFERENCES

Mohammad K. Azarian, The Generating Function for the Fibonacci Sequence, Missouri Journal of Mathematical Sciences, Vol. 2, No. 2, Spring 1990, pp. 7879. Zentralblatt MATH, Zbl 1097.11516.
Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem II, Missouri Journal of Mathematical Sciences, Vol. 16, No. 1, Winter 2004, pp. 1217.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 1.
M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 64.
D. E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, AddisonWesley, Reading, MA, 1998, p. 417. [From Emeric Deutsch, Jun 14 2010]
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 155.
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).
J. L. Yucas, Counting special sets of binary Lyndon words, Ars Combin., 31 (1991), 2129.


LINKS

Christian G. Bower, Table of n, a(n) for n = 1..500
J.L. Baril, J.M. Pallo, Motzkin subposet and Motzkin geodesics in Tamari lattices, 2013.
Andrew M. Baxter, Lara K. Pudwell, Ascent sequences avoiding pairs of patterns, 2014.
Serge Burckel, Syntactical methods for braids of three strands, J. Symbolic Comput. 31 (2001), no. 5, 557564.
A. Burstein and T. Mansour, Counting occurrences of some subword patterns, arXiv:math/0204320 [math.CO], 20022003.
Peter J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
Fan Chung, Ron Graham, Primitive juggling sequences, Am. Math. Monthly 115 (3) (2008) 185194
Michael Dairyko, Samantha Tyner, Lara Pudwell, Casey Wynn, Noncontiguous 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
Emeric Deutsch, Problem Q915, Math. Magazine, vol. 74, No. 5, 2001, p. 404.
Fumio Hazama, Spectra of graphs attached to the space of melodies, Discrete Math., 311 (2011), 23682383. See Table 2.1.
Yasuichi Horibe, An entropy view of Fibonacci trees, Fibonacci Quarterly, 20, No. 2, 1982, 168178. [From Emeric Deutsch, Jun 14 2010]
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 384
R. Lagrange, Quelques résultats dans la métrique des permutations, Annales Scientifiques de l'Ecole Normale Supérieure, Paris, 79 (1962), 199241.
D. A. Lind, On a class of nonlinear binomial sums, Fib. Quart., 3 (1965), 292298.
Rui Liu and FengZhen Zhao, On the Sums of Reciprocal Hyperfibonacci Numbers and Hyperlucas Numbers, Journal of Integer Sequences, Vol. 15 (2012), #12.4.5.  From N. J. A. Sloane, Oct 05 2012
Augustine O. Munagi, Set Partitions with Successions and Separations,IJMMS 2005:3 (2005), 451463.
Sam Northshield, Stern's Diatomic Sequence 0,1,1,2,1,3,2,3,1,4,..., Amer. Math. Month., Vol. 117 (7), pp. 581598, 2010.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.
Lara Pudwell, Pattern avoidance in trees, (slides from a talk, mentions many sequences), 2012.  From N. J. A. Sloane, Jan 03 2013
Stacey Wagner, Enumerating Alternating Permutations with One Alternating Descent, DePaul Discoveries: Vol. 2: Iss. 1, Article 2.
Arthur T. White, Ringing the changes, Math. Proc. Cambridge Philos. Soc. 94 (1983), no. 2, 203215.
Peijun Xu, Growth of positive braids semigroups, Journal of Pure and Applied Algebra, 1992.
LiNa Zheng, Rui Liu, and FengZhen Zhao, On the LogConcavity of the Hyperfibonacci Numbers and the Hyperlucas Numbers, Journal of Integer Sequences, Vol. 17 (2014), #14.1.4.
Index entries for linear recurrences with constant coefficients, signature (2,0,1).


FORMULA

a(n) = A000045(n)  1.
a(0) = 1, a(1) = 0; thereafter a(n) = a(n1) + a(n2) + 1.
a(n) = 2*a(n1)  a(n3).  R. H. Hardin, Apr 02 2011
Partial sums of Fibonacci numbers. G.f.: x^3/((1  x  x^2)*(1  x)).  Wolfdieter Lang
a(n) = 1 + (A*B^n + C*D^n)/10, with A, C = 5 + 3*sqrt(5), B, D = (1 + sqrt(5))/2.  Ralf Stephan, Mar 02 2003
a(1) = 0, a(2) = 0, a(3) = 1, then a(n) = ceiling(phi*a(n1)) where phi is the golden ratio (1 + sqrt(5))/2.  Benoit Cloitre, May 06 2003
Conjecture: for all c such that 2*(2  Phi) <= c < (2 + Phi)*(2  Phi) we have a(n) = floor(Phi*a(n1) + c) for n > 4.  Gerald McGarvey, Jul 22 2004 (n > 3 changed to n > 4, Russell Jay Hendel, Mar 15 2015).
a(n) = sum{k = 0..floor((n2)/2), binomial(nk2, k+1)}.  Paul Barry, Sep 23 2004
a(n+3) = sum{k = 0..floor(n/3), binomial(n2k, k)(1)^k*2^(n3k)}.  Paul Barry, Oct 20 2004
a(n+1) = Sum(binomial(nr, r)), r = 1, 2, ... which is the case t = 2 and k = 2 in the general case of tstrings and k blocks: a(n+1, k, t) = Sum(binomial(nr*(t1), r)*S2(nr*(t1)1, k1)), r = 1, 2, ...  Augustine O. Munagi, Apr 11 2005
a(n) = Sum[k*Fibonacci(n  k  3), {k, 0, n2}].  Ross La Haye, May 31 2006
a(n) = term (3, 2) in the 3x3 matrix [1, 1, 0; 1, 0, 0; 1, 0, 1]^(n1).  Alois P. Heinz, Jul 24 2008
For n >= 4, a(n) = ceiling(phi*a(n1)), where phi is the golden ratio.  Vladimir Shevelev, Jul 04 2010
Closedform without two leading zeros G.f.: 1/(1  2*x  x^3); ((5 + 2*sqrt(5))*((1 + sqrt(5))/2)^n + (5  2*sqrt(5))*((1  sqrt(5))/2)^n  5)/5; closedform with two leading zeros G.f.: x^2/(1  2*x  x^3); ((5 + sqrt(5))*((1 + sqrt(5))/2)^n + (5  sqrt(5))*((1  sqrt(5))/2)^n  10)/10.  Tim Monahan, Jul 10 2011
G.f.: Q(0)*x^2/2, where Q(k) = 1 + 1/(1  x*(4*k + 2  x^2)/( x*(4*k + 4  x^2) + 1/Q(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Aug 30 2013


MAPLE

a[0]:=0:a[1]:=0:for n from 2 to 50 do a[n]:=a[n1]+a[n2]+1 od: seq(a[n], n=0..50); # Kristof
A000071 := proc(n) combinat[fibonacci](n)1 ; end proc; # R. J. Mathar, Apr 07 2011
A000071:=1/(z1)/(z^2+z1); # Simon Plouffe in his 1992 dissertation, dropping initial zeros
a:= n> (Matrix([[1, 1, 0], [1, 0, 0], [1, 0, 1]])^(n1))[3, 2]; seq(a(n), n=1..50); # Alois P. Heinz, Jul 24 2008


MATHEMATICA

Fibonacci[Range[40]]  1 (* or *) LinearRecurrence[{2, 0, 1}, {0, 0, 1}, 40] (* Harvey P. Dale, Aug 23 2013 *)


PROG

(PARI) {a(n) = if( n<1, 0, fibonacci(n)1)};
(MAGMA) [Fibonacci(n)1: n in [1..150]] // Vincenzo Librandi, Apr 04 2011
(Haskell)
a000071 n = a000071_list !! n
a000071_list = map (subtract 1) $ tail a000045_list
 Reinhard Zumkeller, May 23 2013


CROSSREFS

Cf. A000045, A054761, A119282, A001654, A005968, A005969, A098531, A098532, A098533, A128697, A001611, A157725, A001911, A157726, A006327, A157727, A157728, A157729, A167616, A158950, A105488, A105489.
Antidiagonal sums of array A004070.
Righthand column 2 of triangle A011794.
a(n) = A101220(1, 1, n2), for n > 1.
Related to sum of Fibonacci(kn) over n. Cf. A099919, A058038, A138134, A053606.
Subsequence of A226538.
Sequence in context: A126348 A006731 A222036 * A179111 A093607 A005182
Adjacent sequences: A000068 A000069 A000070 * A000072 A000073 A000074


KEYWORD

nonn,easy,nice,hear


AUTHOR

N. J. A. Sloane


EXTENSIONS

Edited by N. J. A. Sloane, Apr 04 2011


STATUS

approved



