cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A000071 a(n) = Fibonacci(n) - 1.

Original entry on oeis.org

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, 63245985, 102334154
Offset: 1

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Keywords

Comments

a(n) is the number of allowable transition rules for passing from one change to the next (on n-1 bells) in the English art of bell-ringing. This is also the number of involutions in the symmetric group S_{n-1} which can be represented as a product of transpositions of consecutive numbers from {1, 2, ..., n-1}. 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, ..., n-1} such that max|p(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 [For a(5) = 4 we have 2143, 1324, 2134 and 1243. - Jon Perry, Sep 14 2013]
Number of 001-avoiding binary words of length n-3. a(n) is the number of partitions of {1, ..., n-1} into two blocks in which only 1- or 2-strings of consecutive integers can appear in a block and there is at least one 2-string. 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 bit-representation is possible and for which the maximal and minimal Fibonacci bit-representations (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, Mar 01 2007
Starting with nonzero terms, a(n) gives the row sums of triangle A158950. - Gary W. Adamson, Mar 31 2009
a(n+2) is the 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 n-1. A Fibonacci tree of order n (n >= 2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; 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 three-strand positive braids of length n (cf. Burckel). - Maxime Bourrigan, Apr 04 2011
a(n+1) is the number of compositions of n with maximal part 2. - Joerg Arndt, May 21 2013
a(n+2) is the number of leafs of great-grandparent DAG (directed acyclic graph) of height n. A great-grandparent DAG of height n is a single node for n = 1; for n > 1 each leaf of ggpDAG(n-1) has two child nodes where pairs of adjacent new nodes are merged into single node if and only if they have disjoint grandparents and same great-grandparent. Consequence: a(n) = 2*a(n-1) - a(n-3). - Hermann Stamm-Wilbrandt, Jul 06 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 establish Gerald McGarvey's conjecture mentioned in the Formula section, however we require n > 4. We need the following 4 prerequisites.
(1) a(n) = F(n) - 1, with {F(n)}A000045.%20(2)%20(Binet%20form)%20F(n)%20=%20(d%5En%20-%20e%5En)/sqrt(5)%20with%20d%20=%20phi%20and%20e%20=%201%20-%20phi,%20de%20=%20-1%20and%20d%20+%20e%20=%201.%20It%20follows%20that%20a(n)%20=%20(d(n)%20-%20e(n))/sqrt(5)%20-%201.%20(3)%20To%20prove%20floor(x)%20=%20y%20is%20equivalent%20to%20proving%20that%20x%20-%20y%20lies%20in%20the%20half-open%20interval%20%5B0,%201).%20(4)%20The%20series%20%7Bs(n)%20=%20c1%20x%5En%20+%20c2%7D">{n >= 1} the Fibonacci numbers A000045. (2) (Binet form) F(n) = (d^n - e^n)/sqrt(5) with d = phi and e = 1 - phi, 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 x - y lies in the half-open 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^(n-1) - e^(n-1))/sqrt(5) - 1) - (d^n - e^n)/sqrt(5) + 1 + c lies in the interval [0, 1). But de = -1, implying de^(n-1) = -e^(n-2). It follows that we must equivalently prove (for all n > 4) that E(n, c) = (e^(n-2) + e^n)/sqrt(5) + 1 - d + c = e^(n-2) (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*(1-d) and c = (1+d)*(1-d). Therefore, the proof is completed by using prerequisite (4). It suffices to verify E(5, 2*(1-d)) = 0, E(6, 2*(1-d)) = 0.236068, E(5, (1-d)*(1+d)) = 0.618034, E(6, (1-d)*(1+d)) = 0.854102, all lie in [0, 1).
(End)
a(n) can be shown to be the number of distinct nonempty matchings on a path with n vertices. (A matching is a collection of disjoint edges.) - Andrew Penland, Feb 14 2017
Also, for n > 3, the lexicographically earliest sequence of positive integers such that {phi*a(n)} is located strictly between {phi*a(n-1)} and {phi*a(n-2)}. - Ivan Neretin, Mar 23 2017
From Eric M. Schmidt, Jul 17 2017: (Start)
Number of sequences (e(1), ..., e(n-2)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) != e(j) <= e(k). [Martinez and Savage, 2.5]
Number of sequences (e(1), ..., e(n-2)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) >= e(j) <= e(k) and e(i) != e(k). [Martinez and Savage, 2.5]
(End)
Numbers whose Zeckendorf (A014417) and dual Zeckendorf (A104326) representations are the same: alternating digits of 1 and 0. - Amiram Eldar, Nov 01 2019
a(n+2) is the length of the longest array whose local maximum element can be found in at most n reveals. See link to the puzzle by Alexander S. Kulikov. - Dmitry Kamenetsky, Aug 08 2020
a(n+2) is the number of nonempty subsets of {1,2,...,n} that contain no consecutive elements. For example, the a(6)=7 subsets of {1,2,3,4} are {1}, {2}, {3}, {4}, {1,3}, {1,4} and {2,4}. - Muge Olucoglu, Mar 21 2021
a(n+3) is the number of allowed patterns of length n in the even shift (that is, a(n+3) is the number of binary words of length n in which there are an even number of 0s between any two occurrences of 1). For example, a(7)=12 and the 12 allowed patterns of length 4 in the even shift are 0000, 0001, 0010, 0011, 0100, 0110, 0111, 1000, 1001, 1100, 1110, 1111. - Zoran Sunic, Apr 06 2022
Conjecture: for k a positive odd integer, the sequence {a(k^n): n >= 1} is a strong divisibility sequence; that is, for n, m >= 1, gcd(a(k^n), a(k^m)) = a(k^gcd(n,m)). - Peter Bala, Dec 05 2022
In general, the sum of a second-order linear recurrence having signature (c,d) will be a third-order recurrence having a signature (c+1,d-c,-d). - Gary Detlefs, Jan 05 2023
a(n) is the number of binary strings of length n-2 whose longest run of 1's is of length 1, for n >= 3. - Félix Balado, Apr 03 2025

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 1.
  • GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See page 28.
  • M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 64.
  • D. E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.
  • 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), 21-29.

Crossrefs

Antidiagonal sums of array A004070.
Right-hand column 2 of triangle A011794.
Related to sum of Fibonacci(kn) over n. Cf. A099919, A058038, A138134, A053606.
Subsequence of A226538. Also a subsequence of A061489.

Programs

  • Haskell
    a000071 n = a000071_list !! n
    a000071_list = map (subtract 1) $ tail a000045_list
    -- Reinhard Zumkeller, May 23 2013
    
  • Magma
    [Fibonacci(n)-1: n in [1..60]]; // Vincenzo Librandi, Apr 04 2011
    
  • Maple
    A000071 := proc(n) combinat[fibonacci](n)-1 ; end proc; # R. J. Mathar, Apr 07 2011
    a:= n-> (Matrix([[1, 1, 0], [1, 0, 0], [1, 0, 1]])^(n-1))[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 *)
    Join[{0}, Accumulate[Fibonacci[Range[0, 39]]]] (* Alonso del Arte, Oct 22 2017, based on Giorgi Dalakishvili's formula *)
  • PARI
    {a(n) = if( n<1, 0, fibonacci(n)-1)};
    
  • SageMath
    [fibonacci(n)-1 for n in range(1,60)] # G. C. Greubel, Oct 21 2024

Formula

a(n) = A000045(n) - 1.
a(0) = -1, a(1) = 0; thereafter a(n) = a(n-1) + a(n-2) + 1.
a(n) = A101220(1, 1, n-2), for n > 1.
G.f.: x^3/((1-x-x^2)*(1-x)). - Simon Plouffe in his 1992 dissertation, dropping initial 0's
a(n) = 2*a(n-1) - a(n-3). - R. H. Hardin, Apr 02 2011
Partial sums of Fibonacci numbers. - 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(n-1)) 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(n-1) + c) for n > 4. - Gerald McGarvey, Jul 22 2004. This is true provided n > 3 is changed to n > 4, see proof in Comments section. - Russell Jay Hendel, Mar 15 2015
a(n) = Sum_{k = 0..floor((n-2)/2)} binomial(n-k-2, k+1). - Paul Barry, Sep 23 2004
a(n+3) = Sum_{k = 0..floor(n/3)} binomial(n-2*k, k)*(-1)^k*2^(n-3*k). - Paul Barry, Oct 20 2004
a(n+1) = Sum(binomial(n-r, r)), r = 1, 2, ... which is the case t = 2 and k = 2 in the general case of t-strings and k blocks: a(n+1, k, t) = Sum(binomial(n-r*(t-1), r)*S2(n-r*(t-1)-1, k-1)), r = 1, 2, ... - Augustine O. Munagi, Apr 11 2005
a(n) = Sum_{k = 0..n-2} k*Fibonacci(n - k - 3). - Ross La Haye, May 31 2006
a(n) = term (3, 2) in the 3 X 3 matrix [1, 1, 0; 1, 0, 0; 1, 0, 1]^(n-1). - Alois P. Heinz, Jul 24 2008
For n >= 4, a(n) = ceiling(phi*a(n-1)), where phi is the golden ratio. - Vladimir Shevelev, Jul 04 2010
Closed-form 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; closed-form with two leading 0's 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
A000119(a(n)) = 1. - Reinhard Zumkeller, Dec 28 2012
a(n) = A228074(n - 1, 2) for n > 2. - Reinhard Zumkeller, Aug 15 2013
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
A083368(a(n+3)) = n. - Reinhard Zumkeller, Aug 10 2014
E.g.f.: 1 - exp(x) + 2*exp(x/2)*sinh(sqrt(5)*x/2)/sqrt(5). - Ilya Gutkovskiy, Jun 15 2016
a(n) = A000032(3+n) - 1 mod A000045(3+n). - Mario C. Enriquez, Apr 01 2017
a(n) = Sum_{i=0..n-2} Fibonacci(i). - Giorgi Dalakishvili (mcnamara_gio(AT)yahoo.com), Apr 02 2005 [corrected by Doug Bell, Jun 01 2017]
a(n+2) = Sum_{j = 0..floor(n/2)} Sum_{k = 0..j} binomial(n - 2*j, k+1)*binomial(j, k). - Tony Foster III, Sep 08 2017
From Peter Bala, Nov 12 2021: (Start)
a(4*n) = Fibonacci(2*n+1)*Lucas(2*n-1) = A081006(n);
a(4*n+1) = Fibonacci(2*n)*Lucas(2*n+1) = A081007(n);
a(4*n+2) = Fibonacci(2*n)*Lucas(2*n+2) = A081008(n);
a(4*n+3) = Fibonacci(2*n+2)*Lucas(2*n+1) = A081009(n). (End)
G.f.: x^3/((1 - x - x^2)*(1 - x)) = Sum_{n >= 0} (-1)^n * x^(n+3) *( Product_{k = 1..n} (k - x)/Product_{k = 1..n+2} (1 - k*x) ) (a telescoping series). - Peter Bala, May 08 2024
Product_{n>=4} (1 + (-1)^n/a(n)) = 3*phi/4, where phi is the golden ratio (A001622). - Amiram Eldar, Nov 28 2024

Extensions

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

A001911 a(n) = Fibonacci(n+3) - 2.

Original entry on oeis.org

0, 1, 3, 6, 11, 19, 32, 53, 87, 142, 231, 375, 608, 985, 1595, 2582, 4179, 6763, 10944, 17709, 28655, 46366, 75023, 121391, 196416, 317809, 514227, 832038, 1346267, 2178307, 3524576, 5702885, 9227463, 14930350, 24157815, 39088167, 63245984
Offset: 0

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Keywords

Comments

This is the sequence A(0,1;1,1;2) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 17 2010
Ternary words of length n - 1 with subwords (0, 1), (0, 2) and (2, 2) not allowed. - Olivier Gérard, Aug 28 2012
For subsets of (1, 2, 3, 5, 8, 13, ...) Fibonacci Maximal terms (Cf. A181631) equals the number of leading 1's per subset. For example, (7-11) in Fibonacci Maximal = (1010, 1011, 1101, 1110, 1111), numbers of leading 1's = (1 + 1 + 2 + 3 + 4) = 11 = a(4) = row 4 of triangle A181631. - Gary W. Adamson, Nov 02 2010
As in our 2009 paper, we use two types of Fibonacci trees: - Ta: Fibonacci analog of binomial trees; Tb: Binary Fibonacci trees. Let D(r(k)) be the sum over all distances of the form d(r, x), across all vertices x of the tree rooted at r of order k. Ignoring r, but overloading, let D(a(k)) and D(b(k)) be the distance sums for the Fibonacci trees Ta and Tb respectively of the order k. Using the sum-of-product form in Equations (18) and (21) in our paper it follows that F(k+4) - 2 = D(a(k+1)) - D(b(k-1)). - K.V.Iyer and P. Venkata Subba Reddy, Apr 30 2011
a(n) is the length of the n-th palindromic prefix of the infinite Fibonacci word A003849. - Dimitri Hendriks, May 19 2014
The first k terms of the infinite Fibonacci word A014675 are palindromic if and only if k is a positive term of this sequence. - Clark Kimberling, Jul 14 2014
Can be expressed in terms of a rule similar to Recamán's sequence (A005132). Instead of following the Recamán rule "subtract if possible, otherwise add", this sequence follows the rule "If subtraction is possible, do nothing; otherwise add." For example when at the fourth term, 6, it is possible to subtract 4 (giving 2 which is not in {0, 1, 3, 6}) so nothing is done with 4. It is not possible to subtract 5 (6-5=1, which is in {0, 1, 3, 6}), so it is added, resulting in 11. - Matthew Malone, Jan 03 2020
For n>=1, a(n) is the maximum number of vertices (Moore bound) of a (1,1)-regular mixed graph with diameter n-1. - Miquel A. Fiol, Jun 24 2024
Repunits in the lazy Fibonacci representation (A104326), and which is the first row of array A372501. - A.H.M. Smeets, Jun 25 2025

Examples

			G.f. = x + 3*x^2 + 6*x^3 + 11*x^4 + 19*x^5 + 32*x^6 + 53*x^7 + 87*x^8 + ...
		

References

  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 233.
  • 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).

Crossrefs

Cf. A001611, A000071, A157725, A001911, A157726, A006327, A157727, A157728, A157729, A167616. [Added by N. J. A. Sloane, Jun 25 2010 in response to a comment from Aviezri S. Fraenkel]
Partial sums of A000045(n+1).
Right-hand column 3 of triangle A011794.
See also A165910.
Subsequence of A226538.
Column k=3 of A261019.

Programs

  • Haskell
    a001911 n = a001911_list !! n
    a001911_list = 0 : 1 : map (+ 2) (zipWith (+) a001911_list $ tail a001911_list)
    -- Reinhard Zumkeller, Jun 18 2013
    
  • Magma
    [(Fibonacci(n+3))-2: n in [0..85]]; // Vincenzo Librandi, Apr 23 2011
    
  • Maple
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+a[n-2]+2 od: seq(a[n],n=0..50); # Miklos Kristof, Mar 09 2005
    A001911:=(1+z)/(z-1)/(z**2+z-1); # Simon Plouffe in his 1992 dissertation with another offset
    a:= n-> (Matrix([[0,-1,1]]). Matrix([[1,1,0], [1,0,0], [2,0,1]])^n)[1,1]: seq(a(n), n=0..50); # Alois P. Heinz, Jul 24 2008
  • Mathematica
    Table[Fibonacci[n+3] -2, {n,0,50}] (* Vladimir Joseph Stephan Orlovsky, Nov 19 2010 *)
    LinearRecurrence[{2,0,-1}, {0,1,3}, 40] (* Harvey P. Dale, Jun 06 2011 *)
    Fibonacci[Range[3,40]]-2 (* Harvey P. Dale, Jun 28 2015 *)
  • PARI
    a(n)=fibonacci(n+3)-2 \\ Charles R Greathouse IV, Mar 14 2012
    
  • SageMath
    [fibonacci(n+3)-2 for n in range(60)] # G. C. Greubel, Oct 21 2024

Formula

From Michael Somos, Jun 09 1999: (Start)
a(n) = A000045(n+3) - 2.
a(n) = a(n-1) + a(n-2) + 2, a(0) = 0, a(1) = 1. (End)
G.f.: x*(1+x)/((1-x)*(1-x-x^2)).
Sum of consecutive pairs of A000071 (partial sums of Fibonacci numbers). - Paul Barry, Apr 17 2004
a(n) = A101220(2, 1, n). - Ross La Haye, Jan 28 2005
a(n) = A108617(n+1, 2) = A108617(n+1, n-1) for n > 0. - Reinhard Zumkeller, Jun 12 2005
a(n) = term (1,1) in the 1 X 3 matrix [0,-1,1].[1,1,0; 1,0,0; 2,0,1]^n. - Alois P. Heinz, Jul 24 2008
a(0) = 0, a(1) = 1, a(2) = 3, a(n) = 2*a(n-1)-a(n-3). - Harvey P. Dale, Jun 06 2011
Eigensequence of an infinite lower triangular matrix with the natural numbers as the left border and (1, 0, 1, 0, ...) in all other columns. - Gary W. Adamson, Jan 30 2012
a(n) = (-2+(2^(-n)*((1-sqrt(5))^n*(-2+sqrt(5))+(1+sqrt(5))^n*(2+sqrt(5))))/sqrt(5)). - Colin Barker, May 12 2016
a(n) = A000032(6+n)-1 mod A000045(6+n). - Mario C. Enriquez, Apr 01 2017
E.g.f.: 2*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 2*sqrt(5)*sinh(sqrt(5)*x/2))/5 - 2*exp(x). - Stefano Spezia, May 08 2022

Extensions

More terms and better description from Michael Somos

A001924 Apply partial sum operator twice to Fibonacci numbers.

Original entry on oeis.org

0, 1, 3, 7, 14, 26, 46, 79, 133, 221, 364, 596, 972, 1581, 2567, 4163, 6746, 10926, 17690, 28635, 46345, 75001, 121368, 196392, 317784, 514201, 832011, 1346239, 2178278, 3524546, 5702854, 9227431, 14930317, 24157781, 39088132, 63245948, 102334116, 165580101
Offset: 0

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Comments

Leading coefficients in certain rook polynomials (for n>=2; see p. 18 of the Riordan paper). - Emeric Deutsch, Mar 08 2004
(1, 3, 7, 14, ...) = row sums of triangle A141289. - Gary W. Adamson, Jun 22 2008
a(n) is the number of nonempty subsets of {1,2,...,n} such that the difference of successive elements is at most 2. See example below. Generally, the o.g.f. for the number of nonempty subsets of {1,2,...,n} such that the difference of successive elements is <= k is: x/((1-x)*(1-2*x+x^(k+1))). Cf. A000217 the case for k=1, A001477 the case for k=0 (counts singleton subsets). - Geoffrey Critzer, Feb 17 2012
-Fibonacci(n-2) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, Dec 31 2012
a(n) is the number of bit strings of length n+1 with the pattern 00 and without the pattern 011, see example. - John M. Campbell, Feb 10 2013
From Jianing Song, Apr 28 2025: (Start)
For n >= 2, a(n-2) is the number of subsets of {1,2,...,n} with 2 or more elements that contain no consecutive elements (i.e., such that the difference of successive elements is at least 2). Note that the number of such subsets with k elements is binomial(n+1-k,k), and Sum_{k=2..floor((n+1)/2)} binomial(n+1-k,k) = F(n+2) - binomial(n+1,0) - binomial(n,1) = F(n+2) - (n+1).
If subsets of {1,2,...,n} are required to contain no consecutive elements module n, then the result is A023548(n-3). (End)

Examples

			a(5) = 26 because there are 31 nonempty subsets of {1,2,3,4,5} but 5 of these have successive elements that differ by 3 or more: {1,4}, {1,5}, {2,5}, {1,2,5}, {1,4,5}. - _Geoffrey Critzer_, Feb 17 2012
From _John M. Campbell_, Feb 10 2013: (Start)
There are a(5) = 26 bit strings with the pattern 00 and without the pattern 011 of length 5+1:
   000000, 000001, 000010, 000100, 000101, 001000,
   001001, 001010, 010000, 010001, 010010, 010100,
   100000, 100001, 100010, 100100, 100101, 101000, 101001,
   110000, 110001, 110010, 110100, 111000, 111001, 111100.
(End)
		

References

  • J. Riordan, Discordant permutations, Scripta Math., 20 (1954), 14-23.
  • 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).

Crossrefs

Right-hand column 4 of triangle A011794.
Cf. A065220.

Programs

  • GAP
    List([0..40], n-> Fibonacci(n+4) -n-3); # G. C. Greubel, Jul 08 2019
  • Haskell
    a001924 n = a001924_list !! n
    a001924_list = drop 3 $ zipWith (-) (tail a000045_list) [0..]
    -- Reinhard Zumkeller, Nov 17 2013
    
  • Magma
    [Fibonacci(n+4)-(n+3): n in [0..40]]; // Vincenzo Librandi, Jun 23 2016
    
  • Maple
    A001924:=-1/(z**2+z-1)/(z-1)**2; # Conjectured by Simon Plouffe in his 1992 dissertation.
    ##
    a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <1|-1|-2|3>>^n.
             <<0, 1, 3, 7>>)[1, 1]:
    seq(a(n), n=0..40);  # Alois P. Heinz, Oct 05 2012
  • Mathematica
    a[n_]:= Fibonacci[n+4] -3-n; Array[a, 40, 0]  (* Robert G. Wilson v *)
    LinearRecurrence[{3,-2,-1,1},{0,1,3,7},40] (* Harvey P. Dale, Jan 24 2015 *)
    Nest[Accumulate,Fibonacci[Range[0,40]],2] (* Harvey P. Dale, Jun 15 2016 *)
  • PARI
    a(n)=fibonacci(n+4)-n-3 \\ Charles R Greathouse IV, Feb 24 2011
    
  • Sage
    [fibonacci(n+4) -n-3 for n in (0..40)] # G. C. Greubel, Jul 08 2019
    

Formula

From Wolfdieter Lang: (Start)
G.f.: x/((1-x-x^2)*(1-x)^2).
Convolution of natural numbers n >= 1 with Fibonacci numbers F(k).
a(n) = Fibonacci(n+4) - (3+n). (End)
From Henry Bottomley, Jan 03 2003: (Start)
a(n) = a(n-1) + a(n-2) + n = a(n-1) + A000071(n+2).
a(n) = A001891(n) - a(n-1) = n + A001891(n-1).
a(n) = A065220(n+4) + 1 = A000126(n+1) - 1. (End)
a(n) = Sum_{k=0..n} Sum_{i=0..k} Fibonacci(i). - Benoit Cloitre, Jan 26 2003
a(n) = (sqrt(5)/2 + 1/2)^n*(7*sqrt(5)/10 + 3/2) + (3/2 - 7*sqrt(5)/10)*(sqrt(5)/2 - 1/2)^n*(-1)^n - n - 3. - Paul Barry, Mar 26 2003
a(n) = Sum_{k=0..n} Fibonacci(k)*(n-k). - Benoit Cloitre, Jun 07 2004
A107909(a(n)) = A000225(n) = 2^n - 1. - Reinhard Zumkeller, May 28 2005
a(n) - a(n-1) = A101220(1,1,n). - Ross La Haye, May 31 2006
F(n) + a(n-3) = A133640(n). - Gary W. Adamson, Sep 19 2007
a(n) = A077880(-3-n) = 2*a(n-1) - a(n-3) + 1. - Michael Somos, Dec 31 2012
INVERT transform is A122595. PSUM transform is A014162. PSUMSIGN transform is A129696. BINOMIAL transform of A039834 with 0,1 prepended is this sequence. - Michael Somos, Dec 31 2012
a(n) = A228074(n+1,3) for n > 1. - Reinhard Zumkeller, Aug 15 2013
a(n) = Sum_{k=0..n} Sum_{i=0..n} i * C(n-k,k-i). - Wesley Ivan Hurt, Sep 21 2017
E.g.f.: exp(x/2)*(15*cosh(sqrt(5)*x/2) + 7*sqrt(5)*sinh(sqrt(5)*x/2))/5 - exp(x)*(3 + x). - Stefano Spezia, Jun 25 2022

Extensions

Description improved by N. J. A. Sloane, Jan 01 1997

A001891 Hit polynomials; convolution of natural numbers with Fibonacci numbers F(2), F(3), F(4), ....

Original entry on oeis.org

0, 1, 4, 10, 21, 40, 72, 125, 212, 354, 585, 960, 1568, 2553, 4148, 6730, 10909, 17672, 28616, 46325, 74980, 121346, 196369, 317760, 514176, 831985, 1346212, 2178250, 3524517, 5702824, 9227400, 14930285, 24157748, 39088098, 63245913, 102334080, 165580064
Offset: 0

Views

Author

Keywords

Comments

a(n) is the sum of the n-th row of the triangle in A119457 for n > 0. - Reinhard Zumkeller, May 20 2006
Convolution of odds (A005408) with Fibonacci numbers (A000045). - Graeme McRae, Jun 06 2006
Equals row sums of triangle A152203. - Gary W. Adamson, Nov 29 2008
Define a triangle by T(n,0) = n*(n+1)+1, T(n,n) = 1, and T(r,c) = T(r-1,c) + T(r-2,c-1). This triangle starts: 1; 3,1; 7,2,1; 13,5,2,1; 21,12,4,2,1; the sum of terms in row n is a(n+1). - J. M. Bergot, Apr 23 2013
a(n) = number of k-tuples (u(1), u(2), ..., u(k)) with 1 <= u(1) < u(2) < ... < u(k) <= n such that u(i) - u(i-1) <= 2 for i = 2,...,k. Changing the bound from 2 to 3, then 4, then 5, yields A356619, A356620, A356621. The patterns suggest that the limiting sequence as the bound increases is A000295. - Clark Kimberling, Aug 24 2022

References

  • J. Riordan, The enumeration of permutations with three-ply staircase restrictions, unpublished memorandum, Bell Telephone Laboratories, Murray Hill, NJ, Oct 1963. (See A001883)
  • 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).

Crossrefs

Partial sums of A001911.
A diagonal of triangle in A080061.
Right-hand column 5 of triangle A011794.

Programs

  • GAP
    List([0..40], n-> Fibonacci(n+5) -2*n-5); # G. C. Greubel, Jul 06 2019
  • Magma
    [Fibonacci(n+5)-(5+2*n): n in [0..40]]; // Vincenzo Librandi, Jun 07 2013
    
  • Mathematica
    LinearRecurrence[{3,-2,-1,1}, {0,1,4,10}, 40] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2012 *)
    Table[Fibonacci[n+5] -(2*n+5), {n,0,40}] (* G. C. Greubel, Jul 06 2019 *)
    maxDiff = 2;
    Map[Length[Select[Map[{#, Max[Differences[#]]} &,
      Drop[Subsets[Range[#]], # + 1]], #[[2]] <= maxDiff &]] &,
      Range[16]] (* Peter J. C. Moses, Aug 14 2022 *)
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 1,-1,-2,3]^n*[0;1;4;10])[1,1] \\ Charles R Greathouse IV, Apr 08 2016
    
  • Sage
    [fibonacci(n+5) -2*n-5 for n in (0..40)] # G. C. Greubel, Jul 06 2019
    

Formula

G.f.: x*(1+x)/((1-x-x^2)*(1-x)^2). - Simon Plouffe in his 1992 dissertation
a(n) = Fibonacci(n+5) - (5+2*n). - Wolfdieter Lang
a(n) = a(n-1) + a(n-2) + (2n+1); a(-x)=0. - Barry E. Williams, Mar 27 2000
a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4). - Sam Lachterman (slachterman(AT)fuse.net), Sep 22 2003
a(n) - a(n-1) = A101220(2,1,n). - Ross La Haye, May 31 2006
a(n) = (-3 + (2^(-1-n)*((1-sqrt(5))^n*(-11+5*sqrt(5)) + (1+sqrt(5))^n*(11+5*sqrt(5)))) / sqrt(5) - 2*(1+n)). - Colin Barker, Mar 11 2017

A014166 Apply partial sum operator 4 times to Fibonacci numbers.

Original entry on oeis.org

0, 1, 5, 16, 41, 92, 189, 365, 674, 1204, 2098, 3588, 6050, 10093, 16703, 27476, 44995, 73440, 119575, 194345, 315460, 511576, 829060, 1342936, 2174596, 3520457, 5698329, 9222440, 14924829, 24151764, 39081553
Offset: 0

Views

Author

Keywords

Crossrefs

Right-hand column 8 of triangle A011794.

Programs

  • GAP
    List([0..30], n-> Fibonacci(n+8)-(n^3+12*n^2+59*n+126)/6); # G. C. Greubel, Sep 06 2019
  • Magma
    [Fibonacci(n+8)-(n^3+12*n^2+59*n+126)/6: n in [0..30]]; // G. C. Greubel, Sep 06 2019
    
  • Maple
    with(combinat); seq(fibonacci(n+8)-(n^3+12*n^2+59*n+126)/6, n = 0..30); # G. C. Greubel, Sep 06 2019
  • Mathematica
    Nest[Accumulate, Fibonacci[Range[0, 30]], 4] (* Jean-François Alcover, Jan 08 2019 *)
  • PARI
    a(n)=fibonacci(n+8)-(n^3+12*n^2+59*n+126)/6 \\ Charles R Greathouse IV, Jun 11 2015
    
  • Sage
    [fibonacci(n+8)-(n^3+12*n^2+59*n+126)/6 for n in (0..30)] # G. C. Greubel, Sep 06 2019
    

Formula

a(n) = Fibonacci(n+8) - (n^3 +12*n^2 +59*n +126)/6.
G.f.: x/((1-x)^4*(1-x-x^2)).

A014162 Apply partial sum operator thrice to Fibonacci numbers.

Original entry on oeis.org

0, 1, 4, 11, 25, 51, 97, 176, 309, 530, 894, 1490, 2462, 4043, 6610, 10773, 17519, 28445, 46135, 74770, 121115, 196116, 317484, 513876, 831660, 1345861, 2177872, 3524111, 5702389, 9226935, 14929789
Offset: 0

Views

Author

Keywords

Comments

With offset 4, number of 132-avoiding two-stack sortable permutations which contain exactly one subsequence of type 51234.

Crossrefs

Right-hand column 6 of triangle A011794.

Programs

  • GAP
    List([0..40], n-> Fibonacci(n+6) - (n^2 + 7*n + 16)/2); # G. C. Greubel, Sep 05 2019
  • Magma
    [Fibonacci(n+6) - (n^2 + 7*n + 16)/2: n in [0..40]]; // G. C. Greubel, Sep 05 2019
    
  • Maple
    with(combinat); seq(fibonacci(n+6)-(n^2+7*n+16)*(1/2), n = 0..40); # G. C. Greubel, Sep 05 2019
  • Mathematica
    Nest[Accumulate,Fibonacci[Range[0,30]],3] (* or *) LinearRecurrence[{4,-5,1,2,-1},{0,1,4,11,25},40] (* Harvey P. Dale, Aug 19 2017 *)
  • PARI
    a(n)=fibonacci(n+6)-n*(n+7)/2-8 \\ Charles R Greathouse IV, Jun 11 2015
    
  • Sage
    [fibonacci(n+6) - (n^2 + 7*n + 16)/2 for n in (0..40)] # G. C. Greubel, Sep 05 2019
    

Formula

a(n) = Sum_{k=0..n} A000045(n-k)*k*(k+1)/2. - Benoit Cloitre, Jan 06 2003
G.f.: x/((1-x)^3*(1-x-x^2)).
From Paul Barry, Oct 07 2004: (Start)
a(n-2) = Sum_{k=0..floor(n/2)} binomial(n-k, k+3).
a(n-2) = Sum_{k=0..n} binomial(k, n-k+3). (End)
Convolution of A000045 and A000217 (Fibonacci and triangular numbers). - Ross La Haye, Nov 08 2004
a(n) = Fibonacci(n+6) - (n^2 + 7*n + 16)/2.

A053808 Partial sums of A001891.

Original entry on oeis.org

1, 5, 15, 36, 76, 148, 273, 485, 839, 1424, 2384, 3952, 6505, 10653, 17383, 28292, 45964, 74580, 120905, 195885, 317231, 513600, 831360, 1345536, 2177521, 3523733, 5701983, 9226500, 14929324, 24156724, 39087009, 63244757, 102332855, 165578768, 267912848
Offset: 0

Views

Author

Barry E. Williams, Mar 27 2000

Keywords

Comments

Antidiagonal sums of the convolution array A213579 and row 1 of the convolution array A213590. - Clark Kimberling, Jun 18 2012
Also number CG(n,2) of complete games with n players of 2 types. - N. J. A. Sloane, Dec 29 2012

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Crossrefs

Convolution of A000290 (squares) with A000045, n >= 1. (Fibonacci) - Wolfdieter Lang, Apr 10 2000
Right-hand column 7 of triangle A011794.

Programs

  • GAP
    List([0..40], n-> Fibonacci(n+8) - (n^2 +8*n+20)); # G. C. Greubel, Jul 06 2019
  • Magma
    [Fibonacci(n+8) - (n^2+8*n+20): n in [0..40]]; // G. C. Greubel, Jul 06 2019
    
  • Mathematica
    Table[Fibonacci[n+8] -(n^2 +8*n+20), {n,0,40}] (* G. C. Greubel, Jul 06 2019 *)
    LinearRecurrence[{4,-5,1,2,-1},{1,5,15,36,76},40] (* Harvey P. Dale, Apr 14 2022 *)
  • PARI
    vector(40, n, n--; fibonacci(n+8) - (n^2 +8*n+20)) \\ G. C. Greubel, Jul 06 2019
    
  • Sage
    [fibonacci(n+8) - (n^2 +8*n+20) for n in (0..20)] # G. C. Greubel, Jul 06 2019
    

Formula

a(n) = a(n-1) + a(n-2) + (n+1)^2, a(-n)=0.
G.f.: (1+x)/((1-x-x^2)*(1-x)^3).
a(n) = Fibonacci(n+6) - (n^2 + 4*n + 8), n >= 2 (see p. 184 of FQ reference).
a(n-2) = Sum_{i=0..n} Fibonacci(i)*(n-i)^2. - Benoit Cloitre, Mar 06 2004

A053295 Partial sums of A053739.

Original entry on oeis.org

1, 7, 29, 92, 247, 591, 1300, 2683, 5270, 9955, 18228, 32551, 56967, 98086, 166681, 280271, 467301, 773906, 1274856, 2091266, 3419252, 5576298, 9076280, 14750858, 23945893, 38839257, 62955061, 101995694
Offset: 0

Views

Author

Barry E. Williams, Mar 04 2000

Keywords

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 189, 194-196.

Crossrefs

Right-hand column 12 of triangle A011794.
Cf. A228074.

Programs

  • Magma
    [(&+[Binomial(n+6-j, n-2*j): j in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, May 24 2018
  • Mathematica
    Table[Sum[Binomial[n+6-j, n-2*j], {j, 0, Floor[n/2]}], {n, 0, 50}] (* G. C. Greubel, May 24 2018 *)
  • PARI
    for(n=0, 30, print1(sum(j=0, floor(n/2), binomial(n+6-j, n-2*j)), ", ")) \\ G. C. Greubel, May 24 2018
    

Formula

a(n) = Sum_{i=0..floor(n/2)} C(n+6-i, n-2i), n >= 0.
a(n) = a(n-1) + a(n-2) + C(n+5,5); n >= 0; a(-1)=0.
G.f.: -1 / ( (x^2 + x - 1)*(x-1)^6 ). - R. J. Mathar, Oct 10 2014

A053739 Partial sums of A014166.

Original entry on oeis.org

1, 6, 22, 63, 155, 344, 709, 1383, 2587, 4685, 8273, 14323, 24416, 41119, 68595, 113590, 187030, 306605, 500950, 816410, 1327986, 2157046, 3499982, 5674578, 9195035, 14893364, 24115804, 39040633, 63192397, 102273950, 165512723, 267839033, 433410661, 701315739, 1134800215
Offset: 0

Views

Author

Barry E. Williams, Feb 13 2000

Keywords

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Crossrefs

Cf. A014166 and A000045.
Right-hand column 10 of triangle A011794.
Cf. A228074.

Programs

  • GAP
    List([0..35], n-> Fibonacci(n+11)-(n^4+22*n^3+203*n^2+974*n + 2112)/24); # G. C. Greubel, Sep 06 2019
  • Magma
    [Fibonacci(n+11) - (n^4+22*n^3+203*n^2+974*n+2112)/24: n in [0..35]]; // G. C. Greubel, Sep 06 2019
    
  • Maple
    with(combinat); seq(fibonacci(n+11)-(n^4 + 22*n^3 + 203*n^2 + 974*n + 2112)/4!, n = 0..35); # G. C. Greubel, Sep 06 2019
  • Mathematica
    Table[Fibonacci[n+11] -(n^4+22*n^3+203*n^2+974*n+2112)/4!, {n,0,35}] (* G. C. Greubel, Sep 06 2019 *)
  • PARI
    vector(35, n, m=n-1; fibonacci(n+10) - (m^4+22*m^3+203*m^2+974*m +2112)/4!) \\ G. C. Greubel, Sep 06 2019
    
  • Sage
    [fibonacci(n+11) - (n^4+22*n^3+203*n^2+974*n+2112)/24 for n in (0..35)] # G. C. Greubel, Sep 06 2019
    

Formula

a(n) = Sum_{i=0..floor(n/2)} binomial(n+5-i, n-2*i) for n >= 0.
a(n) = a(n-1) + a(n-2) + C(n+4,4); n >= 0; a(-1)=0.
G.f.: 1/((1-x-x^2)*(1-x)^5). - R. J. Mathar, May 22 2013
a(n) = Fibonacci(n+11) - (n^4 + 22*n^3 + 203*n^2 + 974*n + 2112)/4!. - G. C. Greubel, Sep 06 2019

Extensions

Terms a(28) onward added by G. C. Greubel, Sep 06 2019

A053296 Partial sums of A053295.

Original entry on oeis.org

1, 8, 37, 129, 376, 967, 2267, 4950, 10220, 20175, 38403, 70954, 127921, 226007, 392688, 672959, 1140260, 1914166, 3189022, 5280288, 8699540, 14275838, 23352118, 38102976, 62048869, 100888126, 163843187, 265838881, 431026972, 698489013, 1131463777, 1832277574, 2966502032, 4802042229
Offset: 0

Views

Author

Barry E. Williams, Mar 04 2000

Keywords

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 189, 194-196.

Crossrefs

Right-hand column 14 of triangle A011794.

Programs

  • Magma
    [(&+[Binomial(n+7-j, n-2*j): j in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, May 24 2018
  • Mathematica
    Table[Sum[Binomial[n+7-j, n-2*j], {j, 0, Floor[n/2]}], {n, 0, 50}] (* G. C. Greubel, May 24 2018 *)
  • PARI
    for(n=0, 30, print1(sum(j=0, floor(n/2), binomial(n+7-j, n-2*j)), ", ")) \\ G. C. Greubel, May 24 2018
    

Formula

a(n) = Sum_{i=0..floor(n/2)} C(n+7-i, n-2i), n >= 0.
a(n) = a(n-1) + a(n-2) + C(n+6,6); n >= 0, with a(-1) = 0.
From G. C. Greubel, Oct 21 2024: (Start)
a(n) = Fibonacci(n+15) - Sum_{j=0..6} Fibonacci(14-2*j)*binomial(n+j,j).
a(n) = Fibonacci(n+15) - (1/6!)*(n^6 + 39*n^5 + 685*n^4 + 7185*n^3 + 48994*n^2 + 209496*n + 438480).
G.f.: 1/((1-x)^7*(1 - x - x^2)). (End)

Extensions

Terms a(28) onward added by G. C. Greubel, May 24 2018
Showing 1-10 of 14 results. Next