A005969 -id:A005969 - cp's OEIS frontend

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

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

N. J. A. Sloane

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

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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.

Christian G. Bower, Table of n, a(n) for n = 1..500
Isha Agarwal, Matvey Borodin, Aidan Duncan, Kaylee Ji, Tanya Khovanova, Shane Lee, Boyan Litchev, Anshul Rastogi, Garima Rastogi, and Andrew Zhao, From Unequal Chance to a Coin Game Dance: Variants of Penney's Game, arXiv:2006.13002 [math.HO], 2020.
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Kassie Archer and Noel Bourne, Pattern avoidance in compositions and powers of permutations, arXiv:2505.05218 [math.CO], 2025. See pp. 6-7.
Kassie Archer and Aaron Geary, Powers of permutations that avoid chains of patterns, arXiv:2312.14351 [math.CO], 2023. See p. 15.
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Andrew M. Baxter and 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, 557-564.
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Fan Chung and R. L. Graham, Primitive juggling sequences, Am. Math. Monthly 115 (3) (2008) 185-194.
Ligia Loretta Cristea, Ivica Martinjak, and Igor Urbiha, Hyperfibonacci Sequences and Polytopic Numbers, arXiv:1606.06228 [math.CO], 2016.
Michael Dairyko, Samantha Tyner, Lara Pudwell, and Casey Wynn, 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
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Christian Ennis, William Holland, Omer Mujawar, Aadit Narayanan, Frank Neubrander, Marie Neubrander, and Christina Simino, Words in Random Binary Sequences I, arXiv:2107.01029 [math.GM], 2021.
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Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000032, A000045, A000119, A001611, A001654, A001911, A005968, A005969.
Cf. A006327, A014417, A054761, A081006, A081007, A081008, A081009, A083368.
Cf. A098531, A098532, A098533, A101220, A104326, A105488, A105489, A119282.
Cf. A128697, A157725, A157726, A157727, A157728, A157729, A158950, A167616.
Cf. A001622, A228074.
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.

a000071 n = a000071_list !! n
a000071_list = map (subtract 1) $ tail a000045_list
-- Reinhard Zumkeller, May 23 2013

[Fibonacci(n)-1: n in [1..60]]; // Vincenzo Librandi, Apr 04 2011

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

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)};
```

[fibonacci(n)-1 for n in range(1,60)] # G. C. Greubel, Oct 21 2024

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

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

A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

Original entry on oeis.org

0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, 12816, 33552, 87841, 229970, 602070, 1576239, 4126648, 10803704, 28284465, 74049690, 193864606, 507544127, 1328767776, 3478759200, 9107509825, 23843770274, 62423800998, 163427632719, 427859097160, 1120149658760

Offset: 0

N. J. A. Sloane

a(n)/A007598(n) ~= golden ratio, especially for larger n. - Robert Happelberg (roberthappelberg(AT)yahoo.com), Jul 25 2005
Let phi be the golden ratio (cf. A001622). Then 1/phi = phi - 1 = Sum_{n>=1} (-1)^(n-1)/a(n), an alternating infinite series consisting solely of unit fractions. - Franz Vrabec, Sep 14 2005
a(n+2) is the Hankel transform of A005807 aerated. - Paul Barry, Nov 04 2008
A more exact name would be: Golden convergents to rectangle numbers. These rectangles are not actually golden (ratio of sides is not phi) but are golden convergents (sides are numerator and denominator of convergents in the continued fraction expansion of phi, whence ratio of sides converges to phi). - Daniel Forgues, Nov 29 2009
The Kn4 sums (see A180662 for definition) of the "Races with Ties" triangle A035317 lead to this sequence. - Johannes W. Meijer, Jul 20 2011
Numbers m such that m(5m+2)+1 or m(5m-2)+1 is a square. - Bruno Berselli, Oct 22 2012
In pairs, these numbers are important in finding binomial coefficients that appear in at least six places in Pascal's triangle. For instance, the pair (m,n) = (40, 104) finds the numbers binomial(n-1,m) = binomial(n,m-1). Two additional numbers are found on the other side of the triangle. The final two numbers appear in row binomial(n-1,m). See A003015. - T. D. Noe, Mar 13 2013
For n>1, a(n) is one-half the area of the trapezoid created by the four points (F(n),L(n)), (L(n),F(n)), (F(n+1), L(n+1)), (L(n+1), F(n+1)) where F(n) = A000045(n) and L(n) = A000032(n). - J. M. Bergot, May 14 2014
[Note on how to calculate: take the two points (a,b) and (c,d) with aa(n) = A067962(n-1) / A067962(n-2), n > 1. - Reinhard Zumkeller, Sep 24 2015
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Fibonacci numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017

			G.f. = x + 2*x^2 + 6*x^3 + 15*x^4 + 40*x^5 + 104*x^6 + 273*x^7 + 714*x^8 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 9.
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).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 17.
Ömer Egecioglu, Elif Saygi, and Zülfükar Saygi, The Mostar index of Fibonacci and Lucas cubes, arXiv:2101.04740 [math.CO], 2021. Mentions this sequence.
Shalosh B. Ekhad and Doron Zeilberger, Automatic Counting of Tilings of Skinny Plane Regions, arXiv preprint arXiv:1206.4864 [math.CO], 2012.
Sergio Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
Dale Gerdemann, Golden Ratio Base Digit Patterns for Columns of the Fibonomial Triangle, "Another interesting pattern is for Golden Rectangle Numbers A001654. I made a short video illustrating this pattern, along with other columns of the Fibonomial Triangle A010048".
Jonny Griffiths and Martin Griffiths, Fibonacci-related sequences via iterated QRT maps, Fib. Q., 51 (2013), 218-227.
James P. Jones and Péter Kiss, Representation of integers as terms of a linear recurrence with maximal index, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae, 25. (1998) pp. 21-37. See Lemma 4.1 p. 34.
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 11.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 6.
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Prapanpong Pongsriiam, Integral Values of the Generating Functions of Fibonacci and Lucas Numbers, College Math. J., 48 (No. 2 2017), pp 97ff.
M. Renault, Dissertation
Wikipedia, Illustration of 273 as a golden rectangle number.
Robert G. Wilson v, Letter to N. J. A. Sloane, circa 1993
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001655, A001656, A001657, A001658, A010048, A067962.
Cf. A005968, A005969, A098531, A098532, A098533, A119283, A128697.
Cf. A000071, A079472, A080145, A239798, A290565.
Bisection of A006498, A070550, A080239.
First differences of A064831.
Partial sums of A007598.

a001654 n = a001654_list !! n
a001654_list = zipWith (*) (tail a000045_list) a000045_list
-- Reinhard Zumkeller, Jun 08 2013

I:=[0,1,2]; [n le 3 select I[n] else 2*Self(n-1) + 2*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 17 2018

with(combinat): A001654:=n->fibonacci(n)*fibonacci(n+1):
seq(A001654(n), n=0..28); # Zerinvary Lajos, Oct 07 2007

LinearRecurrence[{2,2,-1}, {0,1,2}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
Times@@@Partition[Fibonacci[Range[0,30]],2,1] (* Harvey P. Dale, Aug 18 2011 *)
Accumulate[Fibonacci[Range[0, 30]]^2] (* Paolo Xausa, May 31 2024 *)

A001654(n)=fibonacci(n)*fibonacci(n+1);

b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
vector(30, n, b(n-1, 2))  \\ Joerg Arndt, May 08 2016

from sympy import fibonacci as F
def a(n): return F(n)*F(n + 1)
[a(n) for n in range(101)] # Indranil Ghosh, Aug 03 2017

from math import prod
from gmpy2 import fib2
def A001654(n): return prod(fib2(n+1)) # Chai Wah Wu, May 19 2022

a(n) = A010048(n+1, 2) = Fibonomial(n+1, 2).
a(n) = A006498(2*n-1).
a(n) = a(n - 1) + A007598(n) = a(n - 1) + A000045(n)^2 = Sum_{j <= n} Fibonacci(j)^2. - Henry Bottomley, Feb 09 2001 [corrected by Ridouane Oudra, Apr 12 2025]
For n > 0, 1 - 1/a(n+1) = Sum_{k=1..n} 1/(F(k)*F(k+2)) where F(k) is the k-th Fibonacci number. - Benoit Cloitre, Aug 31 2002.
G.f.: x/(1-2*x-2*x^2+x^3) = x/((1+x)*(1-3*x+x^2)). (Simon Plouffe in his 1992 dissertation; see Comments to A055870),
a(n) = 3*a(n-1) - a(n-2) - (-1)^n = -a(-1-n).
Let M = the 3 X 3 matrix [1 2 1 / 1 1 0 / 1 0 0]; then a(n) = the center term in M^n *[1 0 0]. E.g., a(5) = 40 since M^5 * [1 0 0] = [64 40 25]. - Gary W. Adamson, Oct 10 2004
a(n) = Sum{k=0..n} Fibonacci(k)^2. The proof is easy. Start from a square (1*1). On the right side, draw another square (1*1). On the above side draw a square ((1+1)*(1+1)). On the left side, draw a square ((1+2)*(1+2)) and so on. You get a rectangle (F(n)*F(1+n)) which contains all the squares of side F(1), F(2), ..., F(n). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 19 2007
With phi = (1+sqrt(5))/2, a(n) = round((phi^(2*n+1))/5) = floor((1/2) + (phi^(2*n+1))/5), n >= 0. - Daniel Forgues, Nov 29 2009
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), a(1)=1, a(2)=2, a(3)=6. - Sture Sjöstedt, Feb 06 2010
a(n) = (A002878(n) - (-1)^n)/5. - R. J. Mathar, Jul 22 2010
a(n) = 1/|F(n+1)/F(n) - F(n)/F(n-1)| where F(n) = Fibonacci numbers A000045. b(n) = F(n+1)/F(n) - F(n)/F(n-1): 1/1, -1/2, 1/6, -1/15, 1/40, -1/104, ...; c(n) = 1/b(n) = a(n)*(-1)^(n+1): 1, -2, 6, -15, 40, -104, ... (n=1,2,...). - Thomas Ordowski, Nov 04 2010
a(n) = (Fibonacci(n+2)^2 - Fibonacci(n-1)^2)/4. - Gary Detlefs, Dec 03 2010
Let d(n) = n mod 2, a(0)=0 and a(1)=1. For n > 1, a(n) = d(n) + 2*a(n-1) + Sum_{k=0..n-2} a(k). - L. Edson Jeffery, Mar 20 2011
From Tim Monahan, Jul 11 2011: (Start)
a(n+1) = ((2+sqrt(5))*((3+sqrt(5))/2)^n+(2-sqrt(5))*((3-sqrt(5))/2)^n+(-1)^n)/5.
a(n) = ((1+sqrt(5))*((3+sqrt(5))/2)^n+(1-sqrt(5))*((3-sqrt(5))/2)^n-2*(-1)^n)/10. (End)
From Wolfdieter Lang, Jul 21 2012: (Start)
a(n) = (2*A059840(n+2) - A027941(n))/3, n >= 0, with A059840(n+2) = Sum_{k=0..n} F(k)*F(k+2) and A027941(n) = A001519(n+1) - 1, n >= 0, where A001519(n+1) = F(2*n+1). (End)
a(n) = (-1)^n * Sum_{k=0..n} (-1)^k*F(2*k), n >= 0. - Wolfdieter Lang, Aug 11 2012
a(-1-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
a(n) = (L(2*n+1) - (-1)^n)/5 with L(k) = A000032(k). - J. M. Bergot, Apr 15 2016
E.g.f.: ((3 + sqrt(5))*exp((5+sqrt(5))*x/2) - 2*exp((2*x)/(3+sqrt(5))+x) - 1 - sqrt(5))*exp(-x)/(5*(1 + sqrt(5))). - Ilya Gutkovskiy, Apr 15 2016
From Klaus Purath, Apr 24 2019: (Start)
a(n) = A061646(n) - Fibonacci(n-1)^2.
a(n) = (A061646(n+1) - A061646(n))/2. (End)
a(n) = A226205(n+1) + (-1)^(n+1). - Flávio V. Fernandes, Apr 23 2020
Sum_{n>=1} 1/a(n) = A290565. - Amiram Eldar, Oct 06 2020
Product_{n>=2} (1 + (-1)^n/a(n)) = phi^2/2 (A239798). - Amiram Eldar, Dec 02 2024
G.f.: x * exp( Sum_{k>=1} F(3*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extended by Wolfdieter Lang, Jun 27 2000

A005968 Sum of cubes of first n Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 10, 37, 162, 674, 2871, 12132, 51436, 217811, 922780, 3908764, 16558101, 70140734, 297121734, 1258626537, 5331629710, 22585142414, 95672204155, 405273951280, 1716768021816, 7272346018247, 30806152127640, 130496954475672, 552793970116297, 2341672834801754

Offset: 0

N. J. A. Sloane

From Alexander Adamchuk, Aug 07 2006: (Start)
The only two prime terms are a(2) = 2 and a(4) = 37.
The prime p divides a(p-1) iff p is in A045468.
The prime p divides a((p-1)/2) iff p is in A047650.
3^4 divides a(p) iff p is in A003628.
3^5 divides a(p) for p = {37,53,109,181,197,269,397,431,541,...}.
3^6 divides a(p) for p = {109,541,...}.
3^7 divides a(p) for p = {557,...}. (End)

Art Benjamin, Timothy A. Carnes, and Benoit Cloitre, Recounting the Sums of Cubes of Fibonacci Numbers, Congressus Numerantium, Proceedings of the Eleventh International Conference on Fibonacci Numbers and their Applications, (William Webb, ed.), Vol 194, pp. 45-51, 2009.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 14.
A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 18.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Art Benjamin, Timothy A. Carnes, and Benoit Cloitre, Counting the Sums of Cubes of Fibonacci Numbers.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
David Treeby, Hidden Formulas in Fibonacci Tilings, Fibonacci Quart. 54 (2016), no. 1, 23-30.
Index entries for linear recurrences with constant coefficients, signature (4,3,-9,2,1).

Partial sums of A056570.  Cf. A119284 (alternating sum).
Cf. A045468, A047650, A003628.
Sums of other powers: A000071, A001654, A005969, A098531, A098532, A098533, A128697.

[(1/10)*( Fibonacci(3*n+2)-(-1)^(n)*6*Fibonacci(n-1)+5 ): n in [0..30]]; // G. C. Greubel, Jan 17 2018

with(combinat): l[0] := 0: for i from 1 to 50 do l[i] := l[i-1]+fibonacci(i)^3; printf(`%d,`,l[i]) od: # James Sellers, May 29 2000
A005968:=(-1+2*z+z**2)/(z-1)/(z**2+4*z-1)/(z**2-z-1); # conjectured by Simon Plouffe in his 1992 dissertation

f[n_]:=(Fibonacci[n]*Fibonacci[n+1]^2+(-1)^(n-1)*Fibonacci[n-1]+1)/2;Table[f[n],{n,0,5!}] (* Vladimir Joseph Stephan Orlovsky, Nov 22 2010 *)
Accumulate[Fibonacci[Range[0,20]]^3]
CoefficientList[Series[x*(1-2*x-x^2)/((1-x)*(1+x-x^2)*(1-4*x-x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)

a(n)=(fibonacci(n)*fibonacci(n+1)^2+(-1)^(n-1)*fibonacci(n-1)+1)/2

a(n)=(fibonacci(3*n+2)-(-1)^(n)*6*fibonacci(n-1)+5)/10

PARI
```
a(n)=sum(i=1,n,fibonacci(i)^3)
```

G.f.: x*(1-2*x-x^2)/((1-x)*(1+x-x^2)*(1-4*x-x^2)). - Ralf Stephan, Apr 23 2004
a(n) = (1/2)*(F(n)*F(n+1)^2 + (-1)^(n-1)*F(n-1) + 1). - Benoit Cloitre, Aug 06 2004
a(n) = Sum_{i=1..n} A000045(i)^3.
a(n) = (1/10)*(F(3*n+2) - (-1)^(n)*6*F(n-1) + 5). - Art Benjamin and Timothy A. Carnes
a(n+5) = 4*a(n+4) + 3*a(n+3) - 9*a(n+2) + 2*a(n+1) + a(n). - Benoit Cloitre, Sep 12 2004

More terms from James Sellers, May 29 2000

A056571 Fourth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 16, 81, 625, 4096, 28561, 194481, 1336336, 9150625, 62742241, 429981696, 2947295521, 20200652641, 138458410000, 949005240561, 6504586067281, 44583076827136, 305577005139121, 2094455819300625, 14355614096087056

Offset: 0

Wolfdieter Lang, Jul 10 2000

Divisibility sequence; that is, if n divides m, then a(n) divides a(m).

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/4,3/4)-fences and quarter-squares (1/4 X 1 pieces, always placed so that the shorter sides are horizontal). A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/8,3/8)-fences and (1/8,7/8)-fences. - Michael A. Allen, Jan 11 2022

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 31.
Donald E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 85, (exercise 1.2.8. Nr. 30) and p. 492 (solution).
Hilary I. Okagbue, Muminu O. Adamu, Sheila A. Bishop and Abiodun A. Opanuga, Digit and Iterative Digit Sum of Fibonacci numbers, their identities and powers, International Journal of Applied Engineering Research ISSN 0973-4562 Volume 11, Number 6 (2016) pp. 4623-4627.

Vincenzo Librandi, Table of n, a(n) for n = 0..151
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.
Alfred Brousseau, A sequence of power formulas, Fib. Quart., Vol. 6, No. 1 (1968), pp. 81-83.
Andrej Dujella, A bijective proof of Riordan's theorem on powers of Fibonacci numbers, Discrete Math., Vol. 199, No. 1-3 (1999), pp. 217-220. MR1675924 (99k:05016).
Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers cubed, Fib. Q. 58:5 (2020) 128-134.
Hideyuki Ohtsuka, Problem N-1220, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 55, No. 4 (2017), p. 368; Gelin-Cesàro Identity Yields a Telescoping Product, Solution to Problem H-790 by Ramya Dutta, ibid., Vol. 56, No. 4 (2018), p. 372.
John Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J., Vol. 29, No. 1 (1962), pp. 5-12.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A000045, A001622, A007598, A056570, A056588, A055870.
First differences of A005969.
Fourth row of array A103323.

[Fibonacci(n)^4: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011

A056571 := proc(n)
    combinat[fibonacci](n)^4 ;
end proc:
seq(A056571(n),n=0..10) ; # R. J. Mathar, Jan 23 2022

Fibonacci[Range[0, 25]]^4 (* Wesley Ivan Hurt, Jan 11 2022 *)

a(n)=fibonacci(n)^4 \\ Charles R Greathouse IV, Oct 07 2016

a(n) = F(n)^4 = A007598(n)^2, F(n) = A000045(n).
G.f.: x*p(4, x)/q(4, x) with p(4, x) := sum(A056588(3, m)*x^m, m=0..3) = 1 - 4*x - 4*x^2 + x^3 = (1+x)*(1 - 5*x + x^2) and q(4, x) := Sum_{m=0..5} A055870(5, m)*x^m = 1 - 5*x - 15*x^2 + 15*x^3 + 5*x^4 - x^5 = (1-x)*(1 + 3*x + x^2)*(1 - 7*x + x^2) (denominator factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): 1*a(n) - 5*a(n-1) - 15*a(n-2) + 15*a(n-3) + 5*a(n-4) - 1*a(n-5) = 0, n >= 5; inputs: a(n), n=0..4.
(1/25)*((-1)^n*(2*F(2*n-2) - 6*F(2*n+1)) + 2*F(4*n-1) + F(4*n) + 6). - Ralf Stephan, May 14 2004
a(n) = F(n-2)*F(n-1)*F(n+1)*F(n+2) + 1 = A244855(n)+1.
Sum_{j=0..n} binomial(n,j)*a(j) = (3^n*A005248(n) - 4*(-1)^n*A000032(n) + 6*2^n)/25. Sum_{j=0..n} (-1)^j*binomial(n,j)*a(j) = -5^((n+1)/2-2)*(A001906(n) + 4*A000045(n)) if n odd. - R. J. Mathar, Oct 16 2006
a(n) = (F(n)*F(3n) - 3*F(n)^2*(-1)^n)/5. - Gary Detlefs, Dec 26 2010
Product_{n>=3} (1 - 1/a(n)) = phi^5/12, where phi is the golden ratio (A001622)(Ohtsuka, 2017). - Amiram Eldar, Dec 02 2021

A098531 Sum of fifth powers of first n Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 34, 277, 3402, 36170, 407463, 4491564, 49926988, 553211363, 6137270812, 68054635036, 754774491429, 8370420537086, 92830050637086, 1029498223070793, 11417322172518550, 126619992693837974, 1404237451180502875, 15573231068749231000

Offset: 0

Benoit Cloitre, Sep 12 2004

Prime p divides a((p-1)/2) for p = {29,89,101,181,229,...} = A047650[n]. Primes for which golden mean tau is a quadratic residue or Primes of the form x^2 + 20y^2. - Alexander Adamchuk, Aug 07 2006

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (9,32,-100,20,48,-7,-1).

Cf. A000071, A001654, A005968, A005969, A047650, A056572, A098532, A098533, A119286, A128697.

[(&+[Fibonacci(k)^5:k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 17 2018

Accumulate[Fibonacci[Range[0,20]]^5]  (* Harvey P. Dale, Jan 14 2011 *)
CoefficientList[Series[x*(1-7*x-16*x^2+7*x^3+x^4)/((1-x)*(1+4*x-x^2)*(1-x-x^2)*(1-11*x-x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 13 2012 *)

PARI
```
a(n)=sum(i=0,n,fibonacci(i)^5)
```

a(n) = -7/22 + 2*F(n+2)/5 + (F(5*(n+1)) + F(5*n))/(5*55) - (-1)^n*(F(3*(n+1)) - F(3*n))/(2*10), where F=A000045. One may use F(5*(n+1)) + F(5*n) = F(5*n+1) + 4*F(5*n+2) (due to the Binet-de Moivre formula).

G.f.: x*(1-7*x-16*x^2+7*x^3+x^4)/((1-x)*(1+4*x-x^2)*(1-x-x^2)*(1-11*x-x^2)). - Bruno Berselli, Oct 12 2012

Formula corrected, with the author's consent, by Wolfdieter Lang, Oct 12 2012

A098532 Sum of sixth powers of first n Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 66, 795, 16420, 278564, 5105373, 90871494, 1635675910, 29316316535, 526297607496, 9442398055752, 169448124595321, 3040546683808010, 54560921044808010, 979052407236876819, 17568407254504944748

Offset: 0

Benoit Cloitre, Sep 12 2004

G. C. Greubel, Table of n, a(n) for n = 0..795
Index entries for linear recurrences with constant coefficients, signature(13,104,-260,-260,104,13,-1).

Cf. A001654, A005968, A005969, A098531, A098533.

Cf. A119287, A000071, A001654, A005968, A005969, A098531, A098533, A128697.

[(Fibonacci(n)^5*Fibonacci(n+3) + Fibonacci(2*n))/4: n in [0..30]]; // G. C. Greubel, Jan 17 2018

Table[(Fibonacci[n]^5*Fibonacci[n+3] + Fibonacci[2*n])/4, {n,0,30}] (* G. C. Greubel, Jan 17 2018 *)

PARI
```
a(n)=sum(i=0,n,fibonacci(i)^6);
```

for(n=0,30, print1((fibonacci(n)^5*fibonacci(n+3) + fibonacci(2*n))/4, ", ")) \\ G. C. Greubel, Jan 17 2018

a(n) = (1/500)*(F(6*n+1) +3*F(6*n+2) -(-1)^n*(16*F(4*n+1)+8*F(4*n+2))-60*F(2*n+1) +120*F(2*n+2) -(-1)^n*40 ) where F(n)=A000045(n).
G.f.: x*(1-11*x-64*x^2-11*x^3+x^4)/((x+1)*(1-18*x+x^2)*(1-3*x+x^2)*(1+7*x+x^2)). - R. J. Mathar, Feb 26 2012
a(n) = -6*(-1)^n*A049685(n)/125 +3*A002878(n)/25 +A049629(n)/125 -2*(-1)^n/25. - R. J. Mathar, Feb 26 2012
a(n)= (F(n)^5 * F(n+3) + F(2*n))/4. - Gary Detlefs, Jan 05 2013

A098533 Sum of seventh powers of first n Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 130, 2317, 80442, 2177594, 64926111, 1866014652, 54389364796, 1576824599171, 45808159494700, 1329726624043564, 38611060907763141, 1121015217730946894, 32548443577940946894, 945021540449512861377

Offset: 0

Benoit Cloitre, Sep 12 2004

G. C. Greubel, Table of n, a(n) for n = 0..680
Index entries for linear recurrences with constant coefficients, signature (22,252,-1365,-728,2912,-819,-294,20,1).

Cf. A001654, A005968, A005969, A098531, A098532.

Cf. A128696, A000071, A001654, A005968, A005969, A098531, A098532, A128697.

[(&+[Fibonacci(k)^7:k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 17 2018

Accumulate[Fibonacci[Range[0,20]]^7] (* Harvey P. Dale, Jul 16 2017 *)

PARI
```
a(n)=sum(i=0,n,fibonacci(i)^7)
```

a(n) = (1/79750)*( 88*F(7n+1) +198*F(7n+2) -(-1)^n*(1218*F(5n+1) + 812*F(5n+2)) +6699*F(3n+2) -(-1)^n*(44660*F(n+1) -22330*F(n+2)) + 17375) where F(n)=A000045(n).
a(n) = 22 a(n-1) +252 a(n-2) -1365 a(n-3) -728 a(n-4) +2912 a(n-5) -819 a(n-6) -294 a(n-7) +20 a(n-8) +1 a(n-9).
G.f.: x*(x^6+20*x^5-166*x^4-318*x^3+166*x^2+20*x-1)/((x-1)*(x^2-11*x-1)*(x^2-x-1)*(x^2+4*x-1)*(x^2+29*x-1)). - R. J. Mathar, Feb 26 2012

Typo in data corrected by D. S. McNeil, Aug 17 2010

A119285 Alternating sum of the fourth powers of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -16, 65, -560, 3536, -25025, 169456, -1166880, 7983745, -54758496, 375223200, -2572072321, 17628580320, -120829829680, 828175410881, -5676410656400, 38906666170736, -266670338968385, 1827785480332240, -12527828615754816, 85867013279034625, -588541268397840576, 4033921854875707200, -27648911743562183425

Offset: 0

Stuart Clary, May 13 2006

Natural bilateral extension (brackets mark index 0): ..., -3536, 560, -65, 16, 0, 1, 0, [0], -1, 0, -16, 65, -560, 3536, -25025, ... This is (-A119285)-reversed followed by A119285.

Kunle Adegoke, Sums of fourth powers of Fibonacci and Lucas numbers, arXiv:1706.00407 [math.NT], 2017.
Index entries for linear recurrences with constant coefficients, signature (-5,15,15,-5,-1).

Cf. A005969, A119282, A119283, A119284, A119286, A119287, A128696, A128698.

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^4, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^4, {k, 1, -n - 1} ] ]
LinearRecurrence[{-5,15,15,-5,-1},{0,-1,0,-16,65},30] (* Harvey P. Dale, Apr 02 2018 *)

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
a(n) = Sum_{k=1..n} (-1)^k F(k)^4.
Closed form: a(n) = (-1)^n L(4n+2)/75 - (4/25) L(2n+1) + (-1)^n 3/25.
Factored closed form: a(n) = (-1)^n (1/3) F(n-2) F(n) F(n+1) F(n+3).
Recurrence: a(n) + 5 a(n-1) - 15 a(n-2) - 15 a(n-3) + 5 a(n-4) + a(n-5) = 0.
G.f.: A(x) = (-x - 5 x^2 - x^3)/(1 + 5 x - 15 x^2 - 15 x^3 + 5 x^4 + x^5) = -x(1 + 5 x + x^2)/((1 + x)(1 - 3 x + x^2)(1 + 7 x + x^2)).

A128697 Sum of the eighth powers of the first n Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 258, 6819, 397444, 17174660, 832905381, 38655764742, 1824449669638, 85558387560263, 4022147193262344, 188906406088298760, 8875457294194960201, 416941824416535235082, 19587673124144635235082, 920198619736386114829803, 43229838526402491973562764, 2030880577900713476799525260, 95408186647695095521364177901, 4482153365649947417785489568526

Offset: 0

Stuart Clary, Mar 23 2007

Natural bilateral extension (brackets mark index 0): ..., -17174660, -397444, -6819, -258, -2, -1, 0, [0], 1, 2, 258, 6819, 397444, 17174660, ... This is (-A128697)-reversed followed by A128697.

G. C. Greubel, Table of n, a(n) for n = 0..595
Index entries for linear recurrences with constant coefficients, signature (35,680,-5355,-7735,24752,-7735,-5355,680,35,-1).

Cf. A128698 (alternating sum).

Sums of other powers: A000071, A001654, A005968, A005969, A098531, A098532, A098533.

[(&+[Fibonacci(k)^8: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 17 2018

a[ n_Integer ] := If[ n >= 0, Sum[ Fibonacci[ k ]^8, {k, 1, n} ], Sum[ -Fibonacci[ -k ]^8, {k, 1, -n - 1} ] ]
Accumulate[Fibonacci[Range[0,20]]^8] (* Harvey P. Dale, Oct 26 2011 *)

a(n) = sum(k=1, n, fibonacci(k)^8); \\ Michel Marcus, Dec 10 2016

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=1..n} F(k)^8.
Closed form: a(n) = F(8n+4)/1875 - (-1)^n 4 F(6n+3)/625 + 28 F(4n+2)/625 - (-1)^n 56 F(2n+1)/625 + 7(2 n + 1)/125.
Recurrence: a(n) - 35 a(n-1) - 680 a(n-2) + 5355 a(n-3) + 7735 a(n-4) - 24752 a(n-5) + 7735 a(n-6) + 5355 a(n-7) - 680 a(n-8) - 35 a(n-9) + a(n-10) = 0.
G.f.: A(x) = (x - 33 x^2 - 492 x^3 + 1784 x^4 + 1784 x^5 - 492 x^6 - 33 x^7 + x^8)/(1 - 35 x - 680 x^2 + 5355 x^3 + 7735 x^4 - 24752 x^5 + 7735 x^6 + 5355 x^7 - 680 x^8 - 35 x^9 + x^10) = x*(1 + x)*(1 - 34 x - 458 x^2 + 2242 x^3 - 458 x^4 - 34 x^5 + x^6)/((1 - x)^2*(1 + 3 x + x^2)*(1 - 7 x + x^2)*(1 + 18 x + x^2)*(1 - 47 x + x^2)).

A203001 Symmetric matrix based on A007598, by antidiagonals.

Original entry on oeis.org

1, 1, 1, 4, 2, 4, 9, 5, 5, 9, 25, 13, 18, 13, 25, 64, 34, 41, 41, 34, 64, 169, 89, 113, 99, 113, 89, 169, 441, 233, 290, 266, 266, 290, 233, 441, 1156, 610, 765, 689, 724, 689, 765, 610, 1156, 3025, 1597, 1997, 1811, 1866, 1866, 1811, 1997, 1597, 3025

Offset: 1

Clark Kimberling, Dec 27 2011

Let s=A007598 (squared Fibonacci numbers), and let T be the infinite square matrix whose n-th row is formed by putting n-1 zeros before the terms of s. Let T' be the transpose of T. Then A203001 represents the matrix product M=T'*T. M is the self-fusion matrix of s, as defined at A193722. See A203002 for characteristic polynomials of principal submatrices of M, with interlacing zeros.

			Northwest corner:
1...1...4....9....25....64
1...2...5....13...34....89
4...5...18...41...113...290
9...13..41...99...266...724

Cf. A203002, A202453.

s[k_] := Fibonacci[k]^2;
U = NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[Table[s[k], {k, 1, 15}]];
L = Transpose[U]; M = L.U; TableForm[M]
m[i_, j_] := M[[i]][[j]];
Flatten[Table[m[i, n + 1 - i], {n, 1, 12}, {i, 1, n}]]
f[n_] := Sum[m[i, n], {i, 1, n}] + Sum[m[n, j], {j, 1, n - 1}]
Table[f[n], {n, 1, 12}]
Table[Sqrt[f[n]], {n, 1, 12}]   (* A001654 *)
Table[m[1, j], {j, 1, 12}]      (* A007598 *)
Table[m[2, j], {j, 1, 12}]      (* A001519 *)
Table[m[j, j], {j, 1, 12}]      (* A005969 *)

Showing 1-10 of 10 results.

cp's OEIS Frontend

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

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A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

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A005968 Sum of cubes of first n Fibonacci numbers.

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A056571 Fourth power of Fibonacci numbers A000045.

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A098531 Sum of fifth powers of first n Fibonacci numbers.

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