A119283 -id:A119283 - cp's OEIS frontend

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

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

A119282 Alternating sum of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -2, 1, -4, 4, -9, 12, -22, 33, -56, 88, -145, 232, -378, 609, -988, 1596, -2585, 4180, -6766, 10945, -17712, 28656, -46369, 75024, -121394, 196417, -317812, 514228, -832041, 1346268, -2178310, 3524577, -5702888, 9227464, -14930353, 24157816, -39088170, 63245985, -102334156, 165580140, -267914297, 433494436, -701408734, 1134903169, -1836311904, 2971215072, -4807526977, 7778742048

Offset: 0

Stuart Clary, May 13 2006

Apart from signs, same as A008346.
Natural bilateral extension (brackets mark index 0): ..., 88, 54, 33, 20, 12, 7, 4, 2, 1, 0, [0], -1, 0, -2, 1, -4, 4, -9, 12, -22, 3, ... This is A000071-reversed followed by A119282.
Alternating sums of rows of the triangle in A141169. - Reinhard Zumkeller, Mar 22 2011

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Kyu-Hwan Lee, Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.
Index entries for linear recurrences with constant coefficients, signature (0,2,-1)

Cf. A000071, A008346, A119283, A119284, A119285, A119286, A119287.

Cf. A128696, A128698.

[0] cat [(&+[(-1)^k*Fibonacci(k):k in [1..n]]): n in [1..30]]; // G. C. Greubel, Jan 17 2018

FoldList[#1 - Fibonacci@ #2 &, -Range@ 50] (* Michael De Vlieger, Jan 27 2016 *)
Accumulate[Table[(-1)^n Fibonacci[n], {n, 0, 49}]] (* Alonso del Arte, Apr 25 2017 *)

a(n) = sum(k=1, n, (-1)^k*fibonacci(k)); \\ Michel Marcus, Jan 27 2016

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k = 1..n} (-1)^k F(k).
Closed form: a(n) = (-1)^n F(n-1) - 1 = (-1)^n A008346(n-1).
Recurrence: a(n) - 2 a(n-2) + a(n-3)= 0.
G.f.: A(x) = -x/(1 - 2 x^2 + x^3) = -x/((1 - x)(1 + x - x^2)).
Another recurrence: a(n) = a(n-2) - a(n-1) - 1. - Rick L. Shepherd, Aug 12 2009

A119284 Alternating sum of the cubes of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -8, 19, -106, 406, -1791, 7470, -31834, 134541, -570428, 2415556, -10233781, 43348852, -183632148, 777872655, -3295130518, 13958382186, -59128679555, 250473067570, -1061021002966, 4494556993465, -19039249115928, 80651553232104, -341645462408521, 1447233402276936, -6130579072469696, 25969549690613035, -110008777837417954, 466004661036246046

Offset: 0

Stuart Clary, May 13 2006

Natural bilateral extension (brackets mark index 0): ..., 674, 162, 37, 10, 2, 1, 0, [0], -1, 0, -8, 19, -106, 406, -1791, ... This is A005968-reversed followed by A119284.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-2,9,-3,-4,1).

Cf. A005968, A119282, A119283, A119285, A119286, A119287, A128696, A128698.

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^3, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^3, {k, 1, -n - 1} ] ]
Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0,30]]^3,{1,-1},{2,-1,2}],2]] (* or *) LinearRecurrence[{-2,9,-3,-4,1},{0,-1,0,-8,19},40] (* Harvey P. Dale, Aug 23 2020 *)

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=1..n} (-1)^k F(k)^3.
Closed form: a(n) = (-1)^n F(3n+1)/10 - 3 F(n+2)/5 + 1/2.
Recurrence: a(n) + 2 a(n-1) - 9 a(n-2) + 3 a(n-3) + 4 a(n-4) - a(n-5) = 0.
G.f.: A(x) = (-x - 2 x^2 + x^3)/(1 + 2 x - 9 x^2 + 3 x^3 + 4 x^4 - x^5) = x(-1 - 2 x + x^2)/((1 - x)(1 - x - x^2 )(1 + 4 x - x^2)).

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)).

A119286 Alternating sum of the fifth powers of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -32, 211, -2914, 29854, -341439, 3742662, -41692762, 461591613, -5122467836, 56794896388, -629924960005, 6985721085652, -77473909014348, 859194263419359, -9528629686028398, 105674040835291026, -1171943417651373875, 12997050199917354250, -144139501695851560726, 1598531543102764228825, -17727986584911448406232, 196606383515036414871336, -2180398207207766329269289

Offset: 0

Stuart Clary, May 13 2006

Natural bilateral extension (brackets mark index 0): ..., 3402, 277, 34, 2, 1, 0, [0], -1, 0, -32, 211, -2914, 29854, ... This is A098531-reversed followed by A119286.

Index entries for linear recurrences with constant coefficients, signature (-7,48,20,-100,32,9,-1)

Cf. A098531, A119282, A119283, A119284, A119285, A119287, A128696, A128698.

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^5, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^5, {k, 1, -n - 1} ] ]
LinearRecurrence[{-7,48,20,-100,32,9,-1},{0,-1,0,-32,211,-2914,29854},30] (* Harvey P. Dale, Jun 24 2018 *)

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=1..n} (-1)^k F(k)^5.
Closed form: a(n) = (-1)^n (1/275)(F(5n+1) + 2 F(5n+3)) - (1/10) F(3n+2) + (-1)^n (2/5) F(n-1) - 7/22; here F(5n+1) + 2 F(5n+3) = A001060(5n+1) = A013655(5n+2).
Recurrence: a(n) + 7 a(n-1) - 48 a(n-2) - 20 a(n-3) + 100 a(n-4) - 32 a(n-5) - 9 a(n-6) + a(n-7) = 0.
G.f.: A(x) = (-x - 7 x^2 + 16 x^3 + 7 x^4 - x^5)/(1 + 7 x - 48 x^2 - 20 x^3 + 100 x^4 - 32 x^5 - 9 x^6 + x^7) = -x(1 + 7 x - 16 x^2 - 7 x^3 + x^4)/((1 - x)(1 + x - x^2)(1 - 4 x - x^2)(1 + 11 x - x^2)).

A119287 Alternating sum of the sixth powers of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -64, 665, -14960, 247184, -4579625, 81186496, -1463617920, 26217022705, -470764268256, 8445336180000, -151560390359569, 2719538168853120, -48800836192146880, 875690649999921929, -15713664197268146000, 281970036429821245616, -5059748557502924705465, 90793493265349521060160, -1629223203785737022267136, 29235223670642547226470625

Offset: 0

Stuart Clary, May 13 2006

Natural bilateral extension (brackets mark index 0): ..., 14960, -665, 64, 0, 1, 0, [0], -1, 0, -64, 665, -14960, 247184, ... This is (-A119287)-reversed followed by A119287.

Index entries for linear recurrences with constant coefficients, signature (-12,117,156,-520,156,117,-12,-1).

Cf. A098532, A119282, A119283, A119284, A119285, A119286, A128696, A128698.

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^6, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^6, {k, 1, -n - 1} ] ]
Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0,30]]^6,{1,-1},{2,-1,2}],2]] (* Harvey P. Dale, Jul 23 2013 *)

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=1..n} (-1)^k F(k)^6.
a(n) = (-1)^n (1/250) F(6n+3) - (6/125) F(4n+2) + (-1)^n (3/25) F(2n+1) - (2/25)(2 n + 1).
Recurrence: a(n) + 12 a(n-1) - 117 a(n-2) - 156 a(n-3) + 520 a(n-4) - 156 a(n-5) - 117 a(n-6) + 12 a(n-7) + a(n-8) = 0.
G.f.: A(x) = (-x - 12 x^2 + 53 x^3 + 53 x^4 - 12 x^5 - x^6)/(1 + 12 x - 117 x^2 - 156 x^3 + 520 x^4 - 156 x^5 - 117 x^6 + 12 x^7 + x^8) = -x(1 + x)(1 + 11 x - 64 x^2 + 11 x^3 + x^4)/((1 - x)^2 (1 + 3 x + x^2)(1 - 7 x + x^2)(1 + 18 x + x^2)).

A128696 Alternating sum of the seventh powers of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -128, 2059, -76066, 2021086, -60727431, 1740361110, -50782989034, 1471652245341, -42759682650188, 1241158781898676, -36040175501820901, 1046363981321362852, -30381064378888637148, 882092032492683277335, -25611107658594421205278, 743603574761804566730466, -21590121866471006254739195, 626857059065125789349713930

Offset: 0

Stuart Clary, Mar 23 2007

Natural bilateral extension (brackets mark index 0): ..., 2177594, 80442, 2317, 130, 2, 1, 0, [0], -1, 0, -128, 2059, -76066, 2021086 ... This is A098533-reversed followed by A128696.

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

Cf. A098533, A119282, A119283, A119284, A119285, A119286, A119287, A128698.

[(&+[(-1)^k*Fibonacci(k)^7: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 17 2018

a[ n_Integer ] := If[ n >= 0, Sum[ (-1)^k Fibonacci[ k ]^7, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k ]^7, {k, 1, -n - 1} ] ]
Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0,30]]^7,{1,-1}], 2]] (* Harvey P. Dale, May 11 2012 *)

a(n) = sum(k=1, n, (-1)^k*fibonacci(k)^7); \\ Michel Marcus, Dec 10 2016

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=1..n} (-1)^k F(k)^7.
Closed form: a(n) = (-1)^n (F(7n+7) - F(7n))/3625 + 7(F(5n+1) - 2 F(5n+4))/1375 + (-1)^n 21 F(3n+1)/250 - 7 F(n+2)/25 + 139/638.
Recurrence: a(n) + 20 a(n-1) - 294 a(n-2) - 819 a(n-3) + 2912 a(n-4) - 728 a(n-5) - 1365 a(n-6) + 252 a(n-7) + 22 a(n-8) - a(n-9) = 0.
G.f.: A(x) = (-x - 20 x^2 + 166 x^3 + 318 x^4 - 166 x^5 - 20 x^6 + x^7)/(1 + 20 x - 294 x^2 - 819 x^3 + 2912 x^4 - 728 x^5 - 1365 x^6 + 252 x^7 + 22 x^8 - x^9) = -x*(1 + 20 x - 166 x^2 - 318 x^3 + 166 x^4 + 20 x^5 - x^6)/ ((1 - x)*(1 - x - x^2)*(1 + 4 x - x^2)*(1 - 11 x - x^2)*(1 + 29 x - x^2)).

A128698 Alternating sum of the eighth powers of the first n Fibonacci numbers.

Original entry on oeis.org

0, -1, 0, -256, 6305, -384320, 16392896, -799337825, 37023521536, -1748770383360, 81985167507265, -3854603638194816, 181029655256841600, -8505521232849819841, 399560845889490455040, -18771170453838609544960, 881839776158402870049761, -41427800130507702988683200, 1946222939243803281837279296, -91431083130550578762727373345, 4295314095871701743501398017280

Offset: 0

Stuart Clary, Mar 23 2007

Natural bilateral extension (brackets mark index 0): ..., -16392896, 384320, -6305, 256, 0, 1, 0, [0], -1, 0, -256, 6305, -384320, 16392896, ... This is (-A128698)-reversed followed by A128698.

G. C. Greubel, Table of n, a(n) for n = 0..595

Cf. A128697, A119282, A119283, A119284, A119285, A119286, A119287, A128696.

[(&+[(-1)^k*Fibonacci(k)^8: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 17 2018

a[ n_Integer ] := If[ n >= 0, Sum[ (-1)^k Fibonacci[ k ]^8, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k ]^8, {k, 1, -n - 1} ] ]
Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0,20]]^8,{1,-1},{2,-1,2}],2]] (* Harvey P. Dale, May 04 2016 *)

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

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)^8.
Closed form: a(n) = (-1)^n L(8n+4)/4375 - 2 L(6n+3)/625 + (-1)^n 28 L(4n+2)/1875 - 56 L(2n+1)/625 + (-1)^n 7/125.
Factored closed form: a(n) = (-1)^n (1/21) F(n-2) F(n) F(n+1) F(n+3) (3 F(n)^2 F(n+1)^2 + 4).
Recurrence: a(n) + 34 a(n-1) - 714 a(n-2) - 4641 a(n-3) + 12376 a(n-4) + 12376 a(n-5) - 4641 a(n-6) - 714 a(n-7) + 34 a(n-8) + a(n-9) = 0.
G.f.: A(x) = (-x - 34 x^2 + 458 x^3 + 2242 x^4 + 458 x^5 - 34 x^6 - x^7)/(1 + 34 x - 714 x^2 - 4641 x^3 + 12376 x^4 + 12376 x^5 - 4641 x^6 - 714 x^7 + 34 x^8 + x^9) = -x(1 + 34 x - 458 x^2 - 2242 x^3 - 458 x^4 + 34 x^5 + x^6)/((1 + x)(1 - 3 x + x^2)(1 + 7 x + x^2)(1 - 18 x + x^2)(1 + 47 x + x^2)).

A138238 Alternating sum of the squares of the first n Jacobsthal numbers.

Original entry on oeis.org

0, 1, 0, 9, -16, 105, -336, 1513, -5712, 23529, -92752, 373737, -1489488, 5968873, -23853648, 95458281, -381745744, 1527157737, -6108281424, 24433824745, -97733900880, 390938399721, -1563748006480

Offset: 0

Paul Barry, Mar 07 2008

Index entries for linear recurrences with constant coefficients, signature (-2,9,2,-8).

Cf. A119283.

LinearRecurrence[{-2,9,2,-8},{0,1,0,9},30] (* Harvey P. Dale, Feb 13 2018 *)

G.f. : x(1+2x)/((1-x^2)(1+2x-8x^2));
a(n)=4*2^n/9-4(-4)^n/45-(-1)^n/18-3/10;
a(n)=(13-4*J(2n+1))(-1)^n/30+4*J(n)/3-3/10, J(n)=A001045(n);
a(n)=sum{k=0..n (-1)^(k+1)*J(k)^2};

A214884 a(n) = Sum_{k=0..n} (-1)^kF(k)F(k+2), where F=A000045 (Fibonacci numbers).

Original entry on oeis.org

0, -2, 1, -9, 15, -50, 118, -324, 831, -2195, 5725, -15012, 39276, -102854, 269245, -704925, 1845483, -4831574, 12649186, -33116040, 86698875, -226980647, 594243001, -1555748424, 4073002200, -10663258250, 27916772473, -73087059249, 191344405191

Offset: 0

Wolfdieter Lang, Jul 30 2012

The present sequence is the m=2 member of the m-family of sequences b(m,n):=Sum_{k=0..n} (-1)^k*F(k+2)*F(k) given by b(m,n) = (L(m)*A119283(n) + F(m)*(-1)^n*A001654(n))/2, with A119283(n) = b(0,n) = ((-1)^n*F(2*n+1) - (2*n+1))/5 and A001654(n) = F(n+1)*F(n), where F and L are the Fibonacci and Lucas numbers, A000045 and A000032, respectively.
The o.g.f. of b(m,n) is A(m,x) = -(1/2)*x*(F(m+1) + F(m-1)*x)/((1-x)^2*(1+3*x+x^2)), m >= 0, with F(-1) = 1. For the unsigned sums see a comment on A080144.
b(m, n) = ((-1)^n*F(m + 2*n + 1) - n*L(m) - F(m + 1))/5. - Ehren Metcalfe, Aug 21 2017

Cf. A119283, -A077916(n-1) for the m=0 and m=1 cases. A214885 for m=3.

Table[Sum[(-1)^k*Fibonacci[k]*Fibonacci[k + 2], {k, 0, n}], {n, 0, 28}] (* Michael De Vlieger, Aug 23 2017 *)

a(n) = b(2,n) = (3*A119283(n) + (-1)^n*A001654(n))/2, n >= 0.
O.g.f.: -x*(2+x)/((1-x)^2*(1+3*x+x^2)) (see the comment section).
a(n) = ((-1)^n*Fibonacci(2*n + 3) - 3*n - 2)/5. - Ehren Metcalfe, Aug 21 2017

Showing 1-10 of 11 results. Next

cp's OEIS Frontend

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

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A119282 Alternating sum of the first n Fibonacci numbers.

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A119284 Alternating sum of the cubes of the first n Fibonacci numbers.

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

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A119286 Alternating sum of the fifth powers of the first n Fibonacci numbers.

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A119287 Alternating sum of the sixth powers of the first n Fibonacci numbers.

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A128696 Alternating sum of the seventh powers of the first n Fibonacci numbers.

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A214884 a(n) = Sum_{k=0..n} (-1)^kF(k)F(k+2), where F=A000045 (Fibonacci numbers).