A119286 -id:A119286

A119282

Alternating sum of the first n Fibonacci numbers.

+10
16

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,4

COMMENTS

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

LINKS

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)

FORMULA

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

MATHEMATICA

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 *)

PROG

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

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

CROSSREFS

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

Cf. A128696, A128698.

KEYWORD

sign,easy

AUTHOR

Stuart Clary, May 13 2006

STATUS

approved

A119283

Alternating sum of the squares of the first n Fibonacci numbers.

+10
12

0, -1, 0, -4, 5, -20, 44, -125, 316, -840, 2185, -5736, 15000, -39289, 102840, -269260, 704909, -1845500, 4831556, -12649205, 33116020, -86698896, 226980625, -594243024, 1555748400, -4073002225, 10663258224, -27916772500, 73087059221, -191344405220, 500946156380, -1311494063981, 3433536035500, -8989114042584, 23533806092185, -61612304234040

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,4

COMMENTS

Natural bilateral extension (brackets mark index 0): ..., 840, -316, 125, -44, 20, -5, 4, 0, 1, 0, [0], -1, 0, -4, 5, -20, 44, -125, 316, -840, 2185, ... This is (-A119283)-reversed followed by A119283.

LINKS

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (-1,4,-1,-1)

FORMULA

Let F(n) be the Fibonacci number A000045(n).

a(n) = Sum_{k = 1..n} (-1)^k F(k)^2.

Closed form: a(n) = (-1)^n F(2n+1)/5 - (2 n + 1)/5.

Recurrence: a(n) + a(n-1) - 4 a(n-2) + a(n-3) + a(n-4) = 0.

G.f.: A(x) = (-x - x^2)/(1 + x - 4 x^2 + x^3 + x^4) = -x(1 + x)/((1 - x)^2 (1 + 3 x + x^2)).

a(n) = ( -1-2*n+(-1)^n*( A001906(n+1)-A001906(n) ))/5. - R. J. Mathar, Nov 16 2007

a(n) = (2^(-1-n)*(-5*2^(1+n)+5*(-3+sqrt(5))^n - sqrt(5)*(-3+sqrt(5))^n + (-3-sqrt(5))^n*(5+sqrt(5)) - 5*2^(2+n)*n)) / 25. - Colin Barker, Apr 25 2017

EXAMPLE

The first few squares of Fibonacci numbers are: 0, 1, 1, 4, 9, 25, 64, 169.

a(0) = 0.

a(1) = 0 - 1 = -1.

a(2) = 0 - 1 + 1 = 0.

a(3) = 0 - 1 + 1 - 4 = -4.

a(4) = 0 - 1 + 1 - 4 + 9 = 5.

MATHEMATICA

altFiboSqSum[n_Integer] := If[n >= 0, Sum[(-1)^k Fibonacci[k]^2, {k, n}], Sum[-(-1)^k Fibonacci[-k]^2, {k, 1, -n - 1}]]; altFiboSqSum[Range[0, 39]] (* Clary *)

Accumulate[Table[(-1)^n Fibonacci[n]^2, {n, 0, 39}]] (* Alonso del Arte, Apr 25 2017 *)

Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0, 40]]^2, {1, -1}, {2, -1, 2}], 2]] (* Harvey P. Dale, Jul 29 2024 *)

PROG

(PARI) concat(0, Vec(-x*(1 + x) / ((1 - x)^2*(1 + 3*x + x^2)) + O(x^40))) \\ Colin Barker, Apr 25 2017

CROSSREFS

Cf. A001654, A119282, A119284, A119285, A119286, A119287, A128696, A128698.

KEYWORD

sign,easy

AUTHOR

Stuart Clary, May 13 2006

STATUS

approved

A119284

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

+10
12

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,4

COMMENTS

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.

LINKS

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).

FORMULA

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)).

MATHEMATICA

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 *)

CROSSREFS

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

KEYWORD

sign,easy

AUTHOR

Stuart Clary, May 13 2006

STATUS

approved

A098531

Sum of fifth powers of first n Fibonacci numbers.

+10
10

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

COMMENTS

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

LINKS

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).

FORMULA

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

MATHEMATICA

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 *)

PROG

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

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

CROSSREFS

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

KEYWORD

nonn,easy

AUTHOR

Benoit Cloitre, Sep 12 2004

EXTENSIONS

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

STATUS

approved

A119285

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

+10
9

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,4

COMMENTS

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.

LINKS

Table of n, a(n) for n=0..25.

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).

FORMULA

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)).

MATHEMATICA

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 *)

CROSSREFS

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

KEYWORD

sign,easy

AUTHOR

Stuart Clary, May 13 2006

STATUS

approved

A119287

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

+10
9

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,4

COMMENTS

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.

LINKS

Table of n, a(n) for n=0..22.

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

FORMULA

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)).

MATHEMATICA

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 *)

CROSSREFS

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

KEYWORD

sign,easy

AUTHOR

Stuart Clary, May 13 2006

STATUS

approved

A128696

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

+10
9

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,4

COMMENTS

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.

LINKS

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).

FORMULA

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)).

MATHEMATICA

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 *)

PROG

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

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

CROSSREFS

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

KEYWORD

sign,easy

AUTHOR

Stuart Clary, Mar 23 2007

STATUS

approved

A128698

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

+10
9

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,4

COMMENTS

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.

LINKS

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

FORMULA

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)).

MATHEMATICA

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 *)

PROG

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

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

CROSSREFS

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

KEYWORD

sign,easy

AUTHOR

Stuart Clary, Mar 23 2007

STATUS

approved