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.

Showing 1-8 of 8 results.

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

Views

Author

Stuart Clary, May 13 2006

Keywords

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

Crossrefs

Programs

  • Magma
    [0] cat [(&+[(-1)^k*Fibonacci(k):k in [1..n]]): n in [1..30]]; // G. C. Greubel, Jan 17 2018
  • 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 *)
  • PARI
    a(n) = sum(k=1, n, (-1)^k*fibonacci(k)); \\ Michel Marcus, Jan 27 2016
    

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

A119283 Alternating sum of the squares of the first n Fibonacci numbers.

Original entry on oeis.org

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
Offset: 0

Views

Author

Stuart Clary, May 13 2006

Keywords

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.

Examples

			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.
		

Crossrefs

Programs

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

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

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

Views

Author

Stuart Clary, May 13 2006

Keywords

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.

Crossrefs

Programs

  • 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 *)

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)).

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

Views

Author

Benoit Cloitre, Sep 12 2004

Keywords

Crossrefs

Programs

  • Magma
    [(Fibonacci(n)^5*Fibonacci(n+3) + Fibonacci(2*n))/4: n in [0..30]]; // G. C. Greubel, Jan 17 2018
  • Mathematica
    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);
    
  • PARI
    for(n=0,30, print1((fibonacci(n)^5*fibonacci(n+3) + fibonacci(2*n))/4, ", ")) \\ G. C. Greubel, Jan 17 2018
    

Formula

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

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

Views

Author

Stuart Clary, May 13 2006

Keywords

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.

Crossrefs

Programs

  • 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 *)

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)).

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

Views

Author

Stuart Clary, May 13 2006

Keywords

Comments

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.

Crossrefs

Programs

  • Mathematica
    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 *)

Formula

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)).

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

Views

Author

Stuart Clary, Mar 23 2007

Keywords

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.

Crossrefs

Programs

  • Magma
    [(&+[(-1)^k*Fibonacci(k)^7: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 17 2018
  • 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 *)
  • PARI
    a(n) = sum(k=1, n, (-1)^k*fibonacci(k)^7); \\ Michel Marcus, Dec 10 2016
    

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)).

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

Views

Author

Stuart Clary, Mar 23 2007

Keywords

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.

Crossrefs

Programs

  • Magma
    [(&+[(-1)^k*Fibonacci(k)^8: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 17 2018
  • 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 *)
  • PARI
    a(n) = sum(k=1, n, (-1)^k*fibonacci(k)^8); \\ Michel Marcus, Dec 10 2016
    

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)).
Showing 1-8 of 8 results.