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 1-3 of 3 results.

A156088 Alternating sum of the squares of the first n even-indexed Fibonacci numbers.

Original entry on oeis.org

0, -1, 8, -56, 385, -2640, 18096, -124033, 850136, -5826920, 39938305, -273741216, 1876250208, -12860010241, 88143821480, -604146740120, 4140883359361, -28382036775408, 194533374068496, -1333351581704065, 9138927697859960
Offset: 0

Views

Author

Stuart Clary, Feb 04 2009

Keywords

Comments

Apart from signs, same as A092521.
Natural bilateral extension (brackets mark index 0): ..., 2640, -385, 56, -8, 1, 0, [0], -1, 8, -56, 385, -2640, 18096, ... This is (-a(n))-reversed followed by a(n). That is, a(-n) = -a(n-1).

Links

Crossrefs

Cf. A000032, A000045, A081079, A092521, A103434, A103433, A156089.

Programs

  • Magma
    A156088:= func< n | (-1)^n*(Lucas(4*n+2)-3)/15 >; // G. C. Greubel, Jun 12 2025
  • Mathematica
    a[n_]:= If[n >= 0, Sum[(-1)^k Fibonacci[2k]^2, {k,n}], Sum[ -(-1)^k Fibonacci[-2k]^2, {k,-n-1}]];
LinearRecurrence[{-8,-8,-1}, {0,-1,8}, 41] (* G. C. Greubel, Jun 12 2025 *)
  • SageMath
    def A156088(n): return (-1)^n*(lucas_number2(4*n+2,1,-1) -3)//15 # G. C. Greubel, Jun 12 2025

Formula

Let F(n) be the n-th Fibonacci number, A000045(n), and L(n) be the n-th Lucas number, A000032(n), then: (Start)
a(n) = Sum_{k=1..n} (-1)^k F(2*k)^2.
Closed form: a(n) = (-1)^n * (L(4*n+2) - 3)/15.
Factored closed form: a(n) = (1/3) * (-1)^n * F(n)*L(n)*F(n+1)*L(n+1) = (1/3)*(-1)^n * F(2*n)*F(2*n+2).
Recurrence: a(n) + 8*a(n-1) + 8*a(n-2) + a(n-3) = 0.
G.f.: -x/(1 + 8*x + 8*x^2 + x^3) = -x/((1 + x)(1 + 7*x + x^2)). (End)
From G. C. Greubel, Jun 12 2025: (Start)
a(n) = (-1)^n*A081079(n)/15.
E.g.f.: (1/15)*( exp(-7*x/2)*( 3*cosh(p*x) - sqrt(5)*sinh(p*x) ) - 3*exp(-x) ), where p = 3*sqrt(5)/2. (End)

A156090 Alternating sum of the squares of the first n Fibonacci numbers with index divisible by 3.

Original entry on oeis.org

0, -4, 60, -1096, 19640, -352460, 6324596, -113490320, 2036501104, -36543529620, 655747031980, -11766903046104, 211148507797800, -3788906237314396, 67989163763861220, -1220016041512187680, 21892299583455516896, -392841376460687116580
Offset: 0

Views

Author

Stuart Clary, Feb 04 2009

Keywords

Comments

Natural bilateral extension (brackets mark index 0): ..., -19640, 1096, -60, 4, 0, [0], -4, 60, -1096, 19640, -352460, ... This is (-a(n))-reversed followed by a(n). That is, a(-n) = -a(n-1).

Links

Crossrefs

Cf. A000045, A156084, A156085, A156088, A156089, A156091.

Programs

  • GAP
    a:=[0,-4,60,-1096];; for n in [5..20] do a[n]:=-16*a[n-1]+34*a[n-2]-16*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018
  • Magma
    [(-1)^n*Fibonacci(6*n+3)/10 - (2*n + 1)/5: n in [0..20]]; // Vincenzo Librandi, Sep 12 2018
  • Maple
    with(combinat,fibonacci): a:=n->add((-1)^k*fibonacci(3*k)^2,k=1..n): seq(a(n), n=0..20); # Muniru A Asiru, Sep 12 2018
  • Mathematica
    a[n_]:= If[n >= 0, Sum[(-1)^k Fibonacci[3 k]^2, {k, 1, n}], Sum[-(-1)^k Fibonacci[-3 k]^2, {k, 1, -n-1}]];
LinearRecurrence[{-16,34,-16,-1}, {0,-4,60,-1096}, 30] (* Harvey P. Dale, Oct 24 2016 *)
  • SageMath
    def A156090(n): return ((-1)^n*fibonacci(6*n+3) -2*(2*n+1))//10 # G. C. Greubel, Jun 12 2025

Formula

Let F(n) be the n-th Fibonacci number, A000045(n), then: (Start)
a(n) = Sum_{k=1..n} (-1)^k*F(3*k)^2.
Closed form: a(n) = (-1)^n*F(6*n+3)/10 - (2*n + 1)/5.
Recurrence: a(n) + 17*a(n-1) - 17*a(n-2) - a(n-3) = -8.
Recurrence: a(n) + 16*a(n-1) - 34*a(n-2) + 16*a(n-3) + a(n-4) = 0.
G.f.: -4*x*(1 + x)/(1 + 16*x - 34*x^2 + 16*x^3 + x^4) = -4*x(1 + x)/((1 - x)^2*(1 + 18*x + x^2)). (End)
Limit_{n -> oo} a(n)/a(n-1) = -(9 + sqrt(80)). - A.H.M. Smeets, Sep 11 2018
E.g.f.: (-1/(5*sqrt(5)))*( exp(-9*x)*(2*sinh(p*x) - sqrt(5)*cosh(p*x)) + sqrt(5)*(1 + 2*x)*exp(x) ), where p = 4*sqrt(5). - G. C. Greubel, Jun 12 2025

A203172 Alternating sum of the fourth powers of the first n odd-indexed Fibonacci numbers.

Original entry on oeis.org

0, -1, 15, -610, 27951, -1308385, 61433856, -2885861665, 135572548335, -6369013518946, 299207991620175, -14056406104466881, 660351875572408320, -31022481722865482305, 1457396288941918481871, -68466603097469928960610
Offset: 0

Views

Author

Stuart Clary, Dec 30 2011

Keywords

Comments

Natural bilateral extension (brackets mark index 0): ..., -1308385, 27951, -610, 15, -1, [0], -1, 15, -610, 27951, -1308385, ... That is, a(-n) = a(n).

Links

Crossrefs

Cf. A203169, A203170, A203171.
Cf. A156089, A163202.

Programs

  • Mathematica
    a[n_Integer] := (-1)^n (1/525)(3*LucasL[8n] + 28*LucasL[4n] + 63 - (-1)^n 125); Table[a[n], {n, 0, 20}]
LinearRecurrence[{-54,-330,0,330,54,1},{0,-1,15,-610,27951,-1308385},20] (* Harvey P. Dale, Jun 04 2025 *)

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(2k-1)^4.
Closed form: a(n) = (-1)^n (1/525)(3 L(8n) + 28 L(4n) + 63 - (-1)^n 125).
Alternate closed form: a(n) = (1/21) F(2n)^2 (3 F(2n)^2 + 8) if n is even, a(n) = -(1/21)(3 F(2n)^4 + 8 F(2n)^2 + 10) if n is odd.
Recurrence: a(n) + 54 a(n-1) + 330 a(n-2) - 330 a(n-4) - 54 a(n-5) - a(n-6) = 0.
G.f.: A(x) = -(x + 39 x^2 + 130 x^3 + 39 x^4 + x^5)/(1 + 54 x + 330 x^2 - 330 x^4 - 54 x^5 - x^6) = -x(1 + 39 x + 130 x^2 + 39 x^3 + x^4)/((1 - x)(1 + x)(1 + 7 x + x^2)(1 + 47 x + x^2)).
Showing 1-3 of 3 results.

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