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.

A203169 Sum of the fourth powers of the first n even-indexed Fibonacci numbers.

Original entry on oeis.org

0, 1, 82, 4178, 198659, 9349284, 439330980, 20639983621, 969645224182, 45552722051318, 2140008541351943, 100534850436141384, 4722997973709689160, 221880369994471370761, 10423654392318557192602, 489689876072761951752602
Offset: 0

Views

Author

Stuart Clary, Dec 30 2011

Keywords

Comments

Natural bilateral extension (brackets mark index 0): ..., -9349284, -198659, -4178, -82, -1, 0, [0], 1, 82, 4178, 198659, 9349284, ... That is, a(-n) = -a(n-1).

Crossrefs

Programs

  • Mathematica
    a[n_Integer] := (1/75)(Fibonacci[8n+4] - 12*Fibonacci[4n+2] + 9*(2*n+1)); Table[a[n], {n, 0, 20}]

Formula

Let F(n) be the Fibonacci number A000045(n).
a(n) = sum_{k=1..n} F(2k)^4.
Closed form: a(n) = (1/75)(F(8n+4) - 12 F(4n+2) + 9(2 n + 1)).
Recurrence: a(n) - 56 a(n-1) + 440 a(n-2) - 770 a(n-3) + 440 a(n-4) - 56 a(n-5) + a(n-6) = 0.
G.f.: A(x) = (x + 26 x^2 + 26 x^3 + x^4)/(1 - 56 x + 440 x^2 - 770 x^3 + 440 x^4 - 56 x^5 + x^6) = x(1 + x)(1 + 25 x + x^2)/((1 - x)^2 (1 - 7 x + x^2)(1 - 47 x + x^2)).

A203170 Sum of the fourth powers of the first n odd-indexed Fibonacci numbers.

Original entry on oeis.org

0, 1, 17, 642, 29203, 1365539, 64107780, 3011403301, 141469813301, 6646055880582, 312223061019703, 14667837157106759, 689076118833981960, 32371909717271872585, 1520790680382055836761, 71444790066793903279242
Offset: 0

Views

Author

Stuart Clary, Dec 30 2011

Keywords

Comments

Natural bilateral extension (brackets mark index 0): ..., -1365539, -29203, -642, -17, -1, [0], 1, 17, 642, 29203, 1365539, ... That is, a(-n) = -a(n).

Crossrefs

Programs

  • Mathematica
    a[n_Integer] := (1/75)(Fibonacci[8n] + 12*Fibonacci[4n] + 18 n); Table[a[n], {n, 0, 20}]

Formula

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=1..n} F(2k-1)^4.
Closed form: a(n) = (1/75)(F(8n) + 12 F(4n) + 18 n).
Recurrence: a(n) - 56 a(n-1) + 440 a(n-2) - 770 a(n-3) + 440 a(n-4) - 56 a(n-5) + a(n-6) = 0.
G.f.: A(x) = (x - 39 x^2 + 130 x^3 - 39 x^4 + x^5)/(1 - 56 x + 440 x^2 - 770 x^3 + 440 x^4 - 56 x^5 + x^6) = x(1 - 39 x + 130 x^2 - 39 x^3 + x^4)/((1 - x)^2 (1 - 7 x + x^2)(1 - 47 x + x^2)).

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).

Crossrefs

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.