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.

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A081016 a(n) = (Lucas(4*n+3) + 1)/5, or Fibonacci(2*n+1)*Fibonacci(2*n+2), or A081015(n)/5.

Original entry on oeis.org

1, 6, 40, 273, 1870, 12816, 87841, 602070, 4126648, 28284465, 193864606, 1328767776, 9107509825, 62423800998, 427859097160, 2932589879121, 20100270056686, 137769300517680, 944284833567073, 6472224534451830
Offset: 0

Views

Author

R. K. Guy, Mar 01 2003

Keywords

Comments

a(n-1) is, together with b(n) := A089508(n), n >= 1, the solution to a binomial problem; see A089508.
Numbers k such that 1 - 2*k + 5*k^2 is a square. - Artur Jasinski, Oct 26 2008
Also solution y of Diophantine equation x^2 + 4*y^2 = h^2 for which x = y-1. - Carmine Suriano, Jun 23 2010

References

  • Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 26.

Links

Crossrefs

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081015.
Partial sums of A033889. Bisection of A001654. Equals A003482 + 1.
Cf. A145995, A178898.

Programs

  • GAP
    List([0..30], n-> (Lucas(1,-1,4*n+3)[2] +1)/5 ); # G. C. Greubel, Jul 13 2019
  • Magma
    [(Lucas(4*n+3) +1)/5: n in [0..30]]; // G. C. Greubel, Dec 18 2017
  • Maple
    luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d,`,(luc(4*n+3)+1)/5) od: # James Sellers, Mar 03 2003
  • Mathematica
    LinearRecurrence[{8,-8,1}, {1,6,40}, 30] (* Bruno Berselli, Aug 31 2017 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; 1,-8,8]^n*[1;6;40])[1,1] \\ Charles R Greathouse IV, Sep 28 2015
  • PARI
    first(n) = Vec((1-2*x)/((1-x)*(1-7*x+x^2)) + O(x^n)) \\ Iain Fox, Dec 19 2017
  • Sage
    [(lucas_number2(4*n+3,1,-1) +1)/5 for n in (0..30)] # G. C. Greubel, Jul 13 2019

Formula

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: (1 - 2*x)/((1 - x)*(1 - 7*x + x^2)).
a(n) = F(1) + F(5) + F(9) +...+ F(4*n+1) = F(2*n)*F(2*n+3) + 1, where F(j) = Fibonacci(j).
a(n) = 7*a(n-1) - a(n-2) - 1, n >= 2. - R. J. Mathar, Nov 07 2015

A354336 a(n) is the integer w such that (L(2*n)^2, -L(2*n-1)^2, -w) is a primitive solution to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

Original entry on oeis.org

1, 11, 61, 401, 2731, 18701, 128161, 878411, 6020701, 41266481, 282844651, 1938646061, 13287677761, 91075098251, 624238009981, 4278590971601, 29325898791211, 201002700566861, 1377693005176801, 9442848335670731, 64722245344518301, 443612869075957361
Offset: 0

Views

Author

XU Pingya, Jun 20 2022

Keywords

Comments

Subsequence of A017281.

Examples

			2*(L(4)^2)^3 + 2*(-L(3)^2)^3 + (-61)^3 = 2*(49)^3 + 2*(-1)^3 + (-61)^3 = 125, a(2) = 61.

Links

Crossrefs

Cf. A000032, A002878, A005248, A017281, A056914, A081015, A092521.
Cf. A337928, A354337.

Programs

  • Mathematica
    LucasL[4*Range[22]-3] + 1 - LucasL[2*Range[22]-3]^2

Formula

a(n) = (-125 + 2*A005248(n)^6 - 2*A002878(n-1)^6)^(1/3).
a(n) = Lucas(4*n+1) - Lucas(4*n-2) + 3 = A056914(n) - 15*A092521(n-1), for n > 1.
a(n) = Lucas(4*n+1) + 1 - Lucas(2*n-1)^2.
a(n) = 2*A081015(n-1) + 1.
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: (1 + 3*x - 19*x^2)/((1 - x)*(1 - 7*x + x^2)). - Stefano Spezia, Jun 22 2022
a(n) = (F(2*n+1) + F(2*n-1))^2 + (F(2*n+1) + F(2*n-1)) * (F(2*n-1) + F(2*n-3)) - (F(2*n-1) + F(2*n-3))^2. - XU Pingya, Jul 17 2024
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