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 1-2 of 2 results.

A152191 a(n)=Product_{k=1..floor((n-1)/2)} (1 + 4*cos(k*Pi/n)^2)*(1 - 4*sin(k*Pi/n)^2).

Original entry on oeis.org

1, 1, 1, -4, -3, 5, 0, 13, 21, -68, -55, 89, 0, 233, 377, -1220, -987, 1597, 0, 4181, 6765, -21892, -17711, 28657, 0, 75025, 121393, -392836, -317811, 514229, 0, 1346269, 2178309, -7049156, -5702887, 9227465, 0, 24157817, 39088169, -126491972, -102334155
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Nov 28 2008

Keywords

Crossrefs

Cf. A152189.

Programs

  • Mathematica
    f[n_] = Product[(1 + 4*Cos[k*Pi/n]^2)*(1 - 4*Sin[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}]; Table[N[f[n]], {n, 0, 30}]; Round[%]
  • PARI
    a(n) = round(prod(k=1, floor((n-1)/2), (1+4*cos(k*Pi/n)^2)*(1-4*sin(k*Pi/n)^2))) \\ Colin Barker, Apr 11 2014

Formula

Conjecture: a(2n) = A108196(n-1), n>=2. a(n) = (-1)^(n+1)*A000045(n) *A101675(n-1), n>0. G.f.: 1 -x*(x-1)*(x^2-x+1)*(1+x)^3 / ( (x^4-x^3+2*x^2+x+1)*(x^4+x^3+2*x^2-x+1) ). - R. J. Mathar, Mar 08 2011

Extensions

More terms from Colin Barker, Apr 11 2014

A152192 a(n) = Product_{k=1..floor((n-1)/2)} (1 + 4*cos(2*Pi*k/n)^2).

Original entry on oeis.org

1, 1, 1, 2, 1, 5, 4, 13, 9, 34, 25, 89, 64, 233, 169, 610, 441, 1597, 1156, 4181, 3025, 10946, 7921, 28657, 20736, 75025, 54289, 196418, 142129, 514229, 372100, 1346269, 974169, 3524578, 2550409, 9227465, 6677056, 24157817, 17480761, 63245986, 45765225
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Nov 28 2008

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Product[(1 + 4*Cos[2*Pi*k/n]^2), {k, 1, Floor[(n - 1)/2]}]; a = Table[N[a[n]], {n, 0, 30}]
    Join[{1}, Table[If[EvenQ[n], Fibonacci[(n)/2]^2, Fibonacci[n]], {n, 1, 30}]] (* Greg Dresden, Oct 16 2021 *)
  • PARI
    a(n) = round(prod(k=1, floor((n-1)/2), (1+4*cos(2*Pi*k/n)^2))) \\ Colin Barker, Apr 11 2014

Formula

Lim_{n->infinity} sqrt(a(n+2)/a(n)) = (sqrt(5) + 1)/2.
G.f.: (1+x^6-x^5-3*x^4-x^2+x)/((x^2+1)*(x^2+x-1)*(x^2-x-1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009
For n > 0, a(n) = Fibonacci(n) for n odd, and Fibonacci(n/2)^2 for n even. - Greg Dresden, Oct 16 2021

Extensions

More terms and edited by Colin Barker, Michel Marcus, and Joerg Arndt, Apr 11 2014
Showing 1-2 of 2 results.