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.

A056487 a(n) = 5^(n/2) for n even, a(n) = 3*5^((n-1)/2) for n odd.

Original entry on oeis.org

1, 3, 5, 15, 25, 75, 125, 375, 625, 1875, 3125, 9375, 15625, 46875, 78125, 234375, 390625, 1171875, 1953125, 5859375, 9765625, 29296875, 48828125, 146484375, 244140625, 732421875, 1220703125, 3662109375, 6103515625, 18310546875, 30517578125, 91552734375
Offset: 0

Views

Author

Marks R. Nester

Keywords

Comments

Apparently identical to A111386! Is this a theorem? - Klaus Brockhaus, Jul 21 2009
For n > 1, number of necklaces with n-1 beads and 5 colors that are the same when turned over and hence have reflection symmetry. - Herbert Kociemba, Nov 24 2016

References

  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Links

Crossrefs

Column 5 of A284855.
Cf. A029744, A038754, A056391, A056451, A087205, A087206, A111386.

Programs

Formula

a(n+2) = 5*a(n), a(0)=1, a(2)=3.
Binomial transform of A087205. Binomial transform is A087206. - Paul Barry, Aug 25 2003
G.f.: (1+3*x)/(1-5*x^2); a(n) = 5^(n/2)(1/2 + 3*sqrt(5)/10 + (1/2 - 3*sqrt(5)/10)(-1)^n). - Paul Barry, Mar 19 2004
2nd inverse binomial transform of Fibonacci(3n+2). - Paul Barry, Apr 16 2004
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = 3^((1 - (-1)^n)/2) * 5^((2*n + (-1)^n-1)/4). - Bruno Berselli, Mar 24 2011
a(n+1) = (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 2, where k=5 is the number of possible colors. - Robert A. Russell, Sep 22 2018
E.g.f.: cosh(sqrt(5)*x) + 3*sinh(sqrt(5)*x)/sqrt(5). - Stefano Spezia, Jun 06 2023

Extensions

Changed one 'even' to 'odd' in the definition. - R. J. Mathar, Oct 06 2010

A208343 Triangle of coefficients of polynomials v(n,x) jointly generated with A208342; see the Formula section.

Original entry on oeis.org

1, 0, 2, 0, 1, 3, 0, 1, 2, 5, 0, 1, 2, 5, 8, 0, 1, 2, 6, 10, 13, 0, 1, 2, 7, 13, 20, 21, 0, 1, 2, 8, 16, 29, 38, 34, 0, 1, 2, 9, 19, 39, 60, 71, 55, 0, 1, 2, 10, 22, 50, 86, 122, 130, 89, 0, 1, 2, 11, 25, 62, 116, 187, 241, 235, 144, 0, 1, 2, 12, 28, 75, 150, 267, 392, 468
Offset: 1

Views

Author

Clark Kimberling, Feb 25 2012

Keywords

Comments

u(n,n) = A000045(n+1) (Fibonacci numbers).
n-th row sum: 2^(n-1)
As triangle T(n,k) with 0 <= k <= n, it is (0, 1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, -1/2, -1/2, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 26 2012

Examples

			First five rows:
  1;
  0, 2;
  0, 1, 3;
  0, 1, 2, 5;
  0, 1, 2, 5, 8;
First five polynomials v(n,x):
  1
     2x
      x + 3x^2
      x + 2x^2 + 5x^3
      x + 2x^2 + 5x^3 + 8x^4.

Crossrefs

Cf. A208343.
Cf. A084938, A000045, A000079.

Programs

  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 13;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
v[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A208342 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A208343 *)

Formula

u(n,x) = u(n-1,x) + x*v(n-1,x),
v(n,x) = x*u(n-1,x) + x*v(n-1,x),
where u(1,x)=1, v(1,x)=1.
From Philippe Deléham, Feb 26 2012: (Start)
As triangle T(n,k) with 0 <= k <= n:
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k-2) - T(n-2,k-1), T(0,0) = 1, T(1,0) = 0, T(1,1) = 2, T(n,k) = 0 if k > n or if k < 0.
G.f.: (1-(1-y)*x)/(1-(1+y)*x+y*(1-y)*x^2).
Sum_{k=0..n} T(n,k)*x^k = (-1)*A091003(n+1), A152166(n), A000007(n), A000079(n), A055099(n), A152224(n) for x = -2, -1, 0, 1, 2, 3 respectively.
Sum_{k=0..n} T(n,k)*x^(n-k) = A087205(n), A140165(n+1), A016116(n+1), A000045(n+2), A000079(n), A122367(n), A006012(n), A052961(n), A154626(n) for x = -3, -2, -1, 0, 1, 2, 3, 4 respectively. (End)
T(n,k) = A208748(n,k)/2^k. - Philippe Deléham, Mar 05 2012

A319196 a(n) = 2^(n-1)*Fibonacci(n-3), n >= 0.

Original entry on oeis.org

1, -1, 2, 0, 8, 16, 64, 192, 640, 2048, 6656, 21504, 69632, 225280, 729088, 2359296, 7634944, 24707072, 79953920, 258736128, 837287936, 2709520384, 8768192512, 28374466560, 91821703168, 297141272576, 961569357824, 3111703805952, 10069685043200, 32586185310208, 105451110793216
Offset: 0

Views

Author

Wolfdieter Lang, Oct 09 2018

Keywords

Comments

This sequence gives the elements M^n(2, 2) of the matrix M = [[3, 1], [1, -1]].
Motivation to look into these matrix powers came from A319053. M^n[1, 1] = A063782 and M^n(1, 2) = M^n(2, 1) = A085449(n). Proof by Cayley-Hamilton, using S(n, -I) = (-I)^n*F(n+1), and S(n, x) from A049310 and F = A000045.
For a similar signed sequence see A087205.

Links

Crossrefs

Cf. A000045, A001622, A049310, A085449, A063782, A087205, A319053.

Programs

  • Magma
    [2^(n-1)*Fibonacci(n-3): n in [0..30]]; // Vincenzo Librandi, Oct 09 2018
  • Mathematica
    Table[2^(n-1) Fibonacci[n-3], {n, 0, 40}] (* Vincenzo Librandi, Oct 09 2018 *)
LinearRecurrence[{2,4},{1,-1},40] (* Harvey P. Dale, Mar 29 2020 *)

Formula

a(n) = 2^(n-1)*F(n-3), n >= 0, with F = A000045 with F(-1) = 1, F(-2) = -1 and F(-3) = 1.
G.f.: (1-3*x)/(1-2*x-(2*x)^2).
a(n) = 2*(a(n-1) + 2*a(n-2)), n >= 2, with a(0) = 1 and a(1) = -1.
a(n) = 2^(n-1)*(phi^(n-3) - (1 - phi)^(n-3))/(2*phi - 1) with the golden section phi = A001622.
Showing 1-3 of 3 results.

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