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.

A124400 a(n) = a(n-1) + 3*a(n-2) - a(n-4), with a(0)=1, a(1)=1, a(2)=4, a(3)=7.

Original entry on oeis.org

1, 1, 4, 7, 18, 38, 88, 195, 441, 988, 2223, 4992, 11220, 25208, 56645, 127277, 285992, 642615, 1443946, 3244514, 7290360, 16381287, 36808421, 82707768, 185842671, 417584688, 938304280, 2108350576, 4737420745, 10644887785, 23918845740
Offset: 0

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Author

Philippe Deléham, Dec 14 2006

Keywords

Comments

Unsigned version of A077920.
The sequence is the INVERT transform of the aerated even-indexed Fibonacci numbers (i.e., of (1, 0, 3, 0, 8, 0, ...)). Sequence A131322 is the INVERT transform of the aerated odd-indexed Fibonacci numbers. - Gary W. Adamson, Feb 07 2014

Crossrefs

Cf. A131322.

Programs

  • GAP
    a:=[1,1,4,7];; for n in [5..35] do a[n]:=a[n-1]+3*a[n-2]-a[n-4]; od; a; # G. C. Greubel, Dec 25 2019
  • Magma
    I:=[1,1,4,7]; [n le 2 select I[n] else Self(n-1) +3*Self(n-2) -Self(n-4): n in [1..35]]; // G. C. Greubel, Dec 25 2019
    
  • Maple
    seq(coeff(series(1/(1-x-3*x^2+x^4), x, n+1), x, n), n = 0..35); # G. C. Greubel, Dec 25 2019
  • Mathematica
    LinearRecurrence[{1,3,0,-1}, {1,1,4,7}, 35] (* G. C. Greubel, Dec 25 2019 *)
    CoefficientList[Series[1/(1-x-3x^2+x^4),{x,0,30}],x] (* Harvey P. Dale, Feb 01 2022 *)
  • PARI
    my(x='x+O('x^35)); Vec(1/(1-x-3*x^2+x^4)) \\ G. C. Greubel, Dec 25 2019
    
  • Sage
    def A124400_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( 1/(1-x-3*x^2+x^4) ).list()
    A124400_list(35) # G. C. Greubel, Dec 25 2019
    

Formula

G.f.: 1/(1-x-3*x^2+x^4).

A131321 Triangle read by rows: A168561^2.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 4, 0, 1, 5, 0, 6, 0, 1, 0, 14, 0, 8, 0, 1, 13, 0, 27, 0, 10, 0, 1, 0, 46, 0, 44, 0, 12, 0, 1, 34, 0, 107, 0, 65, 0, 14, 0, 1, 0, 145, 0, 204, 0, 90, 0, 16, 0, 1, 89, 0, 393, 0, 345, 0, 119, 0, 18, 0, 1, 0, 444, 0, 854, 0, 538, 0, 152, 0, 20, 0, 1
Offset: 0

Views

Author

Gary W. Adamson, Jun 28 2007

Keywords

Comments

Left border, nonzero terms = odd indexed Fibonacci numbers: (1, 2, 5, 13, ...). Next column, nonzero terms = A030267: (1, 4, 14, 46, 145, ...). Row sums = A131322: (1, 1, 3, 5, 12, 23, 51, ...).
Riordan array (f(x),x*f(x)) where f(x) = (1-x^2)/(1-3*x^2+x^4). Aerated version of triangle in A188137. - Philippe Deléham, Jan 26 2012

Examples

			First few rows of the triangle are:
   1;
   0,  1;
   2,  0,  1;
   0,  4,  0,  1;
   5,  0,  6,  0,  1;
   0, 14,  0,  8,  0,  1;
  13,  0, 27,  0, 10,  0,  1;
  ...
		

Crossrefs

Programs

  • Maple
    F:= (n, k)-> coeff(combinat[fibonacci](n+1, x), x, k):
    T:= (n, k)-> add(F(n, j)*F(j, k), j=0..n):
    seq(seq(T(n, k), k=0..n), n=0..14);  # Alois P. Heinz, Dec 12 2019

Formula

A168561 squared, as an infinite lower triangular matrix.

A166482 a(n) = Sum_{k=0..n} binomial(n+k,2k)*Fibonacci(2k+1).

Original entry on oeis.org

1, 3, 12, 51, 221, 965, 4227, 18540, 81363, 357145, 1567849, 6883059, 30218028, 132664227, 582428789, 2557009709, 11225925267, 49284687948, 216372426339, 949930508209, 4170438905425, 18309298027683, 80382521554380
Offset: 0

Views

Author

Paul Barry, Oct 14 2009

Keywords

Comments

Conjecture: a(n) is the number of tilings of a 4 X 4n rectangle into L tetrominoes (no reflections, only rotations). - Nicolas Bělohoubek, Feb 12 2022
The conjecture above was confirmed by Nicolas Bělohoubek and Antonín Slavík. (See links.) - Nicolas Bělohoubek, Jan 21 2025

Crossrefs

Cf. A131322.

Programs

  • Mathematica
    CoefficientList[Series[(1-4x+4x^2-x^3)/(1-7x+13x^2-7x^3+x^4), {x,0,30}],x]  (* Harvey P. Dale, Mar 23 2011 *)
    LinearRecurrence[{7, -13, 7, -1}, {1, 3, 12, 51}, 50] (* G. C. Greubel, May 15 2016 *)

Formula

G.f.: (1 - 4x + 4x^2 - x^3)/(1 - 7x + 13x^2 - 7x^3 + x^4).
a(n) = Sum_{k=0..n} binomial(n+k,2k) * Sum_{j=0..k} binomial(k+j,2j).
a(n) ~ (1 + 1/sqrt(5) + 2*sqrt(31/290 + 13/(58*sqrt(5)))) * ((7 + sqrt(5) + sqrt(38 + 14*sqrt(5)))^n / 2^(2*n+2)). - Vaclav Kotesovec, Feb 22 2022
a(n) = A131322(2*n). - Nicolas Bělohoubek, Jan 21 2025
Showing 1-3 of 3 results.