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.

A335711 The number of free polyominoes of width 2 and height n.

Original entry on oeis.org

2, 6, 12, 30, 65, 158, 362, 875, 2064, 4984, 11914, 28764, 69155, 166956, 402372, 971413, 2343518, 5657754, 13654968, 32966010, 79577189, 192116330, 463786190, 1119678911, 2703086892, 6525829036, 15754607062, 38034986040, 91824246215, 221683340568, 535190123592
Offset: 2

Views

Author

R. J. Mathar, Jun 18 2020

Keywords

Comments

The second column of A268371.

Examples

			a(2)=2, bounding box 2 X 2, counts the L-shaped 3-omino and the full block 4-omino.
a(3)=6, bounding box 2 X 3, counts three 4-ominoes, two 5-omioes, and the full 2 X 3 block 6-omino.
a(4)=12, bounding box 2 X 4, counts three 5-ominoes, six 6-ominoes, two 7-ominoes, and the full 2 X 4 block 8-omino.
		

Crossrefs

Cf. A268371, A107769 (asymmetric), A005409 (C_2 symmetry and higher), A352720 (width 2 and size n).

Formula

Conjecture: a(n) = A107769(n-1) + A005409(floor((n+3)/2)).
Conjectures from Colin Barker, Jun 24 2020: (Start)
G.f.: x^2*(2 - 8*x^2 + 2*x^3 - x^4 + x^5 + x^6) / ((1 - x)*(1 - 2*x - x^2)*(1 - 2*x^2 - x^4)).
a(n) = 3*a(n-1) + a(n-2) - 7*a(n-3) + 3*a(n-4) - a(n-5) + a(n-6) + a(n-7) for n>8.
(End)
a(n) = (2*r(n) + 2*m(n) + A078057(n) + 1) / 4, where r(n) = A078057(floor((n-1)/2) - 1)/2, and m(n) = A078057(floor((n+1)/2) - 3)/2. - John Mason, Feb 28 2022

Extensions

a(12)-a(20) from Jean-Luc Manguin, Jun 23 2020
a(21)-a(28) from John Mason, Feb 27 2022
a(29)-a(32) from John Mason, Feb 28 2022

A126877 Numbers of unstrained alkane staggered conformers (acyclic) of the unbranched type for point group C1. See Table 4 of the Cyvin et al. reference for precise definition.

Original entry on oeis.org

1, 2, 8, 19, 54, 130, 334, 804, 1984, 4777, 11615, 27957, 67472, 162276, 390546
Offset: 5

Views

Author

N. J. A. Sloane, Mar 17 2007

Keywords

Comments

For an explanation of the various symmetry groups in Table 4 (p. 131) of the paper by Cyvin et al. (1998), see the wikipedia links.

Crossrefs

Cf. A107769.

A193530 Expansion of (1 - 2*x - 2*x^2 + 3*x^3 + x^5)/((1-x)*(1-2*x-x^2)*(1-2*x^2-x^4)).

Original entry on oeis.org

1, 1, 2, 3, 7, 13, 31, 66, 159, 363, 876, 2065, 4985, 11915, 28765, 69156, 166957, 402373, 971414, 2343519, 5657755, 13654969, 32966011, 79577190, 192116331, 463786191, 1119678912, 2703086893, 6525829037, 15754607063, 38034986041, 91824246216, 221683340569, 535190123593, 1292063254826
Offset: 0

Views

Author

F. Chapoton and N. J. A. Sloane, Jul 29 2011

Keywords

Comments

This sequence was initially confused with A003120, but they are different sequences. The g.f. used here as the definition was found by Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009.

Crossrefs

Programs

  • Magma
    m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-2*x-2*x^2 +3*x^3+x^5)/((1-x)*(1-2*x-x^2)*(1-2*x^2-x^4)) )); // Vincenzo Librandi, Aug 28 2016
    
  • Maple
    f:=n->if n mod 2 = 0 then (1/4)*(A001333(n-2)+A001333((n-2)/2)+A001333((n-4)/2)+1) else (1/4)*(A001333(n-2)+A001333((n-1)/2)+A001333((n-3)/2)+1); fi; # produces the sequence with a different offset
  • Mathematica
    LinearRecurrence[{3,1,-7,3,-1,1,1}, {1,1,2,3,7,13,31}, 40] (* Vincenzo Librandi, Aug 28 2016 *)
    Table[(2 +LucasL[n, 2] +2*(1+(-1)^n)*Fibonacci[(n+2)/2, 2] + 2*(1-(-1)^n)*Fibonacci[(n+1)/2, 2])/8, {n, 0, 40}] (* G. C. Greubel, May 21 2021 *)
  • Sage
    @CachedFunction
    def Pell(n): return n if (n<2) else 2*Pell(n-1) + Pell(n-2)
    def A193530(n): return (1 + Pell(n+1) - Pell(n) + (1 + (-1)^n)*Pell((n+2)/2) + (1-(-1)^n)*Pell((n+1)/2) )/4
    [A193530(n) for n in (0..40)] # G. C. Greubel, May 21 2021

Formula

a(n) = 1 + A005409(floor((n+3)/2)) + A107769(n).
From G. C. Greubel, May 21 2021: (Start)
a(n) = (1 + A001333(n) + A135153(n+2))/4.
a(n) = (2 + Q(n) + 2*(1+(-1)^n)*Pell((n+2)/2) + 2*(1-(-1)^n)*Pell((n+1)/2))/8.
a(2*n) = (2 + Q(2*n) + 4*Pell(n+1))/8.
a(2*n+1) = (2 + Q(2*n+1) + 4*Pell(n+1))/8, where Pell(n) = A000129(n), and Q(n) = A002203. (End)
Showing 1-3 of 3 results.