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 1-6 of 6 results.

A070765 Number of polyiamonds with n cells, without holes.

Original entry on oeis.org

1, 1, 1, 3, 4, 12, 24, 66, 159, 444, 1161, 3226, 8785, 24453, 67716, 189309, 528922, 1484738, 4172185, 11756354, 33174451, 93795220, 265565628, 753060469, 2138206966, 6078931114, 17302380313, 49302121747, 140627400927, 401510058179
Offset: 1

Views

Author

Joseph Myers, May 05 2002

Keywords

Comments

If holes are allowed, we get A000577.

Crossrefs

Equals A000577(n)-A070764(n). Cf. A000104, A018190.

Extensions

More terms from Joseph Myers, Nov 11 2003
a(29) and a(30) from Joseph Myers, Nov 21 2010

A071334 Number of polyiamonds with n cells without holes that do not tile the plane.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 20, 103, 594, 1192, 6290, 18099, 54808, 159048, 502366, 1374593, 4076218, 11378831, 32674779, 93006494, 264720498, 748062099, 2134512296, 6071524897, 17289205132, 49268564671, 140605019208, 401392287316
Offset: 1

Views

Author

Joseph Myers, May 19 2002

Keywords

Comments

From Bernard Schott, Feb 21 2020: (Start)
There exist 112 polyiamonds without holes that have from 1 to 8 cells (A070765), but only one of these polyiamonds, corresponding to a(7)= 1 cannot tile the plane. This polyiamond is called V-shaped heptiamond (see proof in Martin Gardner's link in German).
\ /\ /\ /
\/\/\/
\ /\ /
\/\/
(End)

References

  • M. Gardner, Tiling with Polyominoes, Polyiamonds and Polyhexes. Chap. 14 in Time Travel and Other Mathematical Bewilderments. New York: W. H. Freeman, pp. 175-187, 1988.

Crossrefs

Programs

  • Mathematica
    A[s_Integer] := With[{s6 = StringPadLeft[ToString[s], 6, "0"]}, Cases[ Import["https://oeis.org/A" <> s6 <> "/b" <> s6 <> ".txt", "Table"], {, }][[All, 2]]];
    A000577 = A@000577;
    A070764 = A@070764;
    A071332 = A@071332;
    a[n_] := A000577[[n]] - A070764[[n]] - A071332[[n]];
    a /@ Range[30] (* Jean-François Alcover, Feb 21 2020 *)

Extensions

More terms from Joseph Myers, Nov 11 2003
a(29) and a(30) from Joseph Myers, Nov 21 2010

A071333 Number of polyiamonds with n cells that do not tile the plane.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 21, 107, 619, 1300, 6740, 19812, 61075, 181036, 577551, 1626183, 4904626, 14071461, 41336066, 120630534, 352200161, 1023454347, 2997105957, 8761810505, 25648786717, 75161609591, 220583137840, 647825855933
Offset: 1

Views

Author

Joseph Myers, May 19 2002

Keywords

References

  • M. Gardner, Tiling with Polyominoes, Polyiamonds and Polyhexes. Chap. 14 in Time Travel and Other Mathematical Bewilderments. New York: W. H. Freeman, pp. 175-187, 1988.

Crossrefs

Equals A000577(n)-A071332(n) and A070764(n)+A071334(n). Cf. A054360, A070767.

Programs

  • Mathematica
    A[s_Integer] := With[{s6 = StringPadLeft[ToString[s], 6, "0"]}, Cases[ Import["https://oeis.org/A" <> s6 <> "/b" <> s6 <> ".txt", "Table"], {, }][[All, 2]]];
    A000577 = A@000577;
    A071332 = A@071332;
    a[n_] := A000577[[n]] - A071332[[n]];
    a /@ Range[30] (* Jean-François Alcover, Feb 21 2020 *)

Extensions

More terms from Joseph Myers, Nov 11 2003
a(29) and a(30) from Joseph Myers, Nov 21 2010

A359523 Number of free polyiamonds of n cells with multiply-connected interiors.

Original entry on oeis.org

1, 2, 12, 51, 217, 846, 3211
Offset: 12

Views

Author

John Mason, Jan 05 2023

Keywords

Crossrefs

Formula

a(n) = A070764(n) - A359524(n).

A359524 Number of free holey polyiamonds of n cells with simply-connected interiors.

Original entry on oeis.org

1, 4, 25, 107, 448, 1701, 6216, 21771, 74339, 248379
Offset: 9

Views

Author

John Mason, Jan 05 2023

Keywords

Crossrefs

Formula

a(n) = A070764(n) - A359523(n).

A327896 a(n) is the minimum number of tiles needed for constructing a polyiamond with n holes.

Original entry on oeis.org

9, 14, 19, 23, 27, 31, 35, 39, 43, 47, 51, 54, 58, 62, 65, 69, 73, 76, 80, 83, 87, 90, 94, 97, 101, 104, 108, 111, 115, 118, 122, 125, 129, 132, 135, 139, 142, 146, 149, 152, 156, 159, 163, 166, 169, 173, 176, 179, 183, 186, 189, 193, 196, 199, 203, 206, 209, 213
Offset: 1

Views

Author

Stefano Spezia, Sep 29 2019

Keywords

Comments

For n > 0, it is easy to prove that k(n) = floor((3 + sqrt(3*(3+8*n)))/6) is the unique integer that satisfies the inequalities 3*binomial(k,2) <= n <= 3*binomial(k+1,2) of Theorem 1.1 in Malen and Roldán.
Proof: solving in k the above inequalities for n > 0, one gets that x - 1 <= k <= x, where x = (3 + sqrt(3*(3+8*n)))/6. Since 3*(3+8*n) is never a perfect square, it follows that x is not an integer and k = floor(x). QED.

Crossrefs

Programs

  • Maple
    k:=n->floor((3+sqrt(3*(3+8*n)))/6): a:=n->3*(n+k(n))+1+ceil(2*n/k(n)): seq(a(n), n = 1 .. 58)
  • Mathematica
    k[n_]:=Floor[(3+Sqrt[3*(3+8n)])/6]; a[n_]:=3(n+k[n])+1+Ceiling[2n/k[n]]; Array[a,58]

Formula

a(n) = 3*(n + k(n)) + 1 + ceiling(2*n/k(n)), where k(n) = floor((3 + sqrt(3*(3+8*n)))/6).
Showing 1-6 of 6 results.