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.

User: Drake Thomas

Drake Thomas's wiki page.

Drake Thomas has authored 8 sequences.

A342041 Triangle read by rows: T(n,k) = maximum number of lines of size k on n points so that every two lines intersect in one point.

Original entry on oeis.org

1, 3, 1, 3, 1, 1, 4, 2, 1, 1, 5, 4, 1, 1, 1, 6, 7, 2, 1, 1, 1, 7, 7, 2, 1, 1, 1, 1, 8, 7, 3, 2, 1, 1, 1, 1, 9, 7, 5, 2, 1, 1, 1, 1, 1, 10, 7, 6, 2, 2, 1, 1, 1, 1, 1, 11, 7, 9, 3, 2, 1, 1, 1, 1, 1, 1, 12, 7, 13, 3, 2, 2, 1, 1, 1, 1, 1, 1, 13, 7, 13, 4
Offset: 2

Author

Drake Thomas, Feb 26 2021

Keywords

Comments

Rows start at n = 2, and terms range from k = 2 to k = n. (When k = 1, there can be arbitrarily many lines.)
If a projective plane of order k-1 exists, then for n between k^2-k+1 and k^3-2k^2+3k-2 inclusive, T(n,k) = k^2-k+1. For higher n, T(n,k) = floor((n-1)/(k-1)).

Examples

			For n = 10, k = 4, the unique arrangement with 5 lines (up to symmetry) is
  1111000000
  1000111000
  0100100110
  0010010101
  0001001011
There are no such arrangements with 6 lines. Thus T(10,4) = 5.
These lines are in bijection with the sets of 4 polar axes on a dodecahedron whose endpoints form a cube.
Table begins:
n\k | 2  3  4  5  6  7  8  9
----+-----------------------
  2 | 1;
  3 | 3, 1;
  4 | 3, 1, 1;
  5 | 4, 2, 1, 1;
  6 | 5, 4, 1, 1, 1;
  7 | 6, 7, 2, 1, 1, 1;
  8 | 7, 7, 2, 1, 1, 1, 1;
  9 | 8, 7, 3, 2, 1, 1, 1, 1;
		

A345076 Number of generalized polyforms on the elongated triangular tiling with n cells.

Original entry on oeis.org

1, 2, 3, 5, 13, 32, 96, 283, 907, 2929, 9787, 32939, 112476, 386230, 1336150, 4642930, 16208851, 56786242, 199614651, 703678568, 2487109359, 8811020024, 31281360326, 111272475650
Offset: 0

Author

Drake Thomas, Jun 07 2021

Keywords

Comments

This sequence counts free polyforms where holes are allowed. This means that two polyforms are considered the same if one is a rigid transformation (translation, rotation, reflection or glide reflection) of the other.

Examples

			See the PDF in the links section.
		

Crossrefs

Analogous for other tilings: A000105 (square), A000228 (hexagonal), A000577 (triangular), A197156 (prismatic pentagonal), A197159 (floret pentagonal), A197459 (rhombille), A197462 (kisrhombille), A197465 (tetrakis square), A309159 (snub square), A343398 (trihexagonal), A343406 (truncated hexagonal), A343577 (truncated square).

Extensions

a(15)-a(23) from Bert Dobbelaere, Jun 05 2025

A343909 Number of generalized polyforms on the tetrahedral-octahedral honeycomb with n cells.

Original entry on oeis.org

1, 2, 1, 4, 9, 44, 195, 1186, 7385, 49444, 337504, 2353664, 16608401, 118432965, 851396696, 6163949361, 44896941979
Offset: 0

Author

Drake Thomas and Peter Kagey, May 03 2021

Keywords

Comments

This sequence counts "free" polyforms where holes are allowed. This means that two polyforms are considered the same if one is a rigid transformation (translation, rotation, reflection, or a combination thereof) of the other.

Examples

			For n = 1, the a(1) = 2 polyforms are the tetrahedron and the octahedron.
For n = 2, the a(2) = 1 polyform is a tetrahedron and an octahedron connected at a face.
For n = 3, there are a(3) = 4 polyforms with 3 cells:
  - 3 consisting of one octahedron with two tetrahedra, and
  - 1 consisting of two octahedra and one tetrahedron.
For n = 4, there are a(4) = 9 polyforms with 4 cells:
  - 3 with one octahedron and three tetrahedra,
  - 5 with two octahedra and three octahedra, and
  - 1 with three octahedra and one tetrahedron.
For n = 5, there are a(5) = 44 polyforms with 5 cells:
  - 6 with one octahedron and four tetrahedra,
  - 24 with two octahedra and three tetrahedra,
  - 13 with three octahedra and two tetrahedra, and
  - 1 with four octahedra and one tetrahedron.
		

Crossrefs

Row sums of A365970.
Analogous for other honeycombs/tilings: A000105 (square), A000228 (hexagonal), A000577 (triangular), A038119 (cubical), A068870 (tesseractic), A197156 (prismatic pentagonal), A197159 (floret pentagonal), A197459 (rhombille), A197462 (kisrhombille), A197465 (tetrakis square), A309159 (snub square), A343398 (trihexagonal), A343406 (truncated hexagonal), A343577 (truncated square).

Extensions

a(11)-a(16) from Bert Dobbelaere, Jun 10 2025

A342430 Number of prime polyominoes with n cells.

Original entry on oeis.org

0, 0, 1, 2, 1, 12, 5, 108, 145, 974, 2210, 17073, 31950, 238591, 587036, 3174686, 9236343, 50107909
Offset: 0

Author

Drake Thomas, Mar 11 2021

Keywords

Comments

We say that a free polyomino is prime if it cannot be tiled by any other free polyomino besides the 1 X 1 square and itself.
The tiling of P must be with a single polyomino, and that single polyomino may not be the unique monomino or P itself. For example, decomposing the T-tetromino into a 3 X 1 and a 1 X 1 would use multiple tiles, and this is not permitted.
It can be shown that a(n) > 0 for all n >= 4, by considering the polyomino whose cells are at (0,1), (-1,1), (0,2), and (x,0) for all x = 0, 1, ..., n-4.

Examples

			For n = 4, the T-tetromino cannot be decomposed into smaller congruent polyominoes:
      +---+
      |   |
  +---+   +---+
  |           |
  +-----------+
The other four free tetrominoes can, however:
  +---+
  |   |
  |   |    +---+
  |   |    |   |
  +---+    |   |         +---+---+        +---+---+
  |   |    |   |         |   |   |        |       |
  |   |    +---+---+     |   |   |    +---+---+---+
  |   |    |       |     |   |   |    |       |
  +---+    +-------+     +---+---+    +---+---+
Thus a(4) = 1.
		

Crossrefs

Formula

a(n) = A000105(n) if n is prime.

Extensions

a(14)-a(17) from John Mason, Sep 16 2022
a(1) corrected by John Mason, Feb 27 2023

A340984 Number of prime rectangle tilings with n tiles up to equivalence.

Original entry on oeis.org

1, 1, 0, 0, 1, 0, 2, 6, 29, 119, 600
Offset: 1

Author

Drake Thomas, Feb 01 2021

Keywords

Comments

Say that a tiling of a rectangle by other rectangles is prime if the only sub-rectangles in the tiling are those formed by a single tile. Say that two tilings are equivalent if there exists an inclusion/overlap-preserving bijection between the vertices, edges, and faces of every rectangle in them.
Problem 69 in Hugo Steinhaus's One Hundred Problems In Elementary Mathematics asks the reader to show that a(3) = a(4) = 0, and that there exist prime dissections for 5, 7, and 8 in which the pieces are of equal area. It cites Czesław Ryll-Nardzewski as proving that a(6) = 0, though this is not difficult to show by hand. The book also provides diagrams of both n = 7 solutions and four of the six n = 8 solutions.
Chung et al.'s paper Tiling Rectangles with Rectangles shows that the sequence grows at least as fast as c*2^(n/7) for some positive constant c, and states without proof that it is bounded above by 20000^n.

Examples

			For n = 5 the a(5) = 1 example looks like
   _____
  | |___|
  |_|_| |
  |___|_|
.
For n = 7 the a(7) = 2 examples look like
   _______    _______
  | |_____|  |_____| |
  |_|___| |  |___| | |
  |   |_|_|  | |_|_|_|
  |___|___|  |_|_____|
		

Crossrefs

Extensions

a(9)-a(11) from Benjamin D. Prins, Jun 13 2025

A335586 Number of domino tilings of a 2n X 2n toroidal grid.

Original entry on oeis.org

1, 8, 272, 90176, 311853312, 11203604497408, 4161957566985310208, 15954943354032349049274368, 630665326543010382995142219988992, 256955886436135671144699761794930161483776
Offset: 0

Author

Drake Thomas, Jan 26 2021

Keywords

Comments

For n > 1, number of perfect matchings of the graph C_2n X C_2n.

Examples

			For n = 1, there are a(1) = 8 tilings (see the Links section for a diagram).
		

Crossrefs

Number of perfect matchings of the graph C_2m X C_n: A162484 (m=1), A220864 (m=2), A232804 (m=3), A253678 (m=4), A281679 (m=5), A309018 (m=6).

Programs

  • PARI
    default(realprecision, 120);
    b(n) = round(prod(j=1, n-1, prod(k=1, n, 4*sin(j*Pi/n)^2+4*sin((2*k-1)*Pi/(2*n))^2)));
    c(n) = round(prod(j=1, n, prod(k=1, n, 4*sin((2*j-1)*Pi/(2*n))^2+4*sin((2*k-1)*Pi/(2*n))^2)));
    a(n) = if(n==0, 1, 4*b(n)+c(n)/2); \\ Seiichi Manyama, Feb 13 2021

Formula

a(n) = 4 * Product_{j=1..n-1} Product_{k=1..n} (4*sin(j*Pi/n)^2 + 4*sin((2*k-1)*Pi/(2*n))^2) + 1/2 * Product_{1<=j,k<=n} (4*sin((2*j-1)*Pi/(2*n))^2 + 4*sin((2*k-1)*Pi/(2*n))^2) = 4 * A341478(n)^2 + A341479(n)/2 for n > 0. - Seiichi Manyama, Feb 13 2021
a(n) ~ (1 + sqrt(2)) * exp(4*G*n^2/Pi), where G is Catalan's constant A006752. - Vaclav Kotesovec, Feb 14 2021

Extensions

More terms from Seiichi Manyama, Feb 13 2021

A337518 Number of non-isomorphic graphs on n unlabeled nodes modulo 3.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 0, 0, 1, 0, 2, 2, 0, 0, 1, 2, 0, 1, 0, 2, 1, 2, 0, 0, 2, 0, 0, 1, 1, 1, 0, 0, 2, 0, 1, 0, 2, 1, 0, 0, 2, 2, 0, 2, 0, 0, 2, 1, 2, 1, 0, 1, 0, 1, 1, 1, 0, 1, 2, 2, 1, 0, 1, 2, 0, 1, 2, 1, 1, 1, 1, 0, 0, 2, 1, 1, 2, 0, 2, 0, 0, 0, 2, 1, 2, 2, 1
Offset: 0

Author

Drake Thomas, Nov 21 2020

Keywords

Comments

For the mod-2 case, the sequence is eventually constant, because there are an even number of graphs on n vertices for n>4. (In fact, the number of factors of 2 in A000088(n) is asymptotically n/2; see Cater and Robinson in the Links section.)

Examples

			For n = 4, there are 11 graphs on 4 nodes up to isomorphism, so a(4) = 2 = 11 mod 3.
		

Crossrefs

Programs

  • Python
    from itertools import combinations
    from math import prod, factorial, gcd
    from fractions import Fraction
    from sympy.utilities.iterables import partitions
    def A337518(n): return int(sum(Fraction(1<>1)*r+(q*r*(r-1)>>1) for q, r in p.items()),prod(q**r*factorial(r) for q, r in p.items()))%3 for p in partitions(n))) % 3 # Chai Wah Wu, Jul 02 2024

Formula

a(n) = A000088(n) mod 3.

A262231 First term is 2; each subsequent term is the least number greater than the previous term but not a multiple of the successor of any previous term.

Original entry on oeis.org

2, 4, 7, 11, 13, 17, 19, 22, 26, 29, 31, 34, 37, 41, 43, 47, 49, 52, 58, 61, 67, 71, 73, 77, 79, 82, 86, 89, 91, 94, 97, 101, 103, 107, 109, 113, 116, 119, 121, 127, 131, 133, 137, 139, 142, 146, 149, 151, 157, 163, 167, 169, 172, 178, 181, 187, 191
Offset: 1

Author

Drake Thomas, Sep 15 2015

Keywords

Comments

If the twin prime conjecture is true, this sequence has infinitely many pairs of terms with difference at most 3.

Examples

			25 is excluded because it is a multiple of 4+1; 26 is not a multiple of 3,5,8,...,23 so it remains in the sequence.
		

Crossrefs

Cf. A100464.

Programs

  • PARI
    has(n)=fordiv(n,d, if(mapisdefined(m, d-1), return(0))); 1
    first(n)=local(m=Map(Mat([2,0]))); my(t=3); while(#mCharles R Greathouse IV, Sep 16 2015

Formula

a(n) = A100464(n) - 1.