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.

A232091 Smallest square or promic (oblong) number greater than or equal to n.

Original entry on oeis.org

0, 1, 2, 4, 4, 6, 6, 9, 9, 9, 12, 12, 12, 16, 16, 16, 16, 20, 20, 20, 20, 25, 25, 25, 25, 25, 30, 30, 30, 30, 30, 36, 36, 36, 36, 36, 36, 42, 42, 42, 42, 42, 42, 49, 49, 49, 49, 49, 49, 49, 56, 56, 56, 56, 56, 56, 56, 64, 64, 64, 64, 64, 64, 64, 64, 72, 72, 72, 72, 72, 72, 72, 72, 81
Offset: 0

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Author

L. Edson Jeffery, Nov 18 2013

Keywords

Comments

Result attributed to the students Daring, et al., in the links section.
a(n) appears in floor(sqrt(a(n))) = A000194(n) successive terms.
Counting successive equal terms give sequence: 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ... (see A008619). - Michel Marcus, Jan 10 2014

Crossrefs

Cf. A000290 (squares), A002378 (promic or oblong numbers), A002620 (A000290 union A002378).

Programs

  • Magma
    [(Ceiling(n /Ceiling(Sqrt(n)))*Ceiling(Sqrt(n))): n in [1..80]]; // Vincenzo Librandi, Jun 22 2015
  • Mathematica
    Join[{0}, Table[Ceiling[n/Ceiling[Sqrt[n]]] Ceiling[Sqrt[n]], {n, 100}]] (* Alonso del Arte, Nov 18 2013 *)
  • PARI
    a(n)=my(t=sqrtint(n-1)+1);t*((n-1)\t+1) \\ Charles R Greathouse IV, Nov 18 2013
    

Formula

a(n) = ceiling(n/ceiling(sqrt(n)))*ceiling(sqrt(n)).
a(n) = min(k : k >= n, k in A002620).
a(k^2) = k^2; a(k*(k+1)) = k*(k+1).
It appears that a(n) = A216607(n) + n. (Verified for all n<10^9 by Lars Blomberg, Jan 09 2014.) This conjecture now follows from a proof given by David Applegate, Jan 10 2014 (see [Applegate]).
a(n) = min(A048761(n), A259225(n)). - Michel Marcus, Jun 22 2015
Sum_{n>=1} 1/a(n)^2 = 2 - Pi^2/6 + zeta(3). - Amiram Eldar, Aug 16 2022

Extensions

Extended by Charles R Greathouse IV, Nov 18 2013
a(0)=0 prepended by Michel Marcus, Jun 22 2015

A345205 Minimum number of unit cubes needed to fully enclose n unit cubes in 3D space.

Original entry on oeis.org

8, 26, 34, 42, 44, 52, 54, 56, 56, 64, 66, 68, 68, 76, 78, 80, 80, 82, 82, 90, 92, 94, 94, 96, 96, 98, 98, 98
Offset: 0

Views

Author

Abraham Maxfield, Jun 10 2021

Keywords

Comments

Cubes are assumed to be aligned in a 3D grid. Cubes with an exposed edge or corner are not considered enclosed.

Examples

			For a(1) the solution is the number of neighbors in Moore's neighborhood in 3 dimensions (3^3-1 = 26).
For a(2) the solution is the neighbors in Moore's neighborhood in 3 dimensions plus the number of neighbors in 2 dimensions (3^2-1 = 8).
		

Crossrefs

Cf. A345206, A235382 (2D equivalent), A007395 (1D equivalent), A024023.

A275113 a(n) is the minimal number of squares needed to enclose n squares with a wall so that there is a gap of at least one cell between the wall and the enclosed cells.

Original entry on oeis.org

12, 14, 15, 16, 16, 17, 18, 18, 19, 19, 20, 20, 20, 21, 21, 22
Offset: 1

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Author

Kival Ngaokrajang, Jul 17 2016

Keywords

Comments

Inspired by beehive construction in which wax is used in the most efficient way. This problem is likened to construction of a fence around a house with minimum materials and maximum enclosed area. I conjectured that a specific house pattern shall be selected. See illustration in links.
If the conjecture in A261491 is true (i.e., A261491(n) is the number of squares required to enclose n squares without a gap), then a(n) = A261491(n) + 8. - Charlie Neder, Jul 11 2018
The conjecture in A261491 holds through a(16). - David Consiglio, Jr., Nov 10 2022

Examples

			     a(1) = 12:
     +--+--+--+
     | 1| 2| 3|
  +--+--+--+--+--+
  |12|        | 4|
  +--+  +--+  +--+
  |11|  | 1|  | 5|
  +--+  +--+  +--+
  |10|        | 6|
  +--+--+--+--+--+
     | 9| 8| 7|
     +--+--+--+
.
      a(2) = 14:
     +--+--+--+--+
     | 1| 2| 3| 4|
  +--+--+--+--+--+--+
  |14|           | 5|
  +--+  +--+--+  +--+
  |13|  | 1| 2|  | 6|
  +--+  +--+--+  +--+
  |12|           | 7|
  +--+--+--+--+--+--+
     |11|10| 9| 8|
     +--+--+--+--+
.
     a(3) = 15:
     +--+--+--+
     | 1| 2| 3|
  +--+--+--+--+--+
  |15|        | 4|
  +--+  +--+  +--+--+
  |14|  | 3|     | 5|
  +--+  +--+--+  +--+
  |13|  | 1| 2|  | 6|
  +--+  +--+--+  +--+
  |12|           | 7|
  +--+--+--+--+--+--+
     |11|10| 9| 8|
     +--+--+--+--+
		

Crossrefs

Extensions

a(11)-a(16) from David Consiglio, Jr., Nov 10 2022

A275937 The number of distinct patterns of the smallest number of unit squares required to enclose n units of area, where corner contact is allowed.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 4, 2, 4, 1
Offset: 0

Views

Author

Kival Ngaokrajang, Aug 12 2016

Keywords

Comments

Inspired by A235382 and A261491. The rotations and/or reflections are excluded.

Crossrefs

A290648 a(n) is the smallest number of faces of the triangular lattice required to enclose an area consisting of exactly n faces.

Original entry on oeis.org

6, 12, 14, 16, 18, 19, 18, 20, 22, 23, 22
Offset: 0

Views

Author

Christian Majenz, Aug 08 2017

Keywords

Crossrefs

Triangular version of A235382.

Formula

Conjecture: a(n) = 2*A067628(n) + 6.

A373927 a(n) is the minimum number of hypercubes needed to admit a hole of size n in the 4D tesseractic honeycomb.

Original entry on oeis.org

16, 80, 106, 132
Offset: 0

Views

Author

Abraham Maxfield, Jun 22 2024

Keywords

Examples

			16 hypercubes surround a single vertex so a(0) = 16.
A 3 X 3 X 3 X 3 hypercube will admit a single hole in its center, so a(1) = 3*3*3*3 - 1 = 80.
		

Crossrefs

Cf. A345205 (3D equivalent), A235382 (2D equivalent).
Showing 1-6 of 6 results.