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 1-4 of 4 results.

A001971 Nearest integer to n^2/8.

Original entry on oeis.org

0, 0, 1, 1, 2, 3, 5, 6, 8, 10, 13, 15, 18, 21, 25, 28, 32, 36, 41, 45, 50, 55, 61, 66, 72, 78, 85, 91, 98, 105, 113, 120, 128, 136, 145, 153, 162, 171, 181, 190, 200, 210, 221, 231, 242, 253, 265, 276, 288, 300, 313, 325, 338, 351, 365, 378, 392, 406, 421, 435, 450
Offset: 0

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Author

Keywords

Comments

Restricted partitions.
a(0) = a(1) = 0; a(n) are the partitions of floor((3*n+3)/2) with 3 distinct numbers of the set {1, ..., n}; partitions of floor((3*n+3)/2)-C and ceiling((3*n+3)/2)+C have equal numbers. - Paul Weisenhorn, Jun 05 2009, corrected by M. F. Hasler, Jun 16 2022
Odd-indexed terms are the triangular numbers, even-indexed terms are the midpoint (rounded up where necessary) of the surrounding odd-indexed terms. - Carl R. White, Aug 12 2010
a(n+2) is the number of points one can surround with n stones in Go (including the points under the stones). - Thomas Dybdahl Ahle, May 11 2014
Corollary of above: a(n) is the number of points one can surround with n+2 stones in Go (excluding the points under the stones). - Juhani Heino, Aug 29 2015
From Washington Bomfim, Jan 13 2021: (Start)
For n >= 4, a(n) = A026810(n+2) - A026810(n-4).
Let \n,m\ be the number of partitions of n into m non-distinct parts.
For n >= 1, \n,4\ = round((n-2)^2/8).
For n >= 6, \n,4\ = A026810(n) - A026810(n-6).
(End)

References

  • A. Cayley, Numerical tables supplementary to second memoir on quantics, Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 2, pp. 276-281.
  • M. Jeger, Einfuehrung in die Kombinatorik, Klett, 1975, Bd.2, pages 110 ff. [Paul Weisenhorn, Jun 05 2009]
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

The 4th diagonal of A061857?
Kind of an inverse of A261491 (regarding Go).
Cf. A026810 (partitions with greatest part 4), A001400 (partitions in at most 4 parts), A000217 (a(2n+1): triangular numbers n(n+1)/2), A000982 (a(2n): round(n^2/2)).

Programs

  • Haskell
    a001971 = floor . (+ 0.5) . (/ 8) . fromIntegral . (^ 2)
    -- Reinhard Zumkeller, May 08 2012
  • Magma
    [Round(n^2/8): n in [0..60]]; // Vincenzo Librandi, Jun 23 2011
    
  • Maple
    A001971:=-(1-z+z**2)/((z+1)*(z**2+1)*(z-1)**3); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation [Note that this "generating function" is Sum_{n >= 0} a(n+2)*z^n, not a(n)*z^n. - M. F. Hasler, Jun 16 2022]
  • Mathematica
    LinearRecurrence[{2,-1,0,1,-2,1},{0,0,1,1,2,3},70] (* Harvey P. Dale, Jan 30 2014 *)
  • PARI
    {a(n) = round(n^2 / 8)};
    
  • PARI
    apply( {A001971(n)=n^2\/8}, [0..99]) \\ M. F. Hasler, Jun 16 2022
    

Formula

The listed terms through a(20)=50 satisfy a(n+2) = a(n-2) + n. - John W. Layman, Dec 16 1999
G.f.: x^2 * (1 - x + x^2) / (1 - 2*x + x^2 - x^4 + 2*x^5 - x^6) = x^2 * (1 - x^6) / ((1 - x) * (1 - x^2) * (1 - x^3) * (1 - x^4)). - Michael Somos, Feb 07 2004
a(n) = floor((n^2+4)/8). - Paul Weisenhorn, Jun 05 2009
a(2*n+1) = A000217(n), a(2*n) = floor((A000217(n-1)+A000217(n)+1)/2). - Carl R. White, Aug 12 2010
From Michael Somos, Aug 29 2015: (Start)
Euler transform of length 6 sequence [ 1, 1, 1, 1, 0, -1].
a(n) = a(-n) for all n in Z. (End)
a(2n) = A000982(n). - M. F. Hasler, Jun 16 2022
Sum_{n>=2} 1/a(n) = 2 + Pi^2/12 + tanh(Pi/2)*Pi/2. - Amiram Eldar, Jul 02 2023

Extensions

Edited Feb 08 2004

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

A346958 a(n) is the minimal number of cubes required to make a void of volume n.

Original entry on oeis.org

6, 10, 13, 15, 17, 18, 18, 21, 23, 25, 26, 26
Offset: 1

Views

Author

Mohammed Yaseen, Aug 08 2021

Keywords

Comments

Following is an illustration of the first few voids in the form of polycubes (where an o represents a continuation upwards and an x represents a continuation downwards) each of which can be made by concealing it with a(n) cubes.
.---. .---.
| | | |
.---. .---.---. .---.---. .---.---.
| | | | | | | | | | o |
.---. .---.---. .---.---. .---.---.
n=1 n=2 n=3 n=4
.---. .---. .---.
| | | | | |
.---.---. .---.---.---. .---.---.---.
| | o | | | o | | | | ox| |
.---.---. .---.---.---. .---.---.---.
| | | | | |
.---. .---. .---.
n=5 n=6 n=7
Equivalently, the minimum perimeter size of any polycube of size n. - Sean A. Irvine, Aug 23 2021
Conjecture: When n is in A001845 the void is an octahedral crystal ball of volume n = A001845(m), which is concealed by a(n) = A005899(m+1) cubes. So a(A001845(m)) = A005899(m+1), m>=0. For example, a(1)=6 and a(7)=18. - Mohammed Yaseen, Sep 15 2022

Examples

			A cube-shaped void can be made by concealing it with 6 cubes, which is the minimal number to do so. So a(1)=6.
A dicube-shaped void can be made by concealing it with 10 cubes, which is the minimal number to do so. So a(2)=10.
		

Crossrefs

Cf. A261491 (2D analog).

Formula

a(n) < A193416(n) for n>2.

Extensions

a(8)-a(12) from Sean A. Irvine, Aug 23 2021
Showing 1-4 of 4 results.