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: Maurice Clerc

Maurice Clerc's wiki page.

Maurice Clerc has authored 2 sequences.

A376215 Maximum number of non-overlapping discs with a diameter of 1 that can fit within an n×n square, using hexagonal packing or square packing, whichever is more efficient.

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 49, 68, 85, 105, 126, 150, 175, 216, 247, 279, 314, 350, 389, 429, 492, 538, 585, 635, 686, 740, 822, 880, 941, 1003, 1068, 1134, 1203, 1307, 1380, 1456, 1533, 1613, 1694, 1817, 1904, 1992, 2083, 2175, 2270, 2366, 2511, 2613, 2716, 2822, 2929, 3039
Offset: 1

Author

Maurice Clerc, Sep 15 2024

Keywords

Comments

Maximum error on the first 99 terms = 3.3%. (See the Specht's site).
The first row of discs is placed on the base of the square. Then hexagonal packing is applied, row by row, as much as possible. We find a number N of discs. If n^2>N then the whole packing is replaced by the square one. Example: n=3 => N=8 but since n^2=9 we can indeed place 9 discs.
Note: useful only for n=2, 3, 4, 5, 6, 7.

Crossrefs

Cf. A084617.

Programs

  • Python
    from math import isqrt
    def A376215(n): return max(n**2,n*(m:=1+isqrt(((n-1)**2<<2)//3))-(m>>1)) # Chai Wah Wu, Nov 06 2024

Formula

a(n) = max(n^2,n*floor(1+2*(n-1)/sqrt(3)) - floor(floor(1+2*(n-1)/sqrt(3))/2));

A376066 Minimum number of unit squares needed to cover the circumference of a circle of radius n.

Original entry on oeis.org

4, 9, 14, 18, 23, 27, 32, 36, 40, 45, 49, 54, 58, 63, 67, 72, 76, 80, 85, 89, 94, 98, 103, 107, 112, 116, 120, 125, 129, 134, 138, 143, 147, 152, 156, 160, 165, 169, 174, 178, 183, 187, 192, 196, 200, 205, 209, 214, 218, 223, 227, 232, 236, 240, 245, 249, 254, 258, 263, 267, 272, 276, 280, 285, 289, 294, 298, 303, 307, 311
Offset: 1

Author

Maurice Clerc, Sep 08 2024

Keywords

Comments

For n>=2, a unit square covers the most circumference when it has two diagonally opposite corners on the circumference, forming a chord of length sqrt(2).
A simple upper bound a(n) <= u(n) = ceiling(2*Pi*n/sqrt(2)) would be by sqrt(2) arcs instead of chords, and which is bigger at for instance a(70) = 311 < u(70) = 312 (see A376207).

Crossrefs

Cf. A376207.

Formula

a(n) = ceiling(Pi/arcsin(sqrt(2)/(2*n))).