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.

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A346232 Maximum number of squares in a square grid whose interiors can be touched by a (possibly skew) line segment of length n.

Original entry on oeis.org

3, 5, 7, 8, 9, 11, 12, 14, 15, 17, 18, 19, 21, 22, 24, 25, 27, 28, 29, 31, 32, 34, 35, 36, 38, 39, 41, 42, 43, 45, 46, 48, 49, 51, 52, 53, 55, 56, 58, 59, 60, 62, 63, 65, 66, 68, 69, 70, 72, 73, 75, 76, 77, 79, 80, 82, 83, 85, 86, 87, 89, 90, 92, 93, 94, 96, 97
Offset: 1

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Author

Luis Mendo, Jul 11 2021

Keywords

Comments

In order to derive the formulas below, the following two facts are useful. Let i, j be arbitrary nonnegative integers. (1) A segment with length slightly greater than sqrt(i^2+j^2) can touch i+j+3 squares, if it is aligned from (0,0) to (i,j) and then shifted a little. (2) For a given length, it suffices to consider segments aligned from (0,0) to either (i,i) or (i,i+1) (other orientations result in a smaller number of squares touched).
The sequence is increasing, with a(n+1)-a(n) equal to 1 or 2. There can be no more than two consecutive increments equal to 1. Increments equal to 2 always appear isolated, except at the initial sequence terms (3,5,7). - Luis Mendo, Jul 29 2021

Examples

			For n = 1, a line segment with its center close to a square vertex and oriented at 45 degrees with respect to the grid touches 3 squares (see image linked above).
		

Crossrefs

"Inverse" of A346693.

Programs

  • Mathematica
    Table[Floor[Sqrt[2*n^2-2]]+3,{n,80}] (* Stefano Spezia, Jul 13 2021 *)
  • PARI
    a(n) = sqrtint(2*n^2-2) + 3; \\ Michel Marcus, Aug 09 2022
  • Python
    from math import isqrt
    def A346232(n): return isqrt(n**2-1<<1)+3 # Chai Wah Wu, Aug 09 2022
    

Formula

a(n) = i+j-1 with i = ceiling(n/sqrt(2))+1, j = ceiling(sqrt(n^2-(i-2)^2))+1.
a(n) = i+j-1 with i = ceiling((sqrt(2*n^2-1)+1)/2)+1, j = ceiling(sqrt(n^2-(i-2)^2))+1.
a(n) = max{m with m integer such that m^2 - 6*m + 10 - m mod 2 < 2*n^2}.
a(n) = floor(sqrt(2*n^2-2))+3. - Alex Arkhipov, Jul 11 2021

Extensions

More terms from Stefano Spezia, Jul 13 2021
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