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: Tomas Rigaux

Tomas Rigaux's wiki page.

Tomas Rigaux has authored 2 sequences.

A362016 Maximal number of unmarked cells with at least 3 marked neighboring cells in the n X n kings' graph.

Original entry on oeis.org

0, 1, 4, 8, 13, 20, 28, 38, 50, 61, 75, 90, 108, 124, 139
Offset: 1

Author

Tomas Rigaux, Apr 04 2023

Keywords

Comments

The value of r = lim sup a(n) / n^2 is in the half-open interval [2/3, 8/11).
It appears from the computed terms that r = 2/3.

Examples

			a(2) = 1, as the only pattern is
  .X
  XX
a(9) = 50, with a similar pattern to prove that r >= 2/3:
  X.......X
  .XXXXXXX.
  X.......X
  .........
  XXXXXXXXX
  .........
  X.......X
  .XXXXXXX.
  X.......X
a(10) = 61, and a pattern that reaches that is
  X..X...X..
  XX.X.X.X.X
  ..........
  .X.XX.X.XX
  XX....X...
  ....X....X
  X.XX..XX.X
  X....X....
  ...X....XX
  XX.X.XX.X.
		

Crossrefs

If we only want 1 marked neighbor, we get n^2 - A075561(n).

A330671 Largest number whose base-n expansion cannot be subdivided to form a sequence of numbers which ordered form a multiple of n+1 when using +, *, and ().

Original entry on oeis.org

1, 7, 13, 41, 206, 335, 503, 2746, 9898, 13938, 20588, 28390, 38007, 50366, 1006418, 82650, 1865809, 1738855, 2879137, 4024861, 5135433, 5585431, 7932985
Offset: 2

Author

Tomas Rigaux, Jan 18 2020

Keywords

Comments

Each valid number in a given base n must be a one-digit extension to the right (or left) of another valid number in the same base; otherwise, you could ignore the added digit, make a multiple of n+1 with the rest of the number, then multiply it by the new digit to get a multiple of n+1.
a(n) is always less than n^(n+1).
a(n) also appears to be less than n^(n/2), except at n=3.

Examples

			For n = 2, the binary notation of a number cannot contain any 0, as you could then construct 0 by multiplying all the digits together, so the only candidates are 1, 11, 111, 1111 (or 1, 3, 7, 15, ... in base 10).
Out of those, if you have at least 2 digits, the number contains the substring '11', which can be multiplied by all the other digits to give 11 (or 3 in base 10), which gives a(2) = 1 as the largest and only solution.
For n = 4, a(n) = 13 can be easily checked using the fact that the base-4 expansion of a valid number cannot contain a 2 and a 3 next to each other, as 2+3 = 5 = n+1.
For n = 10, 12345 is not a valid number as 1+2*3*4*5 = 121 = 11*11.
		

Programs

  • Python
    # See Python link