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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A114086 Numbers m such that m XOR 32*m = 33*m.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 65, 68, 69, 72, 73, 76, 77, 80, 81, 84, 85, 88, 89, 92, 93, 96, 100, 104, 108, 112, 116, 120
Offset: 0

Views

Author

Reinhard Zumkeller, Feb 04 2006

Keywords

Comments

A116361(a(n)) <= 5.

Links

Crossrefs

Differs from A001477 for the first time at n=33 (33, 35, 37, 39, etc. are not present in this sequence). Cf. A003714, A048716, A115845, A115847.

Programs

A348484 Maximum number of squares on an n X n chessboard such that no two are two steps apart horizontally or vertically.

Original entry on oeis.org

1, 4, 5, 8, 13, 20, 25, 32, 41, 52
Offset: 1

Views

Author

Yang Hong, Oct 20 2021

Keywords

Comments

The sequence 1, 4, 5, 8, 13, ... with g.f. -x*(1 +2*x -2*x^2 +2*x^3 +x^4)/ ((1+x) *(x^2+1) *(x-1)^3) and a(n)= 2*a(n-1) -a(n-2) +a(n-4) -2*a(n-5) +a(n-6) is a lower bound for a(n) achieved by packing 2x2 squares with 1's and 2x2 squares with 0's in a checkerboard pattern into the chessboard. - R. J. Mathar, Dec 03 2022

Examples

			For n = 1, a(1) = (1^2 + 1)/2 = 1
  1
For n = 2, a(2) = (2^2)/2 + 2 = 4
  11
  11
For n = 3, a(3) = (3^2 + 1)/2 = 5
Starting here the solutions are not unique. We can mix 2X2 blocks from and S shapes along the diagonals.
  110
  110
  001
or
  110
  011
  001
For n = 4, a(4) = (4^2)/2 = 8
  1100
  1100
  0011
  0011
or
  1100
  0110
  0011
  1001
For n = 5, a(5) = (5^2 + 1)/2 = 13
  11001
  11001
  00110
  00110
  11001
or
  11001
  01100
  00110
  10011
  11001
For n = 6, a(6) = (6^6)/2 + 2 = 20
  110011
  110011
  001100
  001100
  110011
  110011

Crossrefs

Cf. A048716.

Formula

Conjectures:
a(n) = n^2/2 for n == 0 (mod 4).
a(n) = (n^2 + 1)/2 for n == 1 or 3 (mod 4).
a(n) = n^2/2 + 2 for n == 2 (mod 4).
Previous Showing 11-12 of 12 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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