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.

A092542 Table whose n-th row is constant and equal to n, read by antidiagonals alternately upwards and downwards.

Original entry on oeis.org

1, 1, 2, 3, 2, 1, 1, 2, 3, 4, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9
Offset: 1

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Author

Sam Alexander, Feb 27 2004

Keywords

Comments

Let A be sequence A092542 (this sequence) and B be sequence A092543 (1, 2, 1, 1, 2, 3, 4, ...). Under upper trimming or lower trimming, A transforms into B and B transforms into A. Also, B gives the number of times each element of A appears. For example, A(7) = 1 and B(7) = 4 because the 1 in A(7) is the fourth 1 to appear in A. - Kerry Mitchell, Dec 28 2005
First inverse function (numbers of rows) for pairing function A056023 and second inverse function (numbers of columns) for pairing function A056011. - Boris Putievskiy, Dec 24 2012
The rational numbers a(n)/A092543(n) can be systematically ordered and numbered in this way, as Georg Cantor first proved in 1873. - Martin Renner, Jun 05 2016

Examples

			The table
  1 1 1 1 1 ...
  2 2 2 2 2 ...
  3 3 3 3 3 ...
  4 4 4 4 4 ...
gives
  1;
  1 2;
  3 2 1;
  1 2 3 4;
  5 4 3 2 1;
  1 2 3 4 5 6;
		

References

  • Amir D. Aczel, "The Mystery of the Aleph, Mathematics, the Kabbalah and the Search for Infinity", Barnes & Noble, NY 2000, page 112.

Crossrefs

Variants of Cantor's enumeration are: A352911, A366191, A319571, A354266.

Programs

  • Mathematica
    Table[ Join[Range[2n - 1], Reverse@ Range[2n - 2]], {n, 8}] // Flatten (* Robert G. Wilson v, Sep 28 2006 *)

Formula

a(n) = ((-1)^t+1)*j/2-((-1)^t-1)*i/2, where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 24 2012

Extensions

Name edited by Michel Marcus, Dec 14 2023