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.

A213927 T(n,k) = (z*(z-1)-(-1+(-1)^(z^2 mod 3))*n+(1+(-1)^(z^2 mod 3))*k)/2, where z=n+k-1; n, k > 0, read by antidiagonals.

Original entry on oeis.org

1, 2, 3, 6, 5, 4, 7, 8, 9, 10, 11, 12, 13, 14, 15, 21, 20, 19, 18, 17, 16, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 45, 44, 43, 42, 41, 40, 39, 38, 37, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 78
Offset: 1

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Author

Boris Putievskiy, Mar 06 2013

Keywords

Comments

Self-inverse permutation of the natural numbers.
a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.
In general, let b(z) be a sequence of integers and denote number of antidiagonal table T(n,k) by z=n+k-1. Natural numbers placed in table T(n,k) by antidiagonals. The order of placement - by antidiagonal downwards, if b(z) is odd; by antidiagonal upwards, if b(z) is even. T(n,k) read by antidiagonals downwards. For A218890 -- the order of placement -- at the beginning m antidiagonals downwards, next m antidiagonals upwards and so on - b(z)=floor((z+m-1)/m). For this sequence b(z)=z^2 mod 3. (This comment should be edited for clarity, Joerg Arndt, Dec 11 2014)

Examples

			The start of the sequence as table.
The direction of the placement denoted by ">" and  "v".
.v.....v       v...v        v....v
.1.....2...6...7..11...21...22...29...45...
.3.....5...8..12..20...23...30...44...47...
>4.....9..13..19..24...31...43...48...58...
.10...14..18..25..32...42...49...59...75...
.15...17..26..33..41...50...60...74...83...
>16...27..34..40..51...61...73...84...97...
.28...35..39..52..62...72...85...98..114...
.36...38..53..63..71...86...99..113..128...
>37...54..64..70..87..100..112..129..145...
...
The start of the sequence as triangle array read by rows:
   1;
   2,  3;
   6,  5,  4;
   7,  8,  9, 10;
  11, 12, 13, 14, 15;
  21, 20, 19, 18, 17, 16;
  22, 23, 24, 25, 26, 27, 28;
  29, 30, 31, 32, 33, 34, 35, 36;
  45, 44, 43, 42, 41, 40, 39, 38, 37;
  ...
Row r consists of r consecutive numbers from r*r/2-r/2+1 to r*r/2+r.
If r is not divisible by 3, rows are increasing.
If r is     divisible by 3, rows are decreasing.
		

Crossrefs

Programs

  • Mathematica
    T[n_, k_] := With[{z = n + k - 1}, (z*(z - 1) - (-1 + (-1)^Mod[z^2, 3])*n + (1 + (-1)^Mod[z^2, 3])*k)/2];
    Table[T[n - k + 1, k], {n, 1, 12}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jul 22 2018 *)
  • Python
    t=int((math.sqrt(8*n-7) - 1)/ 2)
    i=n-t*(t+1)/2
    j=(t*t+3*t+4)/2-n
    z=i+j-1
    result=(z*(z-1)-(-1+(-1)**(z**2%3))*i+(1+(-1)**(z**2%3))*j)/2

Formula

For the general case.
T(n,k) = (z*(z-1)-(-1+(-1)^b(z))*n+(1+(-1)^b(z))*k)/2, where z=n+k-1 (as a table).
a(n) = (z*(z-1)-(-1+(-1)^b(z))*i+(1+(-1)^b(z))*j)/2, where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2), z=i+j-1 (as a linear sequence).
For this sequence b(z)=z^2 mod 3.
T(n,k) = (z*(z-1)-(-1+(-1)^(z^2 mod 3))*n+(1+(-1)^(z^2 mod 3))*k)/2, where z=n+k-1 (as a table).
a(n) = (z*(z-1)-(-1+(-1)^(z^2 mod 3))*i+(1+(-1)^(z^2 mod 3))*j)/2, where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2), z=i+j-1 (as linear sequence).