A081344 Natural numbers in square maze arrangement, read by antidiagonals.

1, 2, 4, 9, 3, 5, 10, 8, 6, 16, 25, 11, 7, 15, 17, 26, 24, 12, 14, 18, 36, 49, 27, 23, 13, 19, 35, 37, 50, 48, 28, 22, 20, 34, 38, 64, 81, 51, 47, 29, 21, 33, 39, 63, 65, 82, 80, 52, 46, 30, 32, 40, 62, 66, 100, 121, 83, 79, 53, 45, 31, 41, 61, 67, 99, 101, 122, 120, 84, 78, 54

Offset: 1

Author

Paul Barry, Mar 19 2003

Keywords

• nonn
• tabl

Comments

Arrange the natural numbers by taking clockwise and counterclockwise turns. Begin (LL) and then repeat (RRR)(LLL).

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. - Boris Putievskiy, Dec 16 2012

For generalizations see A219159, A213928. - Boris Putievskiy, Mar 10 2013

Examples

			The start of the sequence as table T(i,j), i,j > 0:
   1   4    5    16 ...
   2   3    6    15 ...
   9   8    7    14 ...
  10  11   12    13 ...
  ....
		

Links

• Boris Putievskiy, Rows n = 1..100 of triangle, flattened
• Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.
• Eric Weisstein's World of Mathematics, Pairing functions
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A219159, A213928. The main diagonal is A002061. The following appear within interlaced sequences: A016754, A001844, A053755, A004120. The first row is A081345. The first column is A081346. The inverse permutation A194280, the first inverse function (numbers of rows) A220603, the second inverse function (numbers of columns) A220604.

Programs

Mathematica

T[n_, k_] := T[n, k] = Which[OddQ[n] && k==1, n^2, EvenQ[k] && n==1, k^2, EvenQ[n] && k==1, T[n-1, 1]+1, OddQ[k] && n==1, T[1, k-1]+1, k <= n, T[n, k-1]+1 - 2 Mod[n, 2], True, T[n-1, k]-1 + 2 Mod[k, 2]]; Table[T[n-k+1, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Feb 20 2019 *)

Python

t=int((math.sqrt(8*n-7) - 1)/ 2)
i=n-t*(t+1)/2
j=(t*t+3*t+4)/2-n
if j >= i:
     m=(j-1)**2 + j + (j-i)*(-1)**(j-1)
else:
     m=(i-1)**2 + i - (i-j)*(-1)**(i-1)
# Boris Putievskiy, Dec 19 2012

Python

from math import isqrt
def A081344(n):
    t = (k:=isqrt(m:=n<<1))+((m<<2)>(k<<2)*(k+1)+1)-1
    i, j = n-(t*(t+1)>>1), (t*(t+3)>>1)+2-n
    r = max(i,j)
    return (r-1)**2+r+(j-i if r&1 else i-j) # Chai Wah Wu, Nov 04 2024

Formula

From Boris Putievskiy, Dec 19 2012: (Start)

a(n) = (i-1)^2 + i + (i-j)*(-1)^(i-1) if i >= j,

a(n) = (j-1)^2 + j - (j-i)*(-1)^(j-1) if i < j,

where

i = n - t*(t+1)/2,

j = (t*t + 3*t + 4)/2-n,

t = floor((-1 + sqrt(8*n-7))/2). (End)

Enumeration by boustrophedonic ("ox-plowing") method: If i >= j: T(i,j)=(i-1)^2+i + (i-j)*(-1)^(i-1), if i < j: T(i,j)=(j-1)^2+j - (j-i)*(-1)^(j-1). - Boris Putievskiy, Dec 19 2012

T(i,j) = m^2 - m + 1 - (i - j)*(-1)^m where m = max(i,j). - Ziad Ahmed, Jun 09 2025