A213921 Natural numbers placed in table T(n,k) layer by layer. The order of placement: at the beginning filled odd places of layer clockwise, next - even places clockwise. Table T(n,k) read by antidiagonals.

1, 2, 3, 5, 4, 7, 10, 8, 9, 13, 17, 14, 6, 16, 21, 26, 22, 11, 12, 25, 31, 37, 32, 18, 15, 20, 36, 43, 50, 44, 27, 23, 24, 30, 49, 57, 65, 58, 38, 33, 19, 35, 42, 64, 73, 82, 74, 51, 45, 28, 29, 48, 56, 81, 91, 101, 92, 66, 59, 39, 34, 41, 63, 72, 100, 111

Offset: 1

Author

Boris Putievskiy, Mar 05 2013

Keywords

• nonn
• tabl

Comments

A 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.

Layer is pair of sides of square from T(1,n) to T(n,n) and from T(n,n) to T(n,1). Enumeration table T(n,k) is layer by layer. The order of the list:

T(1,1)=1;

T(1,2), T(2,1), T(2,2);

. . .

T(1,n), T(3,n), ... T(n,3), T(n,1), T(2,n), T(4,n), ... T(n,4), T(n,2);

...

Examples

			The start of the sequence as table:
   1   2   5  10  17  26 ...
   3   4   8  14  22  32 ...
   7   9   6  11  18  27 ...
  13  16  12  15  23  33 ...
  21  25  20  24  19  28 ...
  31  36  30  35  29  34 ...
  ...
The start of the sequence as triangle array read by rows:
   1;
   2,  3;
   5,  4,  7;
  10,  8,  9, 13;
  17, 14,  6, 16, 21;
  26, 22, 11, 12, 25, 31;
  ...
		

Links

• Boris Putievskiy, Rows n = 1..140 of triangle, flattened
• Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
• Eric W. Weisstein, MathWorld: Pairing functions
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A060734, A060736; table T(n,k) contains: in rows A002522, A014206, A059100, A027688, A117950, A027689, A087475, A027690, A117951, A027691, A114949, A027692, A117619; in columns A002061, A000290, A002378, A005563, A028387, A008865, A028552, A028872, A014209, A028347, A028875.

Programs

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 i > j:
   result=i*i-(j%2)*i+2-int((j+2)/2)
else:
   result=j*j-((i%2)+1)*j + int((i+3)/2)

Formula

As a table:

T(n,k) = n*n - (k mod 2)*n + 2 - floor((k+2)/2), if n>k;

T(n,k) = k*k - ((n mod 2)+1)*k + floor((n+3)/2), if n<=k.

As a linear sequence:

a(n) = i*i - (j mod 2)*i + 2 - floor((j+2)/2), if i>j;

a(n) = j*j - ((i mod 2)+1)*j + floor((i+3)/2), 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).