A216250 T(n,k) = ((n+k)^2-4*k+3-2*(-1)^n-(-1)^k-(n+k-4)*(-1)^(n+k))/2-3, if k=1 and (n mod 2)=1; T(n,k) = ((n+k)^2-4*k+3-2*(-1)^n-(-1)^k-(n+k-4)*(-1)^(n+k))/2, else. Table T(n,k) read by antidiagonals; n, k > 0.
1, 2, 3, 5, 4, 6, 7, 8, 11, 12, 10, 9, 14, 13, 15, 16, 17, 20, 21, 24, 25, 19, 18, 23, 22, 27, 26, 28, 29, 30, 33, 34, 37, 38, 41, 42, 32, 31, 36, 35, 40, 39, 44, 43, 45, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63, 49, 48, 53, 52, 57, 56, 61, 60, 65, 64, 66, 67
Offset: 1
Examples
The start of the sequence as table: 1....2...5...7..10..16..19... 3....4...8...9..17..18..30... 6...11..14..20..23..33..36... 12..13..21..22..34..35..51... 15..24..27..37..40..54..57... 25..26..38..39..55..56..76... 28..41..44..58..61..79..82... . . . The start of the sequence as triangular array read by rows: 1; 2,3; 5,4,6; 7,8,11,12; 10,9,14,13,15; 16,17,20,21,24,25; 19,18,23,22,27,26,28; . . . The start of the sequence as array read by rows, with length of row r: 4*r-3: First 2*r-2 numbers are from the row number 2*r-2 of above triangle array. Last 2*r-1 numbers are from the row number 2*r-1 of above triangle array. 1; 2,3,5,4,6; 7,8,11,12,10,9,14,13,15; 16,17,20,21,24,25,19,18,23,22,27,26,28; . . . Row number r contains permutation of the 4*r-3 numbers from 2*r*r-5*r+4 to 2*r*r-r: 2*r*r-5*r+4, 2*r*r-5*r+5, ...2*r*r-r-2, 2*r*r-r.
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 Weisstein's World of Mathematics, Pairing functions
- Index entries for sequences that are permutations of the natural numbers
Crossrefs
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 result=((t+2)**2-4*j+3-(-1)**j-2*(-1)**i-(t-2)*(-1)**t)/2 if j==1 and (i%2)==1: result=result-3
Formula
As table
T(n,k) = ((n+k)^2-4*k+3-2*(-1)^n-(-1)^k-(n+k-4)*(-1)^(n+k))/2-3, if k=1 and (n mod 2)=1;
T(n,k) = ((n+k)^2-4*k+3-2*(-1)^n-(-1)^k-(n+k-4)*(-1)^(n+k))/2, else.
As linear sequence
a(n) = ((t+2)^2-4*j+3-2*(-1)^i-(-1)^j-(t-2)*(-1)^t)/2-3, if j=1 and (i mod 2)=1;
a(n) = ((t+2)^2-4*j+3-2*(-1)^i-(-1)^j-(t-2)*(-1)^t)/2, else; where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2).
Comments