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.
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
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.
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
Programs
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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 *)
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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).
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