A214870 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 counterclockwise. T(n,k) read by antidiagonals.
1, 2, 3, 5, 4, 7, 10, 9, 8, 13, 17, 16, 6, 14, 21, 26, 25, 11, 12, 22, 31, 37, 36, 18, 15, 20, 32, 43, 50, 49, 27, 24, 23, 30, 44, 57, 65, 64, 38, 35, 19, 33, 42, 58, 73, 82, 81, 51, 48, 28, 29, 45, 56, 74, 91, 101, 100, 66, 63, 39, 34, 41, 59, 72, 92, 111
Offset: 1
Examples
The start of the sequence as table: 1 2 5 10 17 26 ... 3 4 9 16 25 36 ... 7 8 6 11 18 27 ... 13 14 12 15 24 35 ... 21 22 20 23 19 28 ... 31 32 30 33 29 34 ... ... The start of the sequence as triangle array read by rows: 1; 2, 3; 5, 4, 7; 10, 9, 8, 13; 17, 16, 6, 14, 21; 26, 25, 11, 12, 22, 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 Weisstein's World of Mathematics, Pairing functions
- Index entries for sequences that are permutations of the natural numbers
Crossrefs
Cf. A060734, A060736, A185725, A213921, A213922; table T(n,k) contains: in rows A002522, A000290, A059100, A005563, A117950, A008865, A087475, A028872, A117951, A028347, A114949, A028875, A117619, A028878, A189833, A028881, A189834, A028884, A114948, A028560, A189836; in columns A002061, A014206, A002378, A027688, A028387, A027689, A028552, A027690, A014209, A027691, A027692, A082111, A027693, A028557, A027694, A108195, A187710, A048058, A048840.
Programs
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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 if i > j: result=i*i-i+(j%2)*(2-(j+1)/2)+((j+1)%2)*(j/2+1) else: result=j*j-2*(i%2)*j + (i%2)*((i+1)/2+1) + ((i+1)%2)*(-i/2+1)
Formula
As table
T(n,k) = k*k-2*(n mod 2)*k+(n mod 2)*((n+1)/2+1)+((n+1) mod 2)*(-n/2+1), if n<=k;
T(n,k) = n*n-n+(k mod 2)*(2-(k+1)/2)+((k+1) mod 2)*(k/2+1), if n>k.
As linear sequence
a(n) = j*j-2*(i mod 2)*j+(i mod 2)*((i+1)/2+1)+((i+1) mod 2)*(-i/2+1), if i<=j;
a(n) = i*i-i+(j mod 2)*(2-(j+1)/2)+((j+1) mod 2)*(j/2+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).
Comments