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Table T(n,k) read by antidiagonals. T(n,k)=n-k+1, if n>=k, T(n,k)=k-n+2, if n < k.
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%I #20 Feb 15 2022 13:06:08

%S 1,3,2,4,1,3,5,3,2,4,6,4,1,3,5,7,5,3,2,4,6,8,6,4,1,3,5,7,9,7,5,3,2,4,

%T 6,8,10,8,6,4,1,3,5,7,9,11,9,7,5,3,2,4,6,8,10,12,10,8,6,4,1,3,5,7,9,11

%N Table T(n,k) read by antidiagonals. T(n,k)=n-k+1, if n>=k, T(n,k)=k-n+2, if n < k.

%C In general, let m be natural number. T(n,k)=n-k+1, if n>=k, T(n,k)=k-n+m-1, if n <k. Table T(n,k) read by antidiagonals. The first column of the table T(n,1) is the sequence of the natural numbers A000027. In all columns with number k (k > 1) the segment with the length of (k-1): {m+k-2, m+k-3, ..., m} shifts the sequence A000027. For m=1 the result is A220073, for m=2 the result is A143182. This sequence is the result for m=3.

%H Boris Putievskiy, <a href="/A187760/b187760.txt">Rows n = 1..140 of triangle, flattened</a>

%H Boris Putievskiy, <a href="http://arxiv.org/abs/1212.2732">Transformations Integer Sequences And Pairing Functions</a>, arXiv:1212.2732 [math.CO], 2012.

%F For the general case, a(n) = |(t+1)^2 - 2n| + m*floor((t^2+3t+2-2n)/(t+1)), where t=floor((-1+sqrt(8*n-7))/2).

%F For m=3, a(n) = |(t+1)^2 - 2n| + 3*floor((t^2+3t+2-2n)/(t+1)), where t=floor((-1+sqrt(8*n-7))/2).

%e The start of the sequence as table for the general case:

%e 1....m..m+1..m+2..m+3..m+4..m+5...

%e 2....1....m..m+1..m+2..m+3..m+4...

%e 3....2....1....m..m+1..m+2..m+3...

%e 4....3....2....1....m..m+1..m+2...

%e 5....4....3....2....1....m..m+1...

%e 6....5....4....3....2....1....m...

%e 7....6....5....4....3....2....1...

%e . . .

%e The start of the sequence as triangle array read by rows for the general case:

%e 1;

%e m,2;

%e m+1,1,3;

%e m+2,m,2,4;

%e m+3,m+1,1,3,5;

%e m+4,m+2,m,2,4,6;

%e m+5,m+3,m+1,1,3,5,7;

%e . . .

%e Row number r contains r numbers: m+r-2, m+r-4,...r-2,r.

%t T[n_, k_] := If[1 <= k <= n, n - k + 1, k - n + 2];

%t Table[T[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* _Jean-François Alcover_, Nov 06 2018 *)

%o (Python)

%o t=int((math.sqrt(8*n-7)-1)/2)

%o result=abs((t+1)**2 - 2*n) + 3*int((t**2+3*t+2-2*n)/(t+1))

%Y Cf. A000027, A220073, A143182.

%K nonn,tabl

%O 1,2

%A _Boris Putievskiy_, Jan 04 2013