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Table T(n,k) = k mod n read by antidiagonals (n >= 1, k >= 1).
13

%I #39 May 11 2024 21:19:14

%S 0,0,1,0,0,1,0,1,2,1,0,0,0,2,1,0,1,1,3,2,1,0,0,2,0,3,2,1,0,1,0,1,4,3,

%T 2,1,0,0,1,2,0,4,3,2,1,0,1,2,3,1,5,4,3,2,1,0,0,0,0,2,0,5,4,3,2,1,0,1,

%U 1,1,3,1,6,5,4,3,2,1,0,0,2,2,4,2,0,6,5,4,3,2,1,0,1,0,3,0,3,1,7,6,5,4,3,2,1

%N Table T(n,k) = k mod n read by antidiagonals (n >= 1, k >= 1).

%C Note that the upper right half of this sequence when formatted as a square array is essentially the same as this whole sequence when formatted as an upper right triangle. Sums of antidiagonals are A004125. - _Henry Bottomley_, Jun 22 2001

%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 As a linear array, the sequence is a(n) = A004736(n) mod A002260(n) or a(n) = ((t*t+3*t+4)/2-n) mod (n-(t*(t+1)/2)), where t = floor((-1+sqrt(8*n-7))/2). - _Boris Putievskiy_, Dec 17 2012

%F G.f. for the n-th row: y*Sum_{i=0..n-2} (i + 1)*y^i/(1 - y^n). - _Stefano Spezia_, May 08 2024

%e 0 0 0 0 0 0 0 0 0 0 ...

%e 1 0 1 0 1 0 1 0 1 0 ...

%e 1 2 0 1 2 0 1 2 0 1 ...

%e 1 2 3 0 1 2 3 0 1 2 ...

%e 1 2 3 4 0 1 2 3 4 0 ...

%e 1 2 3 4 5 0 1 2 3 4 ...

%e 1 2 3 4 5 6 0 1 2 3 ...

%e 1 2 3 4 5 6 7 0 1 2 ...

%e 1 2 3 4 5 6 7 8 0 1 ...

%e 1 2 3 4 5 6 7 8 9 0 ...

%e 1 2 3 4 5 6 7 8 9 10 ...

%e 1 2 3 4 5 6 7 8 9 10 ...

%e 1 2 3 4 5 6 7 8 9 10 ...

%t T[n_, m_] = Mod[n - m + 1, m + 1]; Table[Table[T[n, m], {m, 0, n}], {n, 0, 10}]; Flatten[%] (* _Roger L. Bagula_, Sep 04 2008 *)

%o (PARI) T(n, k)=k%n \\ _Charles R Greathouse IV_, Feb 09 2017

%Y Transpose of A051126.

%Y Cf. A048158, A051777, A122750.

%Y Cf. A002260, A004125, A004736.

%K nonn,tabl,easy,nice

%O 1,9

%A _N. J. A. Sloane_

%E More terms from _James A. Sellers_, Dec 11 1999