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%I #26 Nov 07 2016 05:08:12
%S 0,0,1,0,3,3,0,1,5,6,0,7,2,14,10,0,2,9,5,18,15,0,1,3,30,7,33,21,0,3,6,
%T 9,34,12,39,28,0,15,4,11,11,69,15,60,36,0,4,17,13,13,21,75,22,68,45,0,
%U 1,5,62,15,20,24,124,26,95,55,0,5,12,17,66,30,35,38,132,35,105,66
%N Square array A(n,k) (n>=1, k>=0) read by antidiagonals: A(n,0) = 0 and A(n,k) is the least integer > A(n,k-1) that can be expressed as a triangular number divided by n.
%e n\k 0 1 2 3 4 5 6 7
%e ------------------------------------------
%e 1 | 0 1 3 6 10 15 21 28 A000217
%e 2 | 0 3 5 14 18 33 39 60 A074378
%e 3 | 0 1 2 5 7 12 15 22 A001318
%e 4 | 0 7 9 30 34 69 75 124 A154260
%e 5 | 0 2 3 9 11 21 24 38 A057569
%e 6 | 0 1 6 11 13 20 35 46 A154293
%e 7 | 0 3 4 13 15 30 33 54 A057570
%e 8 | 0 15 17 62 66 141 147 252 A157716
%p A193470_rect := proc(n,k) local j,i,L; L := NULL; j := 0; while nops([L]) < k do add(i/n, i=1..j); if type(%,integer) then L := L,% fi; j := j+1 od; L end:
%p seq(print(A193470_rect(n, 12)),n = 1..8);
%t a[_, 0] = 0; a[n_, k_] := a[n, k] = For[j = a[n, k-1]+1, True, j++, If[Reduce[m > 0 && j == m(m+1)/(2n), m, Integers] =!= False, Return[j]]]; Table[a[n-k, k], {n, 1, 12}, {k, 0, n-1}] // Flatten (* _Jean-François Alcover_, Nov 07 2016 *)
%Y Cf. A061782, A011772.
%K nonn,tabl
%O 1,5
%A _Peter Luschny_, Jul 27 2011
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