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%I #23 Oct 06 2020 02:41:31
%S 1,1,0,1,0,1,1,0,0,1,1,0,2,0,1,1,0,0,0,0,1,1,0,3,2,3,0,1,1,0,0,0,0,0,
%T 0,1,1,0,4,0,6,0,4,0,1,1,0,0,3,0,0,3,0,0,1,1,0,5,0,10,2,10,0,5,0,1,1,
%U 0,0,0,0,0,0,0,0,0,0,1,1,0,6,4,15,12,24,12,15,4,6,0,1,1,0,0,0,0
%N Triangle read by rows: T(n,k) = number of k-subsets of the n-th roots of 1 that add to zero (0 <= k <= n).
%C Observe that T(n,k) = binomial(n,k) (mod n). Because the sum of the n n-th roots of unity is 0 for n>1, each row is symmetric for n>1. Hence only k=0..floor(n/2) need to be computed. - _T. D. Noe_, Jan 16 2008
%H Wouter Meeussen and T. D. Noe, <a href="/A103306/b103306.txt">Rows n=0..43 of triangle, flattened</a>
%H Wouter Meeussen, <a href="/A103306/a103306.txt">More terms</a>
%H Gary Sivek, <a href="http://www.emis.de/journals/INTEGERS/papers/k31/k31.Abstract.html">On vanishing sums of distinct roots of unity</a>, #A31, Integers 10 (2010), 365-368.
%e Triangle begins:
%e {1},
%e {1, 0},
%e {1, 0, 1},
%e {1, 0, 0, 1},
%e {1, 0, 2, 0, 1},
%e {1, 0, 0, 0, 0, 1},
%e {1, 0, 3, 2, 3, 0, 1},
%e {1, 0, 0, 0, 0, 0, 0, 1},
%e {1, 0, 4, 0, 6, 0, 4, 0, 1},
%e {1, 0, 0, 3, 0, 0, 3, 0, 0, 1},
%e T(10,4)=10, counting {1,2,6,7}, {1,3,6,8}, {1,4,6,9}, {1,5,6,10}, {2,3,7,8}, {2,4,7,9}, {2,5,7,10}, {3,4,8,9}, {3,5,8,10}, {4,5,9,10}.
%t <<DiscreteMath`Combinatorica`; Table[Count[(KSubsets[Range[n], k]), q_List/;Chop[Plus@@(E^(2.*Pi*I*q/n))]===0], {n, 0, 24}, {k, 0, n}]
%t T[n_, k_] := T[n, k] = Piecewise[{{T[n, n-k], k > n/2 >= 1}}, Count[Subsets[Range[n], {k}], subset_/;PossibleZeroQ[ExpToTrig[Sum[Exp[2*Pi*I*m/n], {m, subset}]]]]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // TableForm (* _David M. Zimmerman_, Sep 23 2020 *)
%Y Row sums give A103314.
%Y Cf. A070894, A322366.
%K nonn,tabl
%O 0,13
%A _Wouter Meeussen_, Mar 11 2005
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