OFFSET
0,13
COMMENTS
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
LINKS
Wouter Meeussen and T. D. Noe, Rows n=0..43 of triangle, flattened
Wouter Meeussen, More terms
Gary Sivek, On vanishing sums of distinct roots of unity, #A31, Integers 10 (2010), 365-368.
EXAMPLE
Triangle begins:
{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, 0, 1},
{1, 0, 4, 0, 6, 0, 4, 0, 1},
{1, 0, 0, 3, 0, 0, 3, 0, 0, 1},
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}.
MATHEMATICA
<<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[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 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Wouter Meeussen, Mar 11 2005
STATUS
approved