dear funsters, when posting stuff like this, I presume it is generally known, or at least could be. Motivation: just fun and beauty. 1/. Intro consider the expansion of (a+b+c+d)^3 it is the sum over M(u,v,w,x) a^u b^v c^w d^x with positive or zero integers u+v+w+x=3 and M(u,v,w,x) is the multinomial u! v! w! x! /(u+v+w+x)! When we drop the multinomial coefficients, then we get the 'compositions' of 4 elements in groups of three. in Mma: Expand[(a+b+c+d)^3 /3! ]/. q_^n_ ->q^n n! gives a^3 + a^2*b + a*b^2 + b^3 + a^2*c + a*b*c + b^2*c + a*c^2 + b*c^2 + c^3 + a^2*d + a*b*d + b^2*d + a*c*d + b*c*d + c^2*d + a*d^2 + b*d^2 + c*d^2 + d^3 this is equivalent to Plus @@ Apply[Times,{a,b,c,d}^#& /@Compositions[3,4],{1}] 2/. Generalise replace the (a+b+c+d) above by Sum(j=1..n, t[ j ] ) , t stands for 'temporary', now replace t[j] with E^(2 Pi I j/n) and it's obvious that this sum equals zero. And that any power of zero is also zero. But now, dropping multinomials, we find: Table[ Plus@@ Apply[Times,(Exp[2 Pi I/n]^Range[n])^#& /@ Compositions[k,n],{1}],{n,5},{k,n}] and we find that it's also zero, except it's 1 when k equals n, that is, we take all n roots of (-1) and put it to the n-th (multinomial-less) power. 3/. an instructive typo : Exp[2 Pi I/k] instead of Exp[2 Pi I/n]: Table[ Plus@@ Apply[Times,(Exp[2 Pi I/k]^Range[n])^#& /@ Compositions[k,n],{1}],{n,5},{k,n}] equals Table[Ceiling[n/m],{n,9},{m,n}] alias A120885 That leads to this silly remark: Table[ Plus@@ (Exp[2 Pi I/k]^Range[n])^k ,{n,16},{k,n}] is integer, except for the following couples of {n,k} : {7, 5}, {8, 5}, {9, 7}, {10, 7}, {10, 8}, {11, 7}, {11, 8}, {11, 9}, {12, 5}, {12, 7}, {12, 8}, {12, 9}, {12, 10}, {13, 5}, {13, 8}, {13, 9}, {13, 10}, {13, 11}, {14, 8}, {14, 9}, {14, 10}, {14, 11}, {14, 12}, {15, 9}, {15, 10}, {15, 11}, {15, 12}, {15, 13}, {16, 7}, {16, 9}, {16, 10}, {16, 11}, {16, 12}, {16, 13}, {16, 14}, ... and, where's the system in *this* madness? Wouter.