Let q[n] represent the n-th root of -1, or exp(2 i pi/n) given that Sum(j=1 to n, q[n]^j) equals zero by symmetry, it is evident that (q[1]+q[2]+..+q[n])^k also equals zero for any k. Now, looking at the expansion of this power, and dropping (setting =1) the multinomial coefficients, it is (was) mildly surprising this expression is still zero, except when k=n, in which case it is 1. reason: use q[n]^z = q[n]^mod(z,n) and find that Sum(j=1 to n, q[n]^j)^k without the usual multinomials, and with the use of q[n]^z = q[n]^mod(z,n) gives oh, darn, now that we've done the mod thing, we can drop the [n] bit, and just write {{1}, {1 + q, 2 + q}, {1 + q + q^2, 2 + 2 q + 2 q^2, 4 + 3 q + 3 q^2}, {1 + q + q^2 + q^3, 3 + 2 q + 3 q^2 + 2 q^3, 5 + 5 q + 5 q^2 + 5 q^3, 10 + 8 q + 9 q^2 + 8 q^3}, {1 + q + q^2 + q^3 + q^4, 3 + 3 q + 3 q^2 + 3 q^3 + 3 q^4, 7 + 7 q + 7 q^2 + 7 q^3 + 7 q^4, 14 + 14 q + 14 q^2 + 14 q^3 + 14 q^4, 26 + 25 q + 25 q^2 + 25 q^3 + 25 q^4}} etc.. so, for each pair (n,k), we got ourselves a polynomial in q, but, a surprisingly simple one, at first sight. I expected a straight 1+q+q^2+..+q^n everywhere, but no! There seems to be a bit more to it. The least we can do, is to look at the coefficients of q^0 and q^n, no? And that, dear reader, if you've not run off by now, severely contorted with abdominal cramps with crap overload, is where two unseemly conjectures get pushed into the open (excuse my french) given that W*q+W*q^2+...+W*q^n equals W*(ziltch)=0 as we very well could and should expect, where does the W factor come from? conjecture 1: the coefficient of q^n is the constant term, q^n being 1, and equals Sum(d | gcd(n+k,k) ; phi(d) C(n/d+k/d,n/d) ) Cf. A004526, A007997, A008610, A008646, A032191, A032192, A032193 (Number of necklaces with n black beads and k white beads check item 1 below. conjecture 2: the coefficient of q^(n-1) is the factor of the highest power of q, and equals Sum(d | gcd(n+k,k) ; mu(d) C(n/d+k/d,k/d) ) A130513: Subtriangle of triangle in A051168 : remove central column of A051168 and all columns to the right; now read by upwards diagonals. check item 2 below. please take a sec to notice their similarities, phi versus mu, C(..,n/d) versus C(..,k/d) the other coefficients? I dun'no, phu versus mi? Wouter. --------------------------------- item 1 : conjecture 1. z=Table[Expand[ Plus@@ Table[q[n,j],{j,n}]^k /k! ],{n,5},{k,n}]/. q[n_,j_]^(e_:1) -> e! q[n]^(e j) a=Table[ Plus@@ Apply[Times,(q[n]^Range[n])^#& /@ Compositions[k,n],{1}],{n,5},{k,n}] i assume z==a 'by nature', .. handwaving...then, b=a/. q[n_]^(e_:1) :> q^Mod[e,n] (* just take 'a', replace q[n]^e with q^mod(e,n) as we should for roots of 1 *) b /. q->0 (* and set q equal to 0 to get the constant term.. *) {{1}, {1, 2}, {1, 2, 4}, {1, 3, 5, 10}, {1, 3, 7, 14, 26}, {1, 4, 10, 22, 42, 80}, {1, 4, 12, 30, 66, 132, 246}, {1, 5, 15, 43, 99, 217, 429, 810}, {1, 5, 19, 55, 143, 335, 715, 1430, 2704}, {1, 6, 22, 73, 201, 504, 1144, 2438, 4862, 9252}} and that looks like (=conjecture) Table[1/(n+k) Plus@@ (EulerPhi[#]Binomial[n/#+k/#,n/#]&/@ Divisors[GCD[n+k, k]]),{n,10},{k,n}] --------------------------------- item2 : conjecture 2. Map[Coefficient[q #,q^Exponent[q #,q]]&,b] (* get the highest exponent's factor *) {{1}, {1, 1}, {1, 2, 3}, {1, 2, 5, 8}, {1, 3, 7, 14, 25}, {1, 3, 9, 20, 42, 75}, {1, 4, 12, 30, 66, 132, 245}, {1, 4, 15, 40, 99, 212, 429, 800}, {1, 5, 18, 55, 143, 333, 715, 1430, 2700}, {1, 5, 22, 70, 200, 497, 1144, 2424, 4862, 9225}} and that looks like (=conjecture) Table[1/(n+k) Plus@@ (MoebiusMu[#]Binomial[(n+k)/#,k/#] &/@ Divisors[GCD[n+k,k]]),{n,12},{k,n}] ---------------------------------- minor OEIS bonus (to maybe submit after the summer recess) A051168: T(h,k)=number of Lyndon words of k 1's and h-k 0's. See http://www.theory.cs.uvic.ca/~cos/inf/neck/NecklaceInfo.html Sum(d | gcd(n,k) ; mu(d)/n C(n/d,k/d) ) Table[1/n Plus@@ (MoebiusMu[#]Binomial[n/#,k/#] &/@ Divisors[GCD[n,k]]),{n,12},{k,0,n}] A130513: Subtriangle of triangle in A051168 : remove central column of A051168 and all columns to the right; now read by upwards diagonals. Column order reversed: Sum( d | gcd(n+k,k) ; mu(d) C(n/d+k/d,k/d) /(n+k) ) or, in Mma dialect: Table[1/(n+k) Plus@@ (MoebiusMu[#]Binomial[n/#+k/#,k/#] &/@ Divisors[GCD[n+k,k]]),{n,12},{k,n}] ------------------------------------ end-of-tale