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%I #18 Aug 28 2019 16:23:04
%S 1,0,1,-1,1,1,-3,5,-3,1,1,-5,13,-17,13,-5,1,1,-9,41,-109,191,-229,191,
%T -109,41,-9,1,1,-11,61,-203,457,-731,853,-731,457,-203,61,-11,1,1,-15,
%U 113,-527,1713,-4111,7537,-10767,12113,-10767,7537,-4111,1713,-527,113,-15,1,1,-17,145,-773,2899,-8117,17587
%N Coefficients of numerator polynomials of g.f.s for a certain necklace problem involving prime numbers.
%C The row polynomials P(n,x) := Sum_{k=0..p(n)-1} a(n,k)*x^k, n >= 1, appear in the numerator of the g.f. G(p(n),x) for the numbers N(p(n),m) of inequivalent m-bead necklaces of two colors with p(n) beads of one color and m-p(n) beads of the other color. Here p(n)=A000040(n) (prime numbers). Equivalently, N(p(n),m) counts inequivalent necklaces with p(n) beads which are labeled with nonnegative numbers, such that the sum of the labels is m. For a proof of this equivalent formulation see a comment in A032191. Inequivalence is meant with respect to the cyclic group C_p(n).
%C This necklace g.f. is G(p(n),x) = P(n,x)/((1-x^p(n))*(1-x)^(p(n)-1)), n >= 1. The row polynomials P(n,x) are defined above. This g.f. is Z(C_p(n),x), the two variable (x[1] and x[p(n)]) cycle index polynomial for the cyclic group of prime order p(n), with substitution x[1]->1/(1-x^1)and x[p(n)]->1/(1-x^p(n)). This follows by Polya enumeration if the above mentioned labeled necklace problem is solved.
%C The row length sequence for this array a(n,k) is A000040(n) (n-th prime number), [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...].
%C The rows of this signed array are symmetric: a(n,k) = a(n,p(n)-1-k), n >= 2, k = 0..(p(n)-1)/2. See the explicit formula below.
%C The formulas for a(n,k), given below, produces in fact integers.
%C G.f. for column k, k>=0 (without leading zeros): sum(A103718(k,m)*p(n)^m,m=0..k)/k! produces for all n> pi(n) integers, where pi(n):=A000720(n), primes not exceeeding n.
%H W. Lang, <a href="/A103728/a103728.txt">Array and more comments.</a>
%F a(n, k) = (1 + ((-1)^k)*(p(n)-1)*binomial(p(n)-1, k))/p(n), with p(n): = A000040(n) (n-th prime).
%F a(n, k) = sum(A103718(k, m)*p(n)^m, m=0..k)/k!, (row polynomials of triangle A103718 with x=p(n), divided by k!).
%e Triangle begins:
%e [1, -0];
%e [1, -1, 1];
%e [1, -3, 5, -3, 1];
%e [1, -5, 13, -17, 13, -5, 1];
%e [1, -9, 41,-109, 191, -229, 191, -109, 41, -9, 1];
%e ...
%e n=3: G(p(3),x)=G(5,x)=(1-3*x+5*x^2-3*x^3+1*x^4)/((1-x^5)*(1-x)^4) generates the necklace sequence A008646.
%e A103718(3,m), m=0..3, is [17,-17,7,-1]. Therefore (17-17*p(n)+7*p(n)^2-1*p(n)^3 )/3! gives, for n>=1, the third column [ -3,-17,-109,...].
%Y The unsigned column sequences are for k=0..10: A000012 (powers of 1), A040976 (primes p(n)-2), A103729 - A103914, A103915.
%K sign,easy,tabf
%O 1,7
%A _Wolfdieter Lang_, Feb 24 2005
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