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 A103728 Coefficients of numerator polynomials of g.f.s for a certain necklace problem involving prime numbers. 12
 1, 0, 1, -1, 1, 1, -3, 5, -3, 1, 1, -5, 13, -17, 13, -5, 1, 1, -9, 41, -109, 191, -229, 191, -109, 41, -9, 1, 1, -11, 61, -203, 457, -731, 853, -731, 457, -203, 61, -11, 1, 1, -15, 113, -527, 1713, -4111, 7537, -10767, 12113, -10767, 7537, -4111, 1713, -527, 113, -15, 1, 1, -17, 145, -773, 2899, -8117, 17587 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,7 COMMENTS 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). 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 and x[p(n)]) cycle index polynomial for the cyclic group of prime order p(n), with substitution x->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. 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, ...]. 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. The formulas for a(n,k), given below, produces in fact integers. 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. LINKS W. Lang, Array and more comments. FORMULA a(n, k) = (1 + ((-1)^k)*(p(n)-1)*binomial(p(n)-1, k))/p(n), with p(n): = A000040(n) (n-th prime). 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!). EXAMPLE Triangle begins:   [1, -0];   [1, -1,  1];   [1, -3,  5,  -3,   1];   [1, -5, 13, -17,  13,   -5,   1];   [1, -9, 41,-109, 191, -229, 191, -109, 41, -9, 1];   ... 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. 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,...]. CROSSREFS The unsigned column sequences are for k=0..10: A000012 (powers of 1), A040976 (primes p(n)-2), A103729 - A103914, A103915. Sequence in context: A096196 A351179 A076824 * A243533 A239730 A287765 Adjacent sequences:  A103725 A103726 A103727 * A103729 A103730 A103731 KEYWORD sign,easy,tabf AUTHOR Wolfdieter Lang, Feb 24 2005 STATUS approved

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