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A218452 Number of ways to factor (1 + x + x^2+ ... + x^(n - 1))^2 as the product of two monic polynomials of degree n - 1 with positive coefficients (counting order). 0

%I

%S 1,1,1,1,1,1,1,1,1,1,1,3,5,7,9,9,13,11,17,19,33

%N Number of ways to factor (1 + x + x^2+ ... + x^(n - 1))^2 as the product of two monic polynomials of degree n - 1 with positive coefficients (counting order).

%C a(n) is the number of ways one can divide the unit square in n possibly irregular lines * n possibly irregular columns (parallel to the sides) so that each of the diagonals of the n X n irregular checkerboard thus constructed has the same area as it would in a regular checkerboard. Alternatively, this is the number of ways to construct a pair of n-sided dice (probability distribution on the n sides, labeled 0 through n-1), no face having probability 0, so that the sum of the two dice follows the expected probability distribution for the sum of two fair n-sided dice. Note that a(n) is always odd because there is always the obvious factorization of (1+x+...+x^(n-1))^2 as 1+x+...+x^(n-1) times itself, and each other factorization counts twice.

%e For n=12 we have a(n)=3 because apart from the obvious factorization of (1+x+...+x^11)^2 as (1+x+...+x^11) times itself, there exist the factorizations p*q and q*p where p = (1-sqrt(3)*x+x^2) * (1-x+x^2) * (1+x^2) * (1+x+x^2)^2 * (1+x) and q = (1-sqrt(3)*x+x^2) * (1-x+x^2) * (1+x^2) * (1+sqrt(3)*x+x^2)^2 * (1+x), both of which have positive coefficients, and those are the only two possible.

%o (Sage)

%o R.<x> = AA['x']

%o def has_positive_coefficients(pol):

%o return not any(c <= 0 for c in pol.coeffs())

%o def trydie(m):

%o results = []

%o tmp = list(factor(sum([x^i for i in range(m)])))

%o facs = [f for (f,_) in tmp]

%o n = len(facs)

%o for i in range((3^n+1)//2):

%o exps = [(i//(3^k))%3 for k in range(n)]

%o coexps = [2-v for v in exps]

%o pol = R(prod([facs[k]^exps[k] for k in range(n)]))

%o copol = R(prod([facs[k]^coexps[k] for k in range(n)]))

%o if pol.degree()<m and copol.degree()<m and has_positive_coefficients(pol) and has_positive_coefficients(copol):

%o pol /= pol.subs({x:1})

%o copol /= copol.subs({x:1})

%o results += [(pol.coeffs(),copol.coeffs())]

%o return 2*len(results)-1

%K nonn

%O 1,12

%A _David A. Madore_, Oct 28 2012

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Last modified November 28 04:31 EST 2021. Contains 349400 sequences. (Running on oeis4.)