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answered 1 year ago

Max Alekseyev gravatar image

Let ck be the coefficient of xk in sum( z^n/(1 - 2*z + z^(n+1))^2 for n in range(N)), then for 0<k<N we have ck=[xk]n1zn(12z+zn+1)2=[xk]n1m0zn+m(2m)(2+zn)m =kn=1knm=0mk(modn)(2m)(mknmn)(2)mknmn. Here is a corresponding code:

def c(k):
    return sum(binomial(-2,m) * binomial(m,(t:=(k-m-n)//n)) * (-2)^(m-t) for n in range(1,k+1) for m in range(k%n,k-n+1,n))

Then, the coefficient of xk in (1-z)^2 * sum( z^n/(1 - 2*z + z^(n+1))^2 for n in range(N)) can be obtained simply as c(k) - 2*c(k-1) + c(k-2).

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No.2 Revision

Let ck be the coefficient of xk in sum( z^n/(1 - 2*z + z^(n+1))^2 for n in range(N)), then for 0<k<N we have ck=[xk]n1zn(12z+zn+1)2=[xk]n1m0zn+m(2m)(2+zn)m =kn=1knm=0mk(modn)(2m)(mknmn)(2)mknmn. Here is a corresponding code:

def c(k):
    return 1 if k==0 else sum(binomial(-2,m) * binomial(m,(t:=(k-m-n)//n)) * (-2)^(m-t) for n in range(1,k+1) for m in range(k%n,k-n+1,n))

Then, the coefficient of xk in (1-z)^2 * sum( z^n/(1 - 2*z + z^(n+1))^2 for n in range(N)) can be obtained simply as c(k) - 2*c(k-1) + c(k-2).

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No.3 Revision

Let ck be the coefficient of xk in sum( z^n/(1 - 2*z + z^(n+1))^2 for n in range(N)), then for 0<k<N we have $$c_k = [x^k] [z^k] \sum_{n\geq 1} z^{n} (1-2z+z^{n+1})^{-2} = [x^k] [z^k] \sum_{n\geq1} \sum_{m\geq0} z^{n+m} \binom{-2}{m} (-2+z^n)^m=\sum_{n=1}^k \sum_{m=0\atop m\equiv k\pmod{n}}^{k-n} \binom{-2}{m} \binom{m}{\frac{k-n-m}n}(-2)^{m-\frac{k-n-m}n}.$$ Here is a corresponding code:

def c(k):
    return 1 if k==0 else sum(binomial(-2,m) * binomial(m,(t:=(k-m-n)//n)) * (-2)^(m-t) for n in range(1,k+1) for m in range(k%n,k-n+1,n))

Then, the coefficient of xk in (1-z)^2 * sum( z^n/(1 - 2*z + z^(n+1))^2 for n in range(N)) can be obtained simply as c(k) - 2*c(k-1) + c(k-2).

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No.4 Revision

Let ck be the coefficient of xk zk in sum( z^n/(1 - 2*z + z^(n+1))^2 for n in range(N)), then for 0<k<N we have ck=[zk]n1zn(12z+zn+1)2=[zk]n1m0zn+m(2m)(2+zn)m =kn=1knm=0mk(modn)(2m)(mknmn)(2)mknmn. Here is a corresponding code:

def c(k):
    return 1 if k==0 else sum(binomial(-2,m) * binomial(m,(t:=(k-m-n)//n)) * (-2)^(m-t) for n in range(1,k+1) for m in range(k%n,k-n+1,n))

Then, the coefficient of xk zk in (1-z)^2 * sum( z^n/(1 - 2*z + z^(n+1))^2 for n in range(N)) can be obtained simply as c(k) - 2*c(k-1) + c(k-2).