![]() | 1 | initial version |
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]∑n≥1zn(1−2z+zn+1)−2=[xk]∑n≥1∑m≥0zn+m(−2m)(−2+zn)m
=k∑n=1k−n∑m=0m≡k(modn)(−2m)(mk−n−mn)(−2)m−k−n−mn.
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)
.
![]() | 2 | 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]∑n≥1zn(1−2z+zn+1)−2=[xk]∑n≥1∑m≥0zn+m(−2m)(−2+zn)m
=k∑n=1k−n∑m=0m≡k(modn)(−2m)(mk−n−mn)(−2)m−k−n−mn.
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)
.
![]() | 3 | 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)
.
![]() | 4 | 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]∑n≥1zn(1−2z+zn+1)−2=[zk]∑n≥1∑m≥0zn+m(−2m)(−2+zn)m
=k∑n=1k−n∑m=0m≡k(modn)(−2m)(mk−n−mn)(−2)m−k−n−mn.
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)
.