Suppose we seek to evaluate the following sum (with a condition on the
binomial coefficient)
$$G(n,m) = \sum_{k=0}^n
\sum_{q=0}^k (-1)^q
{k\choose q} {n-1-qm\choose k-1}.$$
Now when $n-1-qm \lt 0$ we usually get a non-zero value for the
binomial coefficient but this is not wanted here. Therefore we have
$$G(n,m) = \sum_{k=0}^n
\sum_{q=0}^{\lfloor (n-k)/m \rfloor} (-1)^q
{k\choose q} {n-1-qm\choose k-1}.$$
If we have lost any values for $q$ above $\lfloor (n-k)/m \rfloor$
these would render the second binomial coefficient zero. If we have
added in any values for $q$ above $k$ the first binomial coefficient
is zero there.
Now with the integral
$${n-1-qm\choose k-1} = {n-1-qm\choose n-k-qm} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-1-qm}}{z^{n-k-qm+1}}
\; dz$$
we get range control because the pole vanishes when $q\gt (n-k)/m$ and
we may extend $q$ to infinity.
We thus obtain for the inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-1}}{z^{n-k+1}}
\sum_{q\ge 0} (-1)^q {k\choose q} \frac{z^{qm}}{(1+z)^{qm}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-1}}{z^{n-k+1}}
\left(1-\frac{z^m}{(1+z)^m}\right)^k
\; dz$$
This yields for the outer sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-1}}{z^{n+1}}
\left(1-z\left(1-\frac{z^m}{(1+z)^m}\right)\right)^{-1}
\\ \times \left(1 - z^{n+1}
\left(1-\frac{z^m}{(1+z)^m}\right)^{n+1}\right)
\; dz$$
which is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+m-1}}{z^{n+1}}
\left((1-z)(1+z)^m + z^{m+1}\right)^{-1}
\\ \times \left(1 - z^{n+1}
\left(1-\frac{z^m}{(1+z)^m}\right)^{n+1}\right)
\; dz$$
Extracting the second component from the difference we get
$$-\frac{1}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^{n+m-1}
\left((1-z)(1+z)^m + z^{m+1}\right)^{-1}
\left(1-\frac{z^m}{(1+z)^m}\right)^{n+1}
\; dz$$
The pole at zero has vanished. We now have non-zero poles at $z=-1$
and from the inverted term. These depend on $m$ and we can certainly
choose $\epsilon$ small enough so that none of them are inside the
contour. Therefore this term does not contribute, leaving only
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+m-1}}{z^{n+1}}
\frac{1}{(1-z)(1+z)^m + z^{m+1}}
\; dz.$$
The generating function $f(w)$ of these numbers is thus given by
$$f(w) = \sum_{n\ge 0} w^n
\sum_{q=0}^n {n+m-1\choose n-q}
[z^q] \frac{1}{(1-z)(1+z)^m + z^{m+1}}.$$
This is
$$\sum_{q\ge 0} [z^q] \frac{1}{(1-z)(1+z)^m + z^{m+1}}
\sum_{n\ge q} w^n {n+m-1\choose n-q}
\\ = \sum_{q\ge 0} w^q [z^q] \frac{1}{(1-z)(1+z)^m + z^{m+1}}
\sum_{n\ge 0} w^n {n+m-1 + q\choose n}
\\ = \frac{1}{(1-w)^m}
\sum_{q\ge 0} \frac{w^q}{(1-w)^q}
[z^q] \frac{1}{(1-z)(1+z)^m + z^{m+1}}.$$
What we have here is an annihilated coefficient extractor that
simplifies to
$$f(w) = \frac{1}{(1-w)^m}
\frac{1}{(1-w/(1-w))(1+w/(1-w))^m + (w/(1-w))^{m+1}}
\\ = \frac{1}{(1-w)^m}
\frac{1}{(1-2w)/(1-w)/(1-w)^m + w^{m+1}/(1-w)^{m+1}}
\\ = \frac{1-w}{1- 2 w + w^{m+1}}.$$
Now observe that
$$1-2w+w^{m+1} = (1-w) (1-w-w^2-\cdots- w^{m-1} - w^m)$$
so we finally have
$$f(w) = \left(1-\sum_{q=1}^m w^q\right)^{-1}
= \frac{1}{1-w-w^2-\cdots-w^m}.$$
We see that by the basic theory of linear recurrences what we have
here is a Fibonacci, Tribonacci, Tetranacci etc. recurrence. The
question is what are the initial values.
Observe however that $[w^0] f(w) = 1$ and for $1\le q\le m$ we have
$$[w^q] \frac{1-w}{1-2w+w^{m+1}}
= [w^q] \frac{1}{1-2w+w^{m+1}} - [w^{q-1}] \frac{1}{1-2w+w^{m+1}}.$$
But $$\frac{1}{1-2w+w^{m+1}} = \frac{1}{1-2w(1-w^{m}/2)}
= \sum_{n\ge 0} 2^n w^n (1-w^m/2)^n$$
With the condition on $q$ and $n\ge 1$ only the constant term from the
term $(1-w^m/2)^n$ contributes because the degree would be more than
$m$ otherwise. This produces just one matching term with coefficient
$2^q.$
This yields for $f(w)$
$$[w^q] f(w) = 2^{q} - 2^{q-1} = 2^{q-1}.$$
Therefore we get for the intial terms starting at $q=0$
$$1, 1, 2, 4, 8, 16, \ldots, 2^{m-1}
\quad\text{with recurrence}\quad
f_n = \sum_{q=1}^m f_{n-q}.$$
This recurrence also shows (by subtraction) that the sequence may be
produced starting from $m-1$ zero terms followed by one.
The OEIS has the Fibonacci numbers, OEIS A000045
$$1, 2, 3, 5, 8, 13, 21, 34, 55, 89,\ldots$$
and the Tribonacci numbers, OEIS A000073
$$1, 2, 4, 7, 13, 24, 44, 81, 149, 274,\ldots$$
and the Tetranacci numbers, OEIS A000078
$$1, 2, 4, 8, 15, 29, 56, 108, 208, 401,\ldots$$
and more.
There are several more examples of the technique of annihilated coefficient extractors at this MSE link
I and at this MSE
link II and also
here at this MSE link
III.
