In how many solutions of equation: $x_1+x_2+...+x_n\equiv q(\text{ mod }m)$ satisfied: $x_i\in \mathbb{N}(i=\overline{1,n}),1\le x_i\le 26,0\le q\le m,q,m\in \mathbb{N}$.
I tried solve this problem by separating it into many small problems, those are: I solved: $x_1+x_2+...+x_n=q$, $x_1+x_2+...+x_n=q+m$,... And I think that, it's too difficult to compute the whole solution of above problem!! And I hope I can build a form for this problem!!!
a)
You have to look for the number of (strict) integer compositions of $n$ with no part greater than $26$, wherein $n\equiv q\ (\text{mod}\ m)$.
Let $C(n,r)$ be the number of (strict) integer compositions of $n$ with no part greater than $r$.
Its generating function is
$$\sum_{n=0}^\infty C(n,r)\ x^n=\sum_{k=0}^{\infty}(x^{1}+x^{2}+...+x^{r})^{k}=\frac{1}{1-x-x^{2}-...-x^{r}}=\frac{1-x}{1-2x+x^{r+1}}.$$
For $n>0$, the numbers $C(n,r)$ are the shifted $r$-generalized Fibonacci numbers:
$$C(n,r)=F(r;n+r-1)=\sum_{i=1}^{n}\sum_{j=0}^{i}(-1)^{j}\binom{i}{j}\binom{n-j\ r-1}{i-1}^{*}$$
*) The binomial is defined to be $0$ for a negative argument.
Without $r=0$, their OEIS sequence is A126198.
Your numbers are among the numbers $C(n,26)$. For $n\le 26$, that are the numbers of (strict) integer compositions of $n$, $C(n)$. $\forall n>0\colon C(n)=2^{n-1}$. Their OEIS sequence is A0117821.
b)
Collect now all $C(n,26)$ with $n\equiv q\ (\text{mod}\ m)$:
$$\sum_{k=0}^\infty\sum_{i=1}^{q+k\ m}\sum_{j=0}^{i}(-1)^{j}\binom{i}{j}\binom{q+k\ m-26j-1}{i-1}^{*}$$
*) The binomial is defined to be $0$ for a negative argument.
Consider that your set $\{n\ |\ n\equiv q\ (\text{mod}\ m)\}$ is an infinite set.
