The number of functions $$f$$, from the set $$A = \{x \in \mathbb{N}: x^2 - 10x + 9 \leq 0\}$$ to the set $$B = \{n^2 : n \in \mathbb{N}\}$$ such that $$f(x) \leq (x-3)^2 + 1$$, for every $$x \in A$$, is _______.

We need to find the number of functions $$f: A \to B$$ such that $$f(x)\le (x-3)^2 + 1$$ for every $$x\in A$$.

From the inequality $$x^2 - 10x + 9 \le 0$$ we have $$(x-1)(x-9)\le 0$$, which implies $$1\le x\le 9$$. Since $$x\in\mathbb{N}$$, it follows that $$A = \{1,2,3,4,5,6,7,8,9\}$$.

Also, by definition, $$B = \{n^2 : n\in\mathbb{N}\} = \{1,4,9,16,25,36,\ldots\}$$.

For each $$x$$ we require $$f(x)\le (x-3)^2 + 1$$ and $$f(x)\in B$$. We list the bounds and counts:

When $$x=1$$, $$(x-3)^2 + 1 = 4 + 1 = 5$$, so the eligible values in $$B$$ are $$\{1,4\}$$, giving 2 choices.

When $$x=2$$, $$(x-3)^2 + 1 = 1 + 1 = 2$$, so the eligible values in $$B$$ are $$\{1\}$$, giving 1 choice.

When $$x=3$$, $$(x-3)^2 + 1 = 0 + 1 = 1$$, so the eligible values in $$B$$ are $$\{1\}$$, giving 1 choice.

When $$x=4$$, $$(x-3)^2 + 1 = 1 + 1 = 2$$, so the eligible values in $$B$$ are $$\{1\}$$, giving 1 choice.

When $$x=5$$, $$(x-3)^2 + 1 = 4 + 1 = 5$$, so the eligible values in $$B$$ are $$\{1,4\}$$, giving 2 choices.

When $$x=6$$, $$(x-3)^2 + 1 = 9 + 1 = 10$$, so the eligible values in $$B$$ are $$\{1,4,9\}$$, giving 3 choices.

When $$x=7$$, $$(x-3)^2 + 1 = 16 + 1 = 17$$, so the eligible values in $$B$$ are $$\{1,4,9,16\}$$, giving 4 choices.

When $$x=8$$, $$(x-3)^2 + 1 = 25 + 1 = 26$$, so the eligible values in $$B$$ are $$\{1,4,9,16,25\}$$, giving 5 choices.

When $$x=9$$, $$(x-3)^2 + 1 = 36 + 1 = 37$$, so the eligible values in $$B$$ are $$\{1,4,9,16,25,36\}$$, giving 6 choices.

Since the choices at each $$x$$ are independent, the total number of such functions is

$$2 \times 1 \times 1 \times 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 1440$$.

The answer is $$\boxed{1440}$$.