Studying a family of recurrent sequences (generalized from the NSW-numbers) I came to the observation, that certain polynomials (over $\mathbb N$) avoid primefactors below some smallest one. For instance the polynomial
$$ \small f_{11}(x) = x^5 + 11x^4 + 44x^3 + 77x^2 + 55x + 11 \qquad \qquad x \in \mathbb N $$
seems to have never a smaller primefactor than $p_{11,0}=23$ (except for the "trivial" $11$ at every $x=11k,k \in \mathbb N$) which occurs the first time at $x=2$.
Similarly, the polynomial
$$ \small f_{13}(x)= x^6 + 13x^5 + 65x^4 + 156x^3 + 182x^2 + 91x + 13 $$
seems to have never a smaller primefactor than $p_{13,0}=53$ (except for $13$ at every $x =13k $) which occurs the first time at $x=6$.
Similarly, the polynomial $$ \small f_{17}(x)= x^8 + 17x^7 + 119x^6 + 442x^5 + 935x^4 + 1122x^3 + 714x^2 + 204x + 17 $$ seems to have never a smaller primefactor than $p_{17,0}=67$ (except for $17$ at every $x =17k $) which occurs the first time at $x=11$.
This comes from just eyeballing a table of data and is possibly false, but the heuristic looks very good and so I'm quite confident that this is true.
I have no idea how I could attack the question of proving such type of claims (I want to say a similar statement in general for that whole family of polynomials).
So besides of the basic question,
Q1: are these three observations true?
I'd like to get an idea for
Q2: how could I proceed to prove such types of conjectures?
[update] Answering to the comment of @chu: Consider the sequence of NSW-numbers recursively defined by $$a_6(k) =- a_6(k-2) + 6a_6(k-1) \qquad \qquad a_6(0)=-1,a_6(1) = 1$$ (the initial elements are one index shifted from that in the OEIS). And in general $$a_m(k) =- a_m(k-2) + ma_m(k-1) \qquad \qquad a_m(0)=-1,a_m(1) = 1$$
The first few members of the sequence are
$$ \small a_6(n) =[-1, 1, 7, 41, 239, 1393, 8119, 47321, 275807, 1607521, 9369319, 54608393,...]$$
The element $a_6(6)$ is that with the value $8119$.
Now the sequence $a_5(n)$ is similarly defined, just replace a $5$ instead of the $6$ in the recursion formula. The element $a_5(6) = 3191$. Do the same with sufficiently many sequences $a_k(6)$ We get the sequence of numbers $b_{11}(m)= 1,-1,11,199,989,3191,8119$ for the sequences $a_0$ to $a_6$ I chose the index $11$ at $b_{11}$ because of the occurence of the value $11$ in $a_2(6)$ (There is also some deeper reason for this, but that's not important here) That sequence can arbitrarily be expanded so that we can define an interpolating polynomial $f_{11}(x)$ which generates that numbers.
Completely similar is this done for the general $f_k(x)$.
But note, that relevant for me in this question is, how I could approach a proof for the primefactor-conditions described above; I want to get an idea/understand how this can be done in general.
Here is the left part of the table with the data. The sequence of $a_6()$ is in the row with index $6$, beginning with $-1,1,7,41,...$
The row with index $2$ having the odd natural numbers give $a_2()$ and make the second column-index at the top of the table. For the vertical columns we can find interpolating polynomials which become then my functions $f_m()$
x | 0 1 2 3 4 5 6 7 8 9
m | -1 1 3 5 7 9 11 13 15 17
- + - - - - - - - - - -
-15 | -1 1 -14 209 -3121 46606 -695969 10392929 -155197966 2317576561
-14 | -1 1 -13 181 -2521 35113 -489061 6811741 -94875313 1321442641
-13 | -1 1 -12 155 -2003 25884 -334489 4322473 -55857660 721827107
-12 | -1 1 -11 131 -1561 18601 -221651 2641211 -31472881 375033361
-11 | -1 1 -10 109 -1189 12970 -141481 1543321 -16835050 183642229
-10 | -1 1 -9 89 -881 8721 -86329 854569 -8459361 83739041
-9 | -1 1 -8 71 -631 5608 -49841 442961 -3936808 34988311
-8 | -1 1 -7 55 -433 3409 -26839 211303 -1663585 13097377
-7 | -1 1 -6 41 -281 1926 -13201 90481 -620166 4250681
-6 | -1 1 -5 29 -169 985 -5741 33461 -195025 1136689
-5 | -1 1 -4 19 -91 436 -2089 10009 -47956 229771
-4 | -1 1 -3 11 -41 153 -571 2131 -7953 29681
-3 | -1 1 -2 5 -13 34 -89 233 -610 1597
-2 | -1 1 -1 1 -1 1 -1 1 -1 1
-1 | -1 1 0 -1 1 0 -1 1 0 -1
0 | -1 1 1 -1 -1 1 1 -1 -1 1
1 | -1 1 2 1 -1 -2 -1 1 2 1
2 | -1 1 3 5 7 9 11 13 15 17
3 | -1 1 4 11 29 76 199 521 1364 3571
4 | -1 1 5 19 71 265 989 3691 13775 51409
5 | -1 1 6 29 139 666 3191 15289 73254 350981
6 | -1 1 7 41 239 1393 8119 47321 275807 1607521
7 | -1 1 8 55 377 2584 17711 121393 832040 5702887
8 | -1 1 9 71 559 4401 34649 272791 2147679 16908641
9 | -1 1 10 89 791 7030 62479 555281 4935050 43860169
10 | -1 1 11 109 1079 10681 105731 1046629 10360559 102558961
11 | -1 1 12 131 1429 15588 170039 1854841 20233212 220710491
12 | -1 1 13 155 1847 22009 262261 3125123 37239215 443745457
13 | -1 1 14 181 2339 30226 390599 5047561 65227694 842912461
14 | -1 1 15 209 2911 40545 564719 7865521 109552575 1525870529
15 | -1 1 16 239 3569 53296 795871 11884769 177475664 2650250191
16 | -1 1 17 271 4319 68833 1097009 17483311 278635967 4440692161
17 | -1 1 18 305 5167 87534 1482911 25121953 425590290 7209912977
18 | -1 1 19 341 6119 109801 1970299 35355581 634430159 11384387281
19 | -1 1 20 379 7181 136060 2577959 48845161 925480100 17535276739
20 | -1 1 21 419 8359 166761 3326861 66370459 1324082319 26415275921
- + - - - - - - - - - -
The following table shows the factorizations; they seem to be cyclic not only in the rows but also in the columns. The columns are in a way multiplicative: the value at some entry can only be prime if the column-index $m$ is prime; for composite column-indexes $m$ the entry has all the primefactors from the other columns whose $m$-indexes are divisors of the "current" $m$ and seemingly one more - just like in the theorem of Szygmondi about the compositions of the Mersenne-numbers. (Note: the minus signs have mostly vanished, but seem irrelevant here)
x | 0 1 2 3 4 5 6 7 8 9
m | -1 1 3 5 7 9 11 13 15 17
- + - - - - - - - - - -
-15 | -1 1 2.7 11.19 3121 2.7.3329 307.2267 53.157.1249 2.7.11.19.29.31.59 17.136328033
-14 | -1 1 13 181 2521 13.37.73 489061 6811741 13.61.181.661 1321442641
-13 | -1 1 2^2.3 5.31 2003 2^2.3^2.719 23.14543 4322473 2^2.3.5.31.59.509 137.5268811
-12 | -1 1 11 131 7.223 11.19.89 23^2.419 157.16823 11.131.21841 375033361
-11 | -1 1 2.5 109 29.41 2.5.1297 141481 13.118717 2.5^2.109.3089 183642229
-10 | -1 1 3^2 89 881 3^3.17.19 131.659 854569 3^2.59.89.179 4691.17851
-9 | -1 1 2^3 71 631 2^3.701 11.23.197 442961 2^3.29.71.239 34988311
-8 | -1 1 7 5.11 433 7.487 26839 131.1613 5.7.11.29.149 101.103.1259
-7 | -1 1 2.3 41 281 2.3^2.107 43.307 90481 2.3.41.2521 67.63443
-6 | -1 1 5 29 13^2 5.197 5741 33461 5^2.29.269 137.8297
-5 | -1 1 2^2 19 7.13 2^2.109 2089 10009 2^2.19.631 229771
-4 | -1 1 3 11 41 3^2.17 571 2131 3.11.241 67.443
-3 | -1 1 -2 5 13 2.17 89 233 2.5.61 1597
-2 | -1 1 -1 1 -1 1 -1 1 -1 1
-1 | -1 1 0 -1 1 0 -1 1 0 -1
0 | -1 1 1 -1 -1 1 1 -1 -1 1
1 | -1 1 2 1 -1 -2 -1 1 2 1
2 | -1 1 3 5 7 3^2 11 13 3.5 17
3 | -1 1 2^2 11 29 2^2.19 199 521 2^2.11.31 3571
4 | -1 1 5 19 71 5.53 23.43 3691 5^2.19.29 101.509
5 | -1 1 2.3 29 139 2.3^2.37 3191 15289 2.3.29.421 350981
6 | -1 1 7 41 239 7.199 23.353 79.599 7.31^2.41 103.15607
7 | -1 1 2^3 5.11 13.29 2^3.17.19 89.199 233.521 2^3.5.11.31.61 1597.3571
8 | -1 1 3^2 71 13.43 3^3.163 34649 53.5147 3^2.71.3361 16908641
9 | -1 1 2.5 89 7.113 2.5.19.37 43.1453 53.10477 2.5^2.89.1109 307.142867
10 | -1 1 11 109 13.83 11.971 23.4597 937.1117 11.109.8641 373.274957
11 | -1 1 2^2.3 131 1429 2^2.3^2.433 23.7393 53.79.443 2^2.3.61.131.211 1531.144161
12 | -1 1 13 5.31 1847 13.1693 262261 103.30341 5.13.31.18481 1973.224909
13 | -1 1 2.7 181 2339 2.7.17.127 11.35509 53.131.727 2.7.181.25741 67.12580783
14 | -1 1 3.5 11.19 41.71 3^2.5.17.53 23.43.571 2131.3691 3.5^2.11.19.29.241 67.101.443.509
15 | -1 1 2^4 239 43.83 2^4.3331 795871 13.914213 2^4.239.46411 67.271.145963
16 | -1 1 17 271 7.617 17.4049 1097009 17483311 17.31.271.1951 4440692161
17 | -1 1 2.3^2 5.61 5167 2.3^3.1621 67.22133 25121953 2.3^2.5.61.77521 101.8431.8467
18 | -1 1 19 11.31 29.211 19.5779 199.9901 79.521.859 11.19.31.181.541 919.3469.3571
19 | -1 1 2^2.5 379 43.167 2^2.5.6803 67.109.353 1091.44771 2^2.5^2.379.24419 17.647.1594261
20 | -1 1 3.7 419 13.643 3^2.7.2647 3326861 66370459 3.7.29.419.5189 137.2551.75583
- + - - - - - - - - - -
As chubakuneo comments, if you want to prove that $p$ never divides any $f(m)$ where $f \in \Bbb Z[X]$, it's enough to check that $f$ has no roots modulo $p$, i.e. that $f(0),f(1),f(2),\ldots f(p-1)$ are not multiples of $p$. So a straightforward computation can confirm your claims.
If you consider $m$ as an indeterminate $X$ over $\Bbb Z$, you have a sequence $-1,1,X+1,X^2+X-1,\ldots$ of polynomials $f_n(X)$ satisfying the recurrent relation $f_{n+1} = Xf_n - f_{n-1}$ and the initialisation $f_0 = -1, f_1 = 1$
It turns out that for $n\ge 2$, the roots of $f_n(X)$ are the $2\cos(\frac{2k\pi}{2n-1})$ for $1 \le k \le n-1$, so that $(X-2)f_n^2 = 2T_{2n-1}(X/2) -2$ where $T_n$ is the Chebyshev polynomial. You can probably prove this equality by induction using the recurrence relations for Chebyshev polynomials.
Now that we know the roots of $f_n$, we also know the Galois group of $f_n$ : it is isomorphic to $G_n = (\Bbb Z/(2n-1)\Bbb Z)^*/\{\pm 1\}$.
Furthermore, the map $\rho : R_n = (\Bbb Z/(2n-1)\Bbb Z \setminus \{0\})/\{\pm 1\} \to \text{the roots of } f_n$ given by $\rho_{\pm k} = 2\cos(2k\pi/(2n-1))$ is compatible with the usual action of $G_n$ on $R_n$, as $\sigma_{\pm u}(\rho_{\pm k}) = \rho_{\pm uk}$.
In the following, I will always assume that $p$ is a prime that doesn't divide $2n-1$ (the discriminant of $f_n$), so that $p$ doesn't ramify.
Then, you can count the number of roots of $f_n$ modulo $p$ by counting the number of fixpoints of the Artin symbol $[p]_n \in G_n$ (when looking at its action on $R_n$). In particular, to see what primes don't appear in the factorizations of $f_n(m)$, we need to point out what are the elements of $G_n$ without fixpoints on $R_n$.
If $2n-1 = \prod_{i=1}^d p_i^{k_i}$, then the roots $\rho_{\pm(2n-1)/p_i}$ are fixed by $\sigma_{\pm u}$ where $u \equiv \pm 1 \pmod {p_i}$, so there are $2^{d-1}$ elements of the Galois group having fixpoints (conversely we check that any automorphism fixing someone fixes one of those roots).
Hence the density of automorphisms having a fixpoint is $1 - \prod_{p\mid 2n-1} (1-2/(p-1))$.
What's more, Artin's reciprocity law tells you that $[p]_n = \sigma_{\pm p \pmod {2n-1}}$. Together with Dirichlet's theorem, this tells you exactly the density of what primes that can appear how often in the factorisations of $f_n(m)$.
In particular the density of primes appearing in the factorisations of $f_n(m)$ is $1 - \prod_{p\mid 2n-1}(1-2/(p-1))$ because they are exactly the primes congruent to $\pm 1$ modulo some prime divisor of $2n-1$.
$f_2 = X+1$. It has trivial Galois group, $R_2 = \{1,2\} = G_2$, so the only element of $G_2$ fixes $R_2$ : $f_2$ has one root mod $p$ forall $p$.
$f_3 = X^2 + X-1$. $R_2 = G_2 = \{\{1,4\},\{2,3\}\}$. If $p \equiv 2,3 \pmod 5$ then it never divides $f_3(m)$, while if $p \equiv 1,4 \pmod 5$ then $f_3$ has 2 roots mod $p$.
$f_4 = X^3+X^2-2X-1$. As in the above, $2n+1 = 7$ is prime, so only the primes $p \equiv 1,6 \pmod 7$ have 3 roots mod $p$, all the others can't appear.
$f_5 = X^4+X^3-3X^2-2X+1$. Now, $2n+1 = 9 = 3^2$. If $p \equiv 1,8 \pmod 9$ then it has 4 roots mod $p$. Otherwise it has 1 roots mod $p$. So every prime appears, but primes like $17,19,37$ especially so.
$f_6 = X^5+X^4-4X^3-3X^2+3X+1$. Again, $2n+1 = 11$ is prime so the only primes that can appear are those congruent to $1,10 \pmod {11}$ and then they have 5 roots.
In general, if $2n+1$ is a prime $q$, then $f_n$ has no root mod $p$ except when $p \equiv \pm 1 \pmod q$, where it splits completely. So since $1,q-1$ and $q+1$ are not primes, the smallest prime that can possibly appear is $2q-1$.