Q:

Is there any pseudoprime that pass this test above tested range, or any prime that does not show these ending patterns?

if the recurrence sequence is defined by the following foormula, $d_{n + 3} = 3d_{n + 2} - d_{n + 1} - 2d_n$ where $d_1 = 1, d_2 = 3$ and $ d_3 = 7$, this produce the  following complex sequence $$1, 3, 7, 16, 35, 75, 158, 329, 679, 1392, ...$$.
if this sequence is evaluated over finite field, i mean modulo some interger, $p$
$$(d_{n + 3} = 3d_{n + 2} - d_{n + 1} - 2d_n) \text{mod}\,p$$ It follows that, if $p$ is prime number, the last three term if evaluated up to $n = p + 2$, last three terms i mean $d_{p}, d_{p + 1}$ and $d_{p + 2}$ fall in one of the following ending sequence $\{...,1, 3, 7\}$ or $\{...,4, 7, 14\}$, if $p$ is smaller than these ending patterns, then modulo by $p$. the only exception for prime numbers less than $7$
i have tested all numbers up to $100,000$, no pseudo prime exist
Prior Studied Sequences
i studied Lucas numbers for existence of pattern like this, but there existed many pseudo prime below $10,000$ e.g $377$. then i studied recussive sequence defined by the following formula $d_{n+3} = d_{n + 2} + 2d_{n + 1} - d_n$. in this sequence pseudo prime exist above $10^6$, may be this sequence can pull up this boundary
Questions

is there any pseudo prime above tested range of this sequence
can anyone explain to me further why this patterns appear to prime only, if mo pseudoprime pass this test

UPDATES ONE DAY AFTER POST
This update i made after reviewing users comments and answers, thanks to @ Alekseyev for discovering pseudo primes and all other contributions
Facts About Pseudo Primes Of This Recurence

all pseudo primes are divisible by primes. $p_1, p_2,..., p_n$, all of these primes are in the form of $10k + 1$. My limitation is computational power, i can not test big numbers
all prime factors of these pseudo primes have period of $p - 1$. prime numbers of the form $10k - 1$ also have periods of the form $p -1$. no counter example for other primes having this form of period. if any found?
if a composite number $n$ have all of its prime factors of this form, $p = 10k + 1$. the resulting period of $n$ get reduced by factor of $10$ and sometimes more than that. the gcd of resultant period $r$ and $n - 1$ is equal to $r$. this explain why $nth$ period occurs at $p - 1$ and the number get evaluated as prime number while not

New Update
as we go to the right of number line toward big numbers, things start to deviate from my expectation and all the statement above seem to be invalid, i have found one pseudo prime whose last digit is not $1$, $6368689$ and some prime factors of pseudo primes are not of form $10k + 1$, $16778881$ 
here is a list of all pseudo prime up to $20 \times 10^6$.
$219781,252601,399001,512461,722261,741751,852841,1024651,1193221,1533601,1690501,1735841,1857241,1909001, 2100901, 2165801,
        2603381, 2704801, 2757241, 3568661, 4504501, 6368689, 6840001, 8341201, 8646121, 8719921, 9863461, 10386241, 10403641, 
        10837321, 12261061, 14609401, 14671801, 16778881$

A:

The sequence of pseudoprimes here starts with:
$$219781, 252601, 399001, 512461, 722261, 741751, 852841, 1024651, 1193221, 1533601, 1690501, 1735841, 1857241, 1909001, \dots$$
UPDATE. Efficient testing of a given positive integer $n$ can be done using the formula:
$$
\begin{bmatrix} d_n\\ d_{n+1}\\ d_{n+2}\end{bmatrix} =
\begin{bmatrix} 
0 & 1 & 0\\
0 & 0 & 1\\
-2& -1& 3
\end{bmatrix}^{~~n-1}
\cdot
\begin{bmatrix} 1\\ 3\\ 7\end{bmatrix},$$
where computations are performed modulo $n$ (in particular, using modular exponentiation of the matrix by squaring).
Here is my PARI/GP code for the test:
test(n) = my(t); t=([0, 1, 0; 0, 0, 1; -2, -1, 3]*Mod(1,n))^(n-1)*[1,3,7]~; t==[4,7,14]~ || t==[1,3,7]~;

