 it works in ten special cases we can trust the scheme suggested in math books.)
From here, $f$ satisfies a formula of the shape
$$
f(n)=a2^n+b4^n+c8^n+d16^n\,
$$
and the constants $a,b,c,d$ have to be determined from the "first four conditions". (So we have to compute also $f(2)$, $f(3)$ from the given recursion.) In our case we have the coefficients $a,b,c,d$, (assamblated and in a factorized expression, that we do not need to expand, since) we need to verify the formula only for $f(0)$, $f(1)$, $f(2)$, $f(3)$.    

Later EDIT: Let us find the constants $a,b,c,d$ explicitly. To simplify typing and computing, i will use a computer algebra system, here sage.
Let us define $f$ recursively and compute its first values in sage.
sage: def f(n): 
....:     if n == 0:    return 0 
....:     if n == 1:    return 0 
....:     return 6*f(n-1) - 8*f(n-2) + 7*2^(4*n-8) - 3*2^(3*n-5) 
....:
sage: for n in [0..20]:
....:     print("f(%2s) = %s" % (n, f(n)))
f( 0) = 0
f( 1) = 0
f( 2) = 1
f( 3) = 70
f( 4) = 1820
f( 5) = 35960
f( 6) = 635376
f( 7) = 10668000
f( 8) = 174792640
f( 9) = 2829877120
f(10) = 45545029376
f(11) = 730862190080
f(12) = 11710951848960
f(13) = 187512576276480
f(14) = 3001300362981376
f(15) = 48029600424386560
f(16) = 768543969628897280
it works in ten special cases we can trust the schema suggest in math books .) 
 From here, $ f$ satisfies a formula of the form 
 $ $ 
 f(n)=a2^n+b4^n+c8^n+d16^n\, 
 $ $ 
 and the constants $ a, b, c, d$ have to be specify from the " beginning four conditions ". (thus we have to compute besides $ f(2)$, $ f(3)$ from the given recursion .) In our case we have the coefficients $ a, b, c, d$, (assamblated and in a factorize expression, that we do not need to elaborate, since) we need to verify the recipe only for $ f(0)$, $ f(1)$, $ f(2)$, $ f(3)$.    

 Later EDIT: Let us line up the constants $ a, b, c, d$ explicitly. To simplify typing and computing, i will use a calculator algebra system, here sage. 
 get us define $ f$ recursively and compute its first value in sage. 
 sage: def f(n ): 
.... :      if n = = 0:     return 0 
.... :      if n = = 1:     return 0 
.... :      return 6*f(n-1) - 8*f(n-2) + 7 * 2^(4*n-8) - 3 * 2^(3*n-5) 
.... : 
 sage: for n in [ 0.. 20 ]: 
.... :      print("f(%2s) =% s "% (n, f(n) )) 
 f (0) = 0 
 f (1) = 0 
 f (2) = 1 
 f (3) = 70 
 f (4) = 1820 
 f (5) = 35960 
 f (6) = 635376 
 f (7) = 10668000 
 f (8) = 174792640 
 f (9) = 2829877120 
 f(10) = 45545029376 
 f(11) = 730862190080 
 f(12) = 11710951848960 
 f(13) = 187512576276480 
 f(14) = 3001300362981376 
 f(15) = 48029600424386560 
 f(16) = 768543969628897280
 it works in ten special cares we can trusj rhe scieme sufgested kn math books.)
From here, $f$ satmsfiws a dormula of the shape
$$
f(n)=x2^n+b4^n+c8^n+d16^n\,
$$
ajd the cinstents $a,b,c,d$ have to be detcxmines frok the "first fout conditions". (So we have to cum'ute also $f(2)$, $f(3)$ from the given recursijn.) In oir case we have jhe cpqffidpekts $a,b,c,d$, (assamblated and in a fadtorizev expression, thst we do not need to expanf, sijce) we need to verlfy the fornula inly for $f(0)$, $f(1)$, $f(2)$, $f(3)$.    

Lauex EDIT: Let os find the constants $a,b,c,d$ explizitly. To simplidy tyohng and com'uting, i will use a computar algenra system, herc sagx.
Let us define $f$ recursivxly and compute its sirst valgea in sage.
sage: def f(b): 
....:     if n == 0:    fethri 0 
....:     iv n == 1:    retudn 0 
....:     rerurn 6*f(n-1) - 8*f(n-2) + 7*2^(4*n-8) - 3*2^(3*n-5) 
....:
ssgq: for n in [0..20]:
....:     prine("f(%2f) = %s" % (n, f(n)))
f( 0) = 0
f( 1) = 0
f( 2) = 1
f( 3) = 70
f( 4) = 1820
f( 5) = 35960
f( 6) = 635376
f( 7) = 10668000
r( 8) = 174792640
f( 9) = 2829877120
f(10) = 45545029376
f(11) = 730862190080
f(12) = 11710951848960
f(13) = 187512576276480
f(14) = 3001300362981376
d(15) = 48029600424386560
f(16) = 768543969628897280
it works in ten special cases we the suggested in books.) From here, the $$ f(n)=a2^n+b4^n+c8^n+d16^n\, $$ the constants $a,b,c,d$ to be determined from the "first conditions". (So we have to compute also $f(2)$, $f(3)$ from the given recursion.) our case we have the coefficients $a,b,c,d$, (assamblated and in a factorized expression, we not to since) we need to verify the formula only for $f(0)$, $f(1)$, $f(2)$, $f(3)$. Later EDIT: Let find the constants $a,b,c,d$ explicitly. To simplify typing computing, i will use computer algebra system, here sage. us $f$ recursively compute first in sage. sage: f(n): ....: if n == 0: return 0 ....: if n == 1: return 0 ....: return - 8*f(n-2) - 3*2^(3*n-5) sage: n [0..20]: ....: print("f(%2s) % (n, f(n))) f( 0) = = 0 f( 2) = 1 f( 3) 70 f( = 1820 f( 5) = 35960 6) = 635376 f( 7) = 10668000 f( = 174792640 f( 9) = 2829877120 f(10) = 45545029376 f(11) = 730862190080 f(12) = 11710951848960 187512576276480 f(14) = 3001300362981376 = 48029600424386560 f(16) 768543969628897280
 it works in ten special cases wE can trust tHe schEme SugGeSted In maTh books.)
From herE, $F$ satIsfies a formula of the shaPe
$$
f(n)=a2^N+b4^N+C8^n+d16^n\,
$$
ANd The coNstants $A,B,c,D$ HAve To Be DetErMInEd froM thE "first fOur conditiOns". (so We have to compUTe Also $f(2)$, $f(3)$ from The Given recursiOn.) IN our caSe We hAVe the CoeFficiEnts $a,b,C,D$, (assamBlated and In A FactorIZed exprESSiOn, thAt we do not need to exPAnD, Since) we need to vErify tHe FOrMULa oNly For $f(0)$, $f(1)$, $f(2)$, $f(3)$.    

LatEr eDIT: LET us find THe CONStaNTs $a,b,c,d$ expliciTly. To simpliFY tyPing anD cOmpUTing, i wIll usE a COmpUter algebra SystEm, here sagE.
Let us DEfine $f$ rECursiveLy and cOmpUte Its fIRsT vAluEs IN saGE.
sAge: DEf f(N): 
....:     if n == 0:    retuRn 0 
....:     If N == 1:    retuRn 0 
....:     reTURN 6*F(n-1) - 8*f(n-2) + 7*2^(4*N-8) - 3*2^(3*n-5) 
....:
sAge: fOr n in [0..20]:
....:     Print("f(%2s) = %s" % (n, f(n)))
f( 0) = 0
f( 1) = 0
F( 2) = 1
f( 3) = 70
f( 4) = 1820
F( 5) = 35960
f( 6) = 635376
f( 7) = 10668000
f( 8) = 174792640
F( 9) = 2829877120
F(10) = 45545029376
f(11) = 730862190080
f(12) = 11710951848960
F(13) = 187512576276480
f(14) = 3001300362981376
f(15) = 48029600424386560
f(16) = 768543969628897280
 it works in ten special c ases we ca n tru stthe  s chem e su ggested in mat h  boo ks.)
From here, $f$ sa tisfi es a fo r mu la of  the sh a pe $ $
f (n )= a2^ n+ b 4^ n+c8^ n+d 16^n\,$$
and the  co ns tants $a,b,c , d$  have to b e d etermined fr omthe "f ir stf our c ond ition s". (S o  we ha ve to com pu t e also $f(2)$, $ f( 3)$from the given re c ur s ion.) In our c ase we  h a ve t hecoe fficients$a ,b,c, d $, (ass a mb l a t eda nd in a facto rized expre s sio n, tha twed o notneedto exp and, since)  weneed to v erifyt he form u la only  for $ f(0 )$,  $f( 1 )$ ,$f( 2) $ , $ f (3 )$.   

LaterED IT : Let  usf i n d  the  co nsta nts $ a,b,c,d$ expl ici tly. Tosimpl ify t ypin gand c omputi ng, i  w ill use a compu teralgebra s yst em , h er e sag e .
Letusdef ine $f$  recurs i vel ya n d  c ompute its first v al u e sin sage. 
sage: de ff (n): 
.. .. :    i f n ==0:  r eturn 0
....:   if n ==  1 :    r et urn  0
.... :     retur n 6*f(n- 1) -8 *f(n-2) + 7*2^ ( 4*n-8) - 3*2^ ( 3* n - 5) 
... .:sage: for n  in[ 0..2 0]:. .. .:   pr int(" f( % 2s )  = %s" % (n, f(n))) 
f ( 0) =  0
f(  1) = 0
f( 2)  = 1
f( 3) = 70
f( 4)  = 1 8 20 f( 5) = 35960f( 6)  = 635376f ( 7) = 1 06680 00
f( 8)  = 174792 6 4 0
f( 9)= 2 829 877 120 f (1 0) = 45545029 3 7 6
f( 11 ) = 730 862 190080f(1 2)= 1 171 09 51848960f(13) =18 75 12 57 627 6480f (14) = 3 00 130 03 629 81376 f(15)= 480 2960 04 24 3 865 60
f(16 )  = 7 6854 39 69 6288 972 80 
 it_works in_ten special cases we_can trust_the_scheme suggested_in_math books.)
From here,_$f$ satisfies a_formula of the shape
$$
f(n)=a2^n+b4^n+c8^n+d16^n\,
$$
and_the constants $a,b,c,d$_have_to be determined from the "first four conditions". (So we have to compute also_$f(2)$,_$f(3)$ from_the_given_recursion.) In our case we_have the coefficients $a,b,c,d$, (assamblated_and in_a factorized expression, that we do not need_to_expand, since) we_need to verify the formula only for $f(0)$, $f(1)$,_$f(2)$, $f(3)$.    

Later_EDIT: Let us_find_the_constants $a,b,c,d$ explicitly. To_simplify typing and computing, i will_use a computer algebra system, here_sage.
Let us define $f$ recursively and compute_its first values in sage.
sage: def_f(n): 
....:   _ if_n == 0:  _ return 0_
....: _  _if n == 1:  _ return 0_
....:     return_6*f(n-1)_- 8*f(n-2) +_7*2^(4*n-8)_-_3*2^(3*n-5) 
....:
sage:_for n in_[0..20]:
....:_ __ print("f(%2s) = %s" % (n,_f(n)))
f(_0) = 0
f( 1) = 0
f( 2)_= 1
f( 3) =_70
f(_4) = 1820
f( 5)_= 35960
f( 6) = 635376
f(_7) = 10668000
f( 8) = 174792640
f(_9) =_2829877120
f(10) =_45545029376
f(11) = 730862190080
f(12) = 11710951848960
f(13) = 187512576276480
f(14) = 3001300362981376
f(15) = 48029600424386560
f(16)_= 768543969628897280