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2015-04-25T07:31:39+05:30

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f(1)=1
f(2)=2
f(3)=2=2*1
f(4)=4=2*2
f(5)=8=4*2
f(6)=32=8*4
\cdots

for n>2, the nth term can be obtained by multiplying previous two terms:
f(1)=1,f(2)=2\\f(n)=f(n-1)*f(n-2)

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2015-04-25T07:51:36+05:30

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Try taking the ratio of successive terms  and difference of successive terms when to find a recursive relation for a function.  Or, try taking the difference of n th term and n-2 nd term  or their ratio.  Check a few combinations. 

Then see the values of the sequence obtained, if that matches with f(n), f(n-1), f(n-2) or their multiples.  Some sequence will match.

n        =          1         2        3        4          5          6
f(n)     =          1         2        2        4          8          32
f(n-1)  =          -          1        2        2          4          8            32
f(n-2)  =          -          -        1        2           2          4            8          32
2 f (n-1) =        -          2       4        4          8          16

f(n)-f(n-1) =       -        1        0        2          4          24
f(n)-2 f(n-1) =    -        0       -2        0          0          16
f(n) - f(n-2) =    -        -        1          2          6          28

f(n) / f(n-1)  =    -        2       1        2          2            4
f(n) / f(n-2) =     -        -         2       2           4            8
f(n) / f(n-3) =    -        -        -          4          4           16


It seems to be that only the data items highlighted are matching. 

Now infer the relationship.  f(n) / f(n-1)  = f(n-2
       f(n) = f(n-1) * f(n-2) ,        for  3 <=  n  <=  6
OR,       f(n+2) =  f(n+1) * f(n)    ,  for    1 <=  n <=  4

If we want the relation to be defined for the first two data items also,  then we may define
f(-1) = 1/2  and  f(0) = 2,  in that case,

       f (n) =  f (n - 1) f (n - 2),    for    1  <=  n  <= 6

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