let a(1)=x, a(2)=y, a(3)=z

then

a(6)=z^2*(x+y)(x+y+1)(y+z)

144=2^4*3^2

z^2(x+y)(x+y+1)(y+z)=144 (1)

divisors:

1,2,3,4,6,8,9,12,16,18,24,36,48,72,144

(x+y) and (x+y+1) are consecutive integers

only pairs of divisors which are consecutive integers is

(1,2) (2,3) (3,4) and (8,9)

case 1:

x+y=1

impossible since x and y are each positive

case 2:

x+y=2

thus x=1 and y=1 leaving (1) to be

z^2*1*2*(z+1)=144

z^2(z+1)=72

which has no integer solutions

case 3:

x+y=3

so either x=1,y=2 or x=2,y=1

subcase 1:

x=1,y=2

giving (1) to be

z^2*3*4*(z+2)=144

again, no integer solutions leaving

subcase 2:

x=2,y=1

giving (1) to be

z^2*3*4*(z+1)=144

only integer solution is z=2

thus x=2,y=1,z=2 which gives a(7)=3456

case 4:

x+y=8

so this gives us

8*9*z^2*(z+y)=144

z^2*(z+y)=2

thus z=1 leaving

y+1=2

y=1

which gives x=7

thus we have x=7,y=1,z=1 which also gives a(7)=3456

thus a(7)=3456

resulting from either starting values (2,1,2) or (7,1,1)

*edited to include the x+y=8 case I originally missed*