Alternate approach for the row 4, 4, 7, X, 103

F(2n+1) = a*F(2n) + b and F(2n) = c*F(2n-1) + d with F(0) = 4

Applied accordingly it results:
F(1) = a*F(0) + b = 4a + b = 4
F(2) = c*F(1) + d = 4c + d = 7
F(3) = a*F(2) + b = 7a + b = X
F(4) = c*F(3) + d = Xc + d = 103

Solving this system of equations delivers the parameters depending on X:
a = (X - 4) / 3
b = (-4X + 28) / 3
c = 96 / (X - 4)
d = (7X - 412) / (X - 4)

So you would get a general solution for each whole-numbered X with whole-numbered continuations after the 103. Interestingly, for even n the sequence does not depend on X. Altogether, in six cases there also exist whole-numbered parameters:

					n
X	a	b	c	d	5	6	7	8	9	10
7	1	0	32	-121	103	3175	3175	101479	101479	3247207
10	2	-4	16	-57	202	3175	6346	101479	202954	3247207
16	4	-12	8	-25	400	3175	12688	101479	405904	3247207
28	8	-24	4	-9	796	3175	25372	101479	811804	3247207
52	16	-60	2	-1	1588	3175	50740	101479	1623604	3247207
100	32	-124	1	3	3172	3175	101476	101479	3247204	3247207

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