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# 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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