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1. Fibonacci-like approach for the row 4, 4, 7, X, 103F(n) = a*(F(n-2) + F(n-1)) + b + c*n with F(0) = 4 and F(1) = 4 Applied accordingly it results: F(2) = a*(F(0) + F(1)) + b + 2c = 8a + b + 2c = 7 F(3) = a*(F(1) + F(2)) + b + 3c = 11a + b 3c = X F(4) = a*(F(2) + F(3)) + b + 4c = (7 + X)a + b + 4c = 103 Solving this system of equations delivers the parameters depending on X: a = (-2X + 110) / (X - 7) b = (-2X2 + 39X - 367) / (X - 7) c = (X2 - 8X - 281) / (X - 7) Anyhow, so you get solutions for eleven whole-numbered X (X = 55 doesn't make any sense, since then due to a=0 there is no link between F(0) and F(1).) with whole-numbered continuations after the 103:
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