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

F(n) = a + b*n + c*n² + d*n³ + e*n⁴

Applied accordingly it results:
F(0) = a + b*0 + c*0² + d*0³ + e*0⁴ = a = 4
F(1) = a + b*1 + c*1² + d*1³ + e*1⁴ = a + b + c + d + e = 4
F(2) = a + b*2 + c*2² + d*2³ + e*2⁴ = a + 2b + 4c + 8d + 16e = 7
F(3) = a + b*3 + c*3² + d*3³ + e*3⁴ = a + 3b + 9c + 27d + 81e = X
F(4) = a + b*4 + c*4² + d*4³ + e*4⁴ = a + 4b + 16c + 64d + 256e = 103

Solving this system of equations delivers the parameters depending on X:
a = 4
b = (16X - 469) / 12
c = (-56X + 1655) / 24
d = (14X - 425) / 12
e = (-4X + 133) / 24

So you would get a general solution for each whole-numbered X you can imagine. Amazingly, with this it seems that all continuations after the 103 are whole-numbered, too.

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