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Polynom approach for the row 4, 4, 7, X, 103F(n) = a + b*n + c*n^{2} + d*n^{3} + e*n^{4} Applied accordingly it results: F(0) = a + b*0 + c*0^{2} + d*0^{3} + e*0^{4} = a = 4 F(1) = a + b*1 + c*1^{2} + d*1^{3} + e*1^{4} = a + b + c + d + e = 4 F(2) = a + b*2 + c*2^{2} + d*2^{3} + e*2^{4} = a + 2b + 4c + 8d + 16e = 7 F(3) = a + b*3 + c*3^{2} + d*3^{3} + e*3^{4} = a + 3b + 9c + 27d + 81e = X F(4) = a + b*4 + c*4^{2} + d*4^{3} + e*4^{4} = 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 wholenumbered X you can imagine. Amazingly, with this it seems that all continuations after the 103 are wholenumbered, too. back to the Mensapage In case of questions or suggestions just write an Email to me. 