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# 1. Fibonacci-like approach for the row 4, 4, 7, X, 103

F(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:
 n X a b c 5 6 7 8 9 10 8 94 -183 -281 8846 839337 79727052 7573238135 719378724866 68333484519101 9 46 -89 -136 4383 205451 9651323 453410427 21300839187 1000695480795 10 30 -59 -87 2896 89389 2767882 85717375 2654556868 82208226361 11 22 -45 -62 2153 49215 1129617 25933763 595393757 13669204775 13 14 -33 -36 1411 20947 312727 4671115 69773431 1042223251 15 10 -29 -22 1041 11279 123017 1342755 14657493 160002231 19 6 -29 -6 673 4591 31513 216547 1488277 10228855 23 4 -33 4 491 2367 11427 55175 266411 1286351 31 2 -45 18 313 895 2497 6883 18877 51655 39 1 -59 29 228 446 818 1437 2457 4125 55 0 -89 48 151 199 247 295 343 391 103 -1 -183 99 106 202 202 205 301 301

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