Seems you could use the first two numbers to determine you have essentially a cube: (37^(1/3)+1)^3 = 81.3, which is pretty close to 83. Next two boxes are predicted 152 and 254, actual 157 and 265. Interesting that the error climbs to 5% then falls as the box gets more and more cube-like. 2% 81, 83 4% 152, 157 4% 254, 265 5% 394, 413 5% 578, 607 5% 813, 853 5% 1103, 1157 5% 1455, 1525 5% 1876, 1963 5% 2370, 2477 4% 2944, 3073 On Thu, Apr 7, 2011 at 8:29 PM, Simon Plouffe <simon.plouffe@gmail.com>wrote:
Hello,
this is the finite differences of Newton,
n^3 = (1,7,12,6) n^2 = (1,3,2) n = (1,1) 1 = (1)
then if we have (37,46,28,6) as coefficients of the differences, a linear combination computation gives ; n^3 + 8*n^2 + 15*n + 13
37, 83, 157, 265, 413, 607, 853, 1157, 1525, 1963, 2477, 3073, ...
the coefficients of n^k are related to Stirling numbers.
ref : The book of Jordan < calculus of finite differences>.
http://books.google.com/books?id=3RfZOsDAyQsC&printsec=frontcover&dq=finite+...
best regards,
Simon Plouffe
Le 08/avr./2011 05:09, Joshua Zucker a écrit :
My method (use fixed-width font for viewing):
37 83 157 ? 46 74 28 6
Has to be a 6 down there, to make the leading coefficient 1.
Thus:
37 83 157 265 46 74 108 28 34 6
Answer: 265.
--Joshua _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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