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