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вернуться в форумa math solution the answer is the coefficient of y^n of A(n)=(1+y+y^2+y^3+...)(1+y^2+y^4+y^6+...) ...(1+y^n+y^2n+y^3n...) if y<1; A(n)=1/((1-y)(1-y^2)(1-y^3)....) Re: a math solution Can you please explain the formula? It is an infinite formula, right? So, How would I apply it? Please give an example Re: a math solution Послано DEAL 29 мар 2011 00:36 It's easy to apply this formula, first of all you should notice that only n brackets must be opened, then you simply count all coefficients, multiplying only terms with the power that is less than n (or equal). (there is also one way to multiply less terms) And of course, you eventually get an answer. < Edited by author 29.03.2011 00:37 Re: a math solution I think it's wrong. The correct answear is the coefficient of A(n) = (1+y)(1+y^2)(1+y^3)...(1+y^n)... less 1, because the staircase with one step must not be counted. |
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