PROBLEM LINK:Contest DIFFICULTY:HardPREREQUISITES:Math, Combinatorics, Pascal TrianglesEXPLANATION:The sequence in the question is the generalisation of Moessner sequencce. In Moessner sequence $d_1=d_2=........d_{m1}=d_{m}.$
Similarly for Second group add a new row of 1’s and a column containing previously marked numbers, Each triangle is obtained from the previous by taking the homogeneous component of degree $diff_k$, and multiplying by $∆(x,y)$ and then putting $y=1$ in the obtained triangle. $h_{k+1}(x,y)=[h_k(x,1)*∆(x,y)]_{d_k}$ and $h_0(x,1)=1$ where $d_k$ is the size of group $k$ and $∆(x,y) =\frac{1}{1(x+y)}=(x+y)^0+(x+y)^1+(x+y)^2+...=\sum_{d=0}^{\infty}(x+y)^d$ (Note:$∆(x,y)$ is the algebraic representation of pascal triangle. which simplifies to $h_k(x,1) = \prod_{i=0}^{k1}((ki)*x+1)^{diff_i})$ where $diff_i=d_id_{i1}$ and $d_0=0$ Now what we require is the sum of all marked numbers in the original group which is:$h_k(x,1)1$ (subtracting 1 as we added the rows of 1’s ,so 1 needed to be subtracted) Final ans: $\sum_{k=1}^{m}(h_k(x,1)1)$ Eg: TIME COMPLEXITY:$O(m*m*log(max(diff[i])))$ SOLUTIONS:
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asked 05 Apr '18, 18:52

For some reason, the note is being rendered incorrectly despite showing correctly in the editing tab, the actual equation is answered 05 Apr '18, 19:08
