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OMCB015

OMCB015(H)

ユーザー解説 by 2_3_5_7

n=137v2(n)v3(n)=n=136(v2(3n2)v3(3n2)+v2(3n1)v3(3n1)+v2(3n)v3(3n))=n=136(0+0+v2(n)(v3(n)+1))=n=136v2(n)v3(n)+n=136v2(n)\begin{aligned} \sum_{n=1}^{3^7} v_2(n) v_3(n) &= \sum_{n=1}^{3^6} (v_2(3n-2) v_3(3n-2) + v_2(3n-1)v_3(3n-1) + v_2(3n) v_3(3n))\\ &= \sum_{n=1}^{3^6} (0 + 0 + v_2(n) (v_3(n) + 1))\\ &= \sum_{n=1}^{3^6} v_2(n) v_3(n) + \sum_{n =1}^{3^6} v_2(n) \end{aligned} より n=36+137v2(n)v3(n)=n=136v2(n)=k=1(362k362k+1)×k=k=1362k=723\begin{aligned} \sum_{n= 3^6 + 1}^{3^7} v_2(n) v_3(n) &= \sum_{n=1}^{3^6} v_2(n)\\ &= \sum_{k=1}^{\infty} \left( \left\lfloor \frac{3^6}{2^k} \right\rfloor - \left\lfloor \frac{3^6}{2^{k+1}} \right\rfloor \right) \times k\\ &= \sum_{k=1}^{\infty} \left\lfloor \frac{3^6}{2^k} \right\rfloor\\ &= \bf{723} \end{aligned} となる.