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本帖最后由 王守恩 于 2021-8-14 19:21 编辑
在 1~n 中任意取10 个不同整数,使得这 10 个数之和能被 10 整除,有几种不同取法?
\(a(n)=\frac{n!/10}{10!(n-10)!}-(\frac{625\lfloor n/10\rfloor^5-125\lfloor n/10\rfloor^3-192\lfloor n/10\rfloor^2+4\lfloor n/10\rfloor\ \ }{240})(\lfloor\frac{(n+4)_{10}+1}{10}\rfloor)\)
\(-(\frac{625\lfloor n/10\rfloor^5-1250\lfloor n/10\rfloor^4+875\lfloor n/10\rfloor^3-442\lfloor n/10\rfloor^2+216\lfloor n/10\rfloor\ \ }{240})(\lfloor\frac{(n+9)_{10}+1}{10}\rfloor+\lfloor\frac{(n+8)_{10}+1}{10}\rfloor)\)
\(-(\frac{625\lfloor n/10\rfloor^5-625\lfloor n/10\rfloor^4+125\lfloor n/10\rfloor^3-167\lfloor n/10\rfloor^2+186\lfloor n/10\rfloor\ \ }{240})(\lfloor\frac{(n+7)_{10}+1}{10}\rfloor+\lfloor\frac{(n+6)_{10}+1}{10}\rfloor)\)
\(-(\frac{625\lfloor n/10\rfloor^5+625\lfloor n/10\rfloor^4+125\lfloor n/10\rfloor^3-217\lfloor n/10\rfloor^2-6\lfloor n/10\rfloor\ \ }{240})(\lfloor\frac{(n+3)_{10}+1}{10}\rfloor+\lfloor\frac{(n+2)_{10}+1}{10}\rfloor)\)
\(-(\frac{625\lfloor n/10\rfloor^5+1250\lfloor n/10\rfloor^4+875\lfloor n/10\rfloor^3+58\lfloor n/10\rfloor^2+24\lfloor n/10\rfloor\ \ }{240})(\lfloor\frac{(n+1)_{10}+1}{10}\rfloor+\lfloor\frac{(n)_{10}+1}{10}\rfloor)\)
\(-(\frac{625\lfloor n/10\rfloor^5-125\lfloor n/10\rfloor^3-192\lfloor n/10\rfloor^2+196\lfloor n/10\rfloor\ \ }{240})(\lfloor\frac{(n+5)_{10}+1}{10}\rfloor)\)
{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 6, 28, 98, 299, 796, 1940, 4364, 9226, 18452, 35248, 64620, 114362, 196048,
326800, 531048,843503, 1312114,2002804, 3004206,4434921, 6450792,9255672, 13112200,18357328,
25417836, 34832164, 47272220, 63573384, 84764512, 112108400,147142272, 191731453, 248123054,
319016108, 407631690, 517803323, 654067352,821778016, 1027222520, 1277765890, 1581995860,...}
注:《整数序列在线百科全书(OEIS)》没有收录。 |
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