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所有自然整数迟早会出现在序列中(根据定义)-但大多是“稍后”!事实上,序列增长非常缓慢:在100000个术语之后,尚未出现的最小术语是32个。
以下是迄今为止在相同范围内的计数{术语,出现次数}示例:
{1,192},{2,396},{3,618},{4,796},{5,1160},{6,1296},{7,2294},{8,2080},{9,2489},{10,2826},{11,3487},{12,1596},{13,2295},{14,1960},{15,2370},{16,2640},{17,4097},{18,2214},{19,4598},{20,2770},{21,3759},{22,4477},{23,5612},{24,4884},{25,5825},{26,6006},{27,6359},{28,4676},{29,5481},{30,3060},{31,1411},{32,0},{33,182},{34,0},{35,315},{36,0},{37,1221},{38,0},{39,214},{40,0},{41,1353},{42,0},{43,1183},{44,0},{45,0},{46,0},{47,1058},{48,0},{49,172},{50,0},{51,0},{52,0},{53,580},...
在100000个术语之后,第一个尚未出现的产品是(质数):59、61、67、71、73、79、83、89、97、101。。。以及(复合材料)118、122、134。。。
以下是迄今为止{乘积,乘积的出现次数}的样本(计算100000项):
{1,1},{2,2},{3,3},{4,4},{5,5},{6,6},{7,7},{8,8},{9,9},{10,10},{11,11},{12,12},{13,13},{14,14},{15,15},{16,16},{17,17},{18,18},{19,19},{20,20},{21,21},{22,22},{23,23},{24,24},{25,25},{26,26},{27,27},{28,28},{29,29},{30,30},{31,31},{32,32},{33,33},{34,34},{35,35},{36,36},{37,37},{38,38},{39,39},{40,40},{41,41},{42,42},{43,43},{44,44},{45,45},{46,46},{47,47},{48,48},{49,49},{50,50},{51,51},{52,52},{53,53},{54,54},{55,55},{56,56},{57,57},{58,58},{59,0},{60,60},{61,0},{62,62},{63,63},{64,64},{65,65},{66,66},{67,0},{68,68},{69,69},{70,70},{71,0},{72,72},{73,0},{74,74},{75,75},{76,76},{77,77},{78,78},{79,0},{80,80},{81,81},{82,82},{83,0},{84,84},{85,85},{86,86},{87,87},{88,88},{89,0},{90,90},{91,91},{92,92},{93,93},{94,94},{95,95},{96,96},{97,0},{98,98},{99,99},{100,100},{101,0},{102,102},{103,0},{104,104},{105,105},{106,106},{107,0},{108,108},{109,0},{110,110},{111,111},{112,112},{113,0},{114114},{115115},{116116},{117117},{118,0},{119119},{120120},{121121},{122,0},{123,123},{124,124},{125,125},{126,126},{127,0},{128,128},{129,129},{130,130},{131,0},{132,132},{133,133},{134,0},{135,135},{136,136},{137,0},{138,138},{139,0},{140,140},{141,141},{142,0},...
定理。这个序列也可以由贪婪算法定义。也就是说,设b(1)=1,对于n>=1,设b是最小的正整数k,使得m=k*b(n)在列表[b(i)*b(i+1):i=1..n-1]中最多出现n-1次。则b(n)=a(n),对于所有n>=1。
(注意,对于n=1,列表是空的,所以我们取k=b(1)=1。)
备注:该定理并不明显,需要在下面的链接中给出证明。“词汇学上最早的”序列通常需要一些回溯,但定理的要点是这里不需要回溯。
证明还表明序列中有无穷多个1,每个k在乘积a(i)*a(i+1)序列中出现k次。(结束)
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