楼上是这样一串数——2, 6, 13, 22, 36, 53, 75, 102, 135, 173, 219, 271, 331, 399, 476, 561, 657, 762, 878, 1005, 1144, 1294, 1458, 1634, 1824, 2028, 2247, 2480, 2730, 2995, ......
Table[Round[(26 n + 19 n^2 + 2 n^3)/24], {n, 40}]——通项公式出奇的好——OEIS是没有的——这串数也是有现实意义。
n=1: a={2} b={1} 和=2= 2(a)*1(b)*1(c)
n=2: a={1,2} b={1,2} 和=6= 1(a)*1(b)*2(c) + 2(a)*2(b)*1(c)
n=3: a={1,2,3} b={1,1,2} 和=13= 1(a)*1(b)*3(c) + 2(a)*1(b)*2(c) + 3(a)*2(b)*1(c)
n=4: a={1,1,2,3} b={1,2,3,1} 和=22=1(a)*1(b)*4(c) + 1(a)*2(b)*3(c) + 2(a)*3(b)*1(c) + 3(a)*1(b)*2(c)
n=5: a={1,1,2,3,4} b={1,2,1,3,1} 和=36= 1(a)*1(b)*5(c) + 1(a)*2(b)*4(c) + 2(a)*1(b)*3(c) + 3(a)*3(b)*1(c) + 4(a)*1(b)*2(c)
n=6: a={1,1,1,2,3,4} b={1,2,3,1,4,1} 和=53=1(a)*1(b)*6(c) + 1(a)*2(b)*5(c) + 1(a)*3(b)*3(c) + 2(a)*1(b)*4(c) + 3(a)*4(b)*1(c) + 4(a)*1(b)*2(c)
n=7: a={1,1,1,2,3,4,5} b={1,2,3,1,4,1,1} 和=75
n=8: a={1,1,1,1,2,3,4,5} b={1,2,3,4,1,5,1,1} 和=102
n=9: a={1,1,1,1,2,3,4,5,6} b={1,2,3,4,1,1,5,1,1} 和=135
n=10: a={1,1,1,1,1,2,3,4,5,6} b={1,2,3,4,6,1,1,5,1,1} 和=173
n=11: a={1,1,1,1,1,2,3,4,5,6,7} b={1,2,3,4,5,1,1,6,1,1,1} 和=219
n=12: a={1,1,1,1,1,1,2,3,4,5,6,7} b={1,2,3,4,5,7,1,1,1,6,1,1} 和=271
n=13: a={1,1,1,1,1,1,2,3,4,5,6,7,8} b={1,2,3,4,5,7,1,1,1,6,1,1,1} 和=331
n=14: a={1,1,1,1,1,1,1,2,3,4,5,6,7,8} b={1,2,3,4,5,6,8,1,1,1,7,1,1,1} 和=399
n=15: a={1,1,1,1,1,1,1,2,3,4,5,6,7,8,9} b={1,2,3,4,5,6,8,1,1,1,1,7,1,1,1} 和=476
SystemException["MemoryAllocationFailure"]——说来不了了。
楼上的解法是来不了了——有一个想法——若能把a与b合并为一个数字串——b是关键——什么规律?谢谢! |