数字串的通项公式

王守恩

1. Table[Solve[{n*x*y*z == (x + n) (y + n) (z + n), 0 < x ≤ y≤  z}, {x, y, z}, Integers], {n, 2, 20}]

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n=2, 有20个解。
n=3, 有17个解。
n=4, 有15个解。
n=5, 有08个解。
n=6, 有18个解。
n=7, 有04个解。
n=8, 有11个解。

得到这样一串数:  20, 17, 15, 8, 18, 4, 11, 5, 13, 1, 22, 2, 10, 13, 4, 1, 15, 1, 15, ......
(1), 什么规律？有最大的 n 吗？
(2), 就想要  20, 17, 15, 8, 18, 4, 11, 5, 13, 1, 22, ...，如何编排？

northwolves

王守恩 发表于 2024-5-29 19:04
n=2, 有20个解。
n=3, 有17个解。
n=4, 有15个解。

1. Table[Length@Solve[{n*x*y*z==(x+n) (y+n) (z+n),0<x<=y<=z},{x,y,z},Integers],{n,2,30}]

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{20,17,15,8,18,4,11,5,13,1,22,2,10,13,4,1,15,1,15,9,6,0,17,3,0,1,8,0,24}

northwolves

n=1-200的解的情况：{0,20,17,15,8,18,4,11,5,13,1,22,2,10,13,4,1,15,1,15,9,6,0,17,3,0,1,8,0,24,0,1,6,2,6,13,0,0,4,11,0,21,0,4,10,0,0,7,0,3,2,4,0,4,1,5,0,0,0,29,0,0,6,0,5,5,0,3,2,16,0,10,0,0,3,2,2,9,0,3,0,0,0,16,1,0,0,7,0,10,1,1,1,0,2,2,0,3,3,3,0,1,0,5,10,0,0,6,0,9,0,4,0,7,1,0,3,0,2,20,0,0,1,0,0,13,0,0,0,10,0,9,2,0,5,4,0,2,0,24,0,0,0,6,2,0,0,0,0,7,0,4,3,4,0,10,0,0,0,2,0,0,0,0,6,0,0,15,0,8,0,1,0,1,4,3,0,0,0,17,0,1,0,3,0,1,0,0,4,1,0,1,0,0,8,0,0,5,0,3}

n 以下数字时无解：{1,23,26,29,31,37,38,41,43,46,47,49,53,57,58,59,61,62,64,67,71,73,74,79,81,82,83,86,87,89,94,97,101,103,106,107,109,111,113,116,118,121,122,124,125,127,128,129,131,134,137,139,141,142,143,146,147,148,149,151,155,157,158,159,161,162,163,164,166,167,169,171,173,177,178,179,181,183,185,187,188,191,193,194,196,197,199}

王守恩

northwolves 发表于 2024-5-30 12:06
n=1-200的解的情况：{0,20,17,15,8,18,4,11,5,13,1,22,2,10,13,4,1,15,1,15,9,6,0,17,3,0,1,8,0,24,0,1,6,2 ...

n=420时有30个解，还有>30个解的n吗？
感觉n从某个数开始都是无解的？
这算法还能提速吗？

王守恩

A062714
a(1)=1: 1,
a(2)=3: 12,1,
a(3)=7: 123,1,2,1,3,
a(4)=12: 1234,1,23,1,42,1,3,
a(5)=19: 12345,1,234,1,523,1,452,1,3,
a(6)=28: 123456,1,2345,1,6234,1,5623,1,4562,1,3,
a(7)=39: 1234567,1,23456,1,72345,1,67234,1,56723,1,45672,1,3,
a(8)=52: 12345678,1,234567,1,823456,1,782345,1,678234,1,567823,1,456782,1,3,
a(9)=67: 123456789,1,2345678,1,9234567,1,8923456,1,7892345,1,6789234,1,5678923,1,4567892,1,3,

A180632       
a(1)=1: 1,
a(2)=3: 1,2,1,
a(3)=9: 1231,2,1321,
a(4)=33: 1234123142312431,2,1342132413214321,
a(5)=153: OEIS没有a(5)=153这串数，好心的网友！能补充一下？谢谢！
OEIS没有a(5)=153这串数，好心的网友！能补充一下？谢谢！

还找这么一串数—— 后面的都要有, 可以不计顺序但要连在一起。
a(1)=1: 1,——{1},
a(2)=2: 12,——{1},{2},{12},
a(3)=4: 1231,——{1},{2},{3},{12},{13},{23},{123},
a(4)=8: 12314234——{1},{2},{3},{4},{12},{13},{14},{23},{24},{34},{123},{124},{134},{234},{1234},
a(5)=17: 12314234512531245——{1},{2},{3},{4},{5},{12},{13},{14},{15},{23},{24},{25},{34},{35},{45},{123},{124},{125}{134},{135},{145},{234},{235},{245},{345},{1234},{1235},{1245},{1345},{2345},{12345},

王守恩

{1, 2, 4, 5, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 63}    可以有通项吗？

1. tmp = 1; Table[tmp = Floor[n+ Sqrt[tmp]], {n, 0, 42}]

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1, 2, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639}    可以有通项吗？

1. tmp = 1; Table[tmp = Floor[n^2 + Sqrt[tmp]], {n, 0, 42}]

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