{27,102030201,{{10000,10201,102010000},{116964,125316,101787921},{350464,395641,101284096},{828100,1002001,100200100},{998001,1232100,99800100},{1368900,1758276,98903025},{2047761,2802276,97180164},{2775556,4036081,95218564},{3168400,4752400,94109401},{3701776,5784025,92544400},{3980025,6350400,91699776},{5433561,9659664,86936976},{5721664,10394176,85914361},{6426225,12320100,83283876},{6708100,13147876,82174225},{8076964,17757796,76195441},{8328996,18740241,74960964},{8803089,20738916,72488196},{9144576,22325625,70560000},{9759376,25603600,66667225},{10310521,29246464,62473216},{10601536,31629376,59799289},{10732176,32867289,58430736},{11088900,37271025,53670276},{11256025,40704400,50069776},{11329956,43996689,46703556},{11336689,45346756,45346756}}}
\(102030201=3^2\times 7^2\times 13^2\times 37^2\)
这个数字选择的比较好。
根据15#，\(r_1+r_2+r_3=(a^2+ab+b^2)^2 h\)
所以我们需要寻找这些数字有更多的上面表达形式即可。
而\(a^2+ab+b^2\)有个特点，其所有素因子要么是3，要么模3余1.
而且记\(\omega\)为三次单位根，那么\(a^2+ab+b^2=(a-\omega b)(a-\bar{\omega}b)\), 也就是是它总是爱森斯坦整环中整数的模长平方形式。
所以由于上面3,7,13,37都是在爱森斯坦整环中可以分解，比如\(3=(1-\omega)(1-\bar{\omega}), 7=(2-\omega)(2-\bar{\omega})=(1-2\omega)(1-2\bar{\omega}), ...\)
由此我们可以选择把部分上面表达形式相乘，得到一些复合形式，如
\(21=(1-\omega)(2-\omega)(1-\bar{\omega})(2-\bar{\omega})=(1-4\omega)(1-4\bar{\omega})\)
也可以表示为
\(21=(1-\omega)(1-2\omega)(1-\bar{\omega})(1-2\bar{\omega})=(-1-5\omega)(-1-5\bar{\omega})=(1+5\omega)(1+5\bar{\omega})\)
其中第一种形式对应a=1,b=4符合都是正数要求，但是第二种a=1,b=-5不符合要求。
另外对于每个\(a-b\omega\)，我们分别将它继续乘上\(\omega,\omega^2\)还可以得到另外两种表达形式，如
\(b+(a+b)\omega, (-a - b)+(-a)\omega\), 再加上它们的之间a,b轮换可以有6种不同形式。
所以最终需要穷举的数目如下
i) 3可以选择\(1-\omega\)或者不参与(h*=9)，两种选择
ii) 7可以选择\(1-2\omega\)或者\(2-\omega\)或者不参与，三种选择
iii)13可以选择\(1-3\omega\)或者\(3-\omega\)或者不参与，三种选择
iv)37可以选择\(3-4\omega\)或者\(4-3\omega\)或者不参与，三种选择。
所以上面共2*3*3*3=54种选择，对每个结果，再可以继续乘上\(\omega,\omega^2\)，所以共需枚举54*3=162种情况，选择其中a,b同号的情况。
按照mathe版主的推理，可以构造任意解的数字，试了一下果然如此。$7^{2n}$正好就有n个解,13,19也可以
{1, 49, {{4, 9, 36}}},
{2, 2401, {{196, 441, 1764}, {225, 576, 1600}}},
{3, 117649, {{324, 361, 116964}, {9604, 21609, 86436}, {11025, 28224,
78400}}},
{4, 5764801, {{15876, 17689, 5731236}, {389376, 774400,
4601025}, {470596, 1058841, 4235364}, {540225, 1382976, 3841600}}},
{5, 282475249, {{777924, 866761, 280830564}, {19079424, 37945600,
225450225}, {23059204, 51883209, 207532836}, {26471025, 67765824,
188238400}, {29095236, 85340644, 168039369}}},
{6, 13841287201, {{38118276, 42471289, 13760697636}, {142826401,
177422400, 13521038400}, {934891776, 1859334400,
11047061025}, {1129900996, 2542277241, 10169108964}, {1297080225,
3320525376, 9223681600}, {1425666564, 4181691556, 8233929081}}},
{7, 678223072849, {{1867795524, 2081093161,
674274184164}, {6998493649, 8693697600,
662530881600}, {35529611049, 62747244036,
579946217764}, {45809697024, 91107385600,
541305990225}, {55365148804, 124571584809,
498286339236}, {63556931025, 162705743424,
451960398400}, {69857661636, 204902886244, 403462524969}}},
{8, 33232930569601, {{91521980676, 101973564889,
33039435024036}, {342926188801, 425991182400,
32464013198400}, {1740950941401, 3074614957764,
28417364670436}, {2244675154176, 4464261894400,
26523993521025}, {2712892291396, 6104007655641,
24416030622564}, {3114289620225, 7972581427776,
22146059521600}, {3423025420164, 10040241425956,
19769663723481}, {3619734553600, 12269502756225, 17343693259776}}},
{9, 1628413597910449, {{4484577053124, 4996704679561,
1618932316177764}, {16803383251249, 20873567937600,
1590736646721600}, {35131723548804, 48718696894884,
1544563177466761}, {85306596128649, 150656132930436,
1392450868851364}, {109989082554624, 218748832825600,
1299675682530225}, {132931722278404, 299096375126409,
1196385500505636}, {152600191391025, 390656489961024,
1085156916558400}, {167728245588036, 491971829871844,
968713522450569}, {177366993126400, 601205635055025,
849840969729024}}},
{10, 79792266297612001, {{219744275603076, 244838529298489,
79327683492710436}, {823365779311201, 1022804828942400,
77946095689358400}, {1721454453891396, 2387216147849316,
75683595695871289}, {2970707817640000, 4675800467030625,
72145758012941376}, {4180023210303801, 7382150513591364,
68230092573716836}, {5389465045176576, 10718692808454400,
63684108443981025}, {6513654391641796, 14655722381194041,
58622889524776164}, {7477409378160225, 19142168008090176,
53172688911361600}, {8218684033813764, 24106619663720356,
47466962600077881}, {8690982663193600, 29459076117696225,
41642207516722176}}}
{p,q,7^p*13^q的解的个数}
{0,0,0} {0,1,0} {0,2,1} {0,3,1} {0,4,2} {0,5,2} {0,6,3} {0,7,3} {0,8,4} {0,9,4}
{1,0,0} {1,1,0} {1,2,1} {1,3,1} {1,4,2} {1,5,2} {1,6,3} {1,7,3} {1,8,4} {1,9,4}
{2,0,1} {2,1,1} {2,2,4} {2,3,4} {2,4,7} {2,5,7} {2,6,10} {2,7,10} {2,8,13} {2,9,13}
{3,0,1} {3,1,1} {3,2,4} {3,3,4} {3,4,7} {3,5,7} {3,6,10} {3,7,10} {3,8,13} {3,9,13}
{4,0,2} {4,1,2} {4,2,7} {4,3,7} {4,4,12} {4,5,12} {4,6,17} {4,7,17} {4,8,22} {4,9,22}
{5,0,2} {5,1,2} {5,2,7} {5,3,7} {5,4,12} {5,5,12} {5,6,17} {5,7,17} {5,8,22} {5,9,22}
{6,0,3} {6,1,3} {6,2,10} {6,3,10} {6,4,17} {6,5,17} {6,6,24} {6,7,24} {6,8,31} {6,9,31}
{7,0,3} {7,1,3} {7,2,10} {7,3,10} {7,4,17} {7,5,17} {7,6,24} {7,7,24} {7,8,31} {7,9,31}
{8,0,4} {8,1,4} {8,2,13} {8,3,13} {8,4,22} {8,5,22} {8,6,31} {8,7,31} {8,8,40} {8,9,40}
{9,0,4} {9,1,4} {9,2,13} {9,3,13} {9,4,22} {9,5,22} {9,6,31} {9,7,31} {9,8,40} {9,9,40}
$7^a*13^b*19^c$的解的个数:
{1,2,3} 4
{1,2,4} 7
{1,2,5} 7
{1,2,6} 10
{1,2,7} 10
{1,2,8} 13
{1,3,4} 7
{1,3,5} 7
{1,3,6} 10
{1,3,7} 10
{1,3,8} 13
{1,4,5} 12
{1,4,6} 17
{1,4,7} 17
{1,4,8} 22
{1,5,6} 17
{1,5,7} 17
{1,5,8} 22
{1,6,7} 24
{1,6,8} 31
{1,7,8} 31
{2,3,4} 22
{2,3,5} 22
{2,3,6} 31
{2,3,7} 31
{2,3,8} 40
{2,4,5} 37
{2,4,6} 52
{2,4,7} 52
{2,4,8} 67
{2,5,6} 52
{2,5,7} 52
{2,5,8} 67
{2,6,7} 73
{2,6,8} 94
{2,7,8} 94
{3,4,5} 37
{3,4,6} 52
{3,4,7} 52
{3,4,8} 67
{3,5,6} 52
{3,5,7} 52
{3,5,8} 67
{3,6,7} 73
{3,6,8} 94
{3,7,8} 94
{4,5,6} 87
{4,5,7} 87
{4,5,8} 112
{4,6,7} 122
{4,6,8} 157
{4,7,8} 157
{5,6,7} 122
{5,6,8} 157
{5,7,8} 157
{6,7,8} 220
假设\(n=3^{u_3} p_1^{\alpha_1}p_2^{\alpha_2}\cdot p_t^{\alpha_t}m\)
其中\(p_1,p_2,\cdots, p_t\)都是模6为1的素数，而m所有素因子都是模6为5, 或是因子2.
于是当\(u_3\ge 2\)解的个数应该是\(\prod_{h=1}^t (\alpha_h|1)\), 当\(u_3\le 1\)解的数目为\(\frac{(\prod_{h=1}^t(\alpha_h|1))-1}2\).
其中\(x|1\)代表将整数x的最后一个二进制位设置为1, 也就是选择不小于x的最小奇数。
northwolves 发表于 2025-3-8 13:54
{3072,27,{{1,4,4},{2,8,8},{3,12,12},{4,16,16},{4,9,36},{5,20,20},{6,24,24},{7,28,28},{8,32,32},{8, ...
小心翼翼的不敢问:已知3个半径(r1,r2,r3)的和=n,A(n)表示具体解, 要如何编排？
王守恩 发表于 2025-3-15 15:37
小心翼翼的不敢问:已知3个半径(r1,r2,r3)的和=n,A(n)表示具体解, 要如何编排？ ...
sol:=(s=Union@Flatten],v[],v[]+v[]},{2}]^2,{h,Select&]},{v,Values@Solve[(a^2+a b+b^2)^2==n/h&&a<=b,{a,b},PositiveIntegers]}],1];
{Length@s,n,s});sol
{22,164648481361,{{404975376,448465329,163795040656},{453434436,505215529,163689831396},{1610497161,1988268100,161049716100},{2099014225,2677441536,159872025600},{3434428816,4732889616,156481162929},{4070184804,5798213316,154780083241},{5547270400,8495124561,150606086400},{6267888900,9929126025,148451466436},{6406401600,10214134225,148027945536},{8014367529,13760697636,142873416196},{8768262321,15588021904,140292197136},{10526760000,20334760000,133786961361},{11120967936,22117638400,131409875025},{11258483236,22545022500,130844975625},{13440692356,30241557801,120966231204},{15429366225,39499177536,109719937600},{16886482704,49140248976,98621749681},{17248619556,52350355204,95049506601},{17894412900,60145110025,86608958436},{18078415936,63556931025,83013134400},{18284177961,71076626404,75287676996},{18289587121,71744693904,74614200336}}}
对于\(n=3^{u_3} p_1^{\alpha_1}p_2^{\alpha_2}\cdot p_t^{\alpha_t}m\)，其中素数\(p_i\equiv 1\pmod 6\).
对于每个\(\alpha_i\ge 2\),我们可以用计算机找到\(p_i=a_i^2+a_ib_i+b_i^2\)
然后我们穷举表达式
\(u+v\omega=\prod_{i=1}^t(a_i-b_i \omega)^{s_i}\), 其中\(\omega\)为三次单位根，而且\(-\lfloor \frac{\alpha_i}2\rfloor \le s_i \le \lfloor \frac{\alpha_i}2\rfloor\)
如果其中u,v异号就是一个结果；如果u,v同号，我们计算\((u+v\omega)\omega\)和\((u+v\omega)\omega^2\), 其中必然有且只有一个展开写成同样形式后两者异号。
如果\(u_3\le 1\), 那么搜索已经完成，而且我们知道所有\(s_i\)都是0的情况对应1，其中v=0需要淘汰，其余情况中会有一半相互共轭(u,v交换），对应所有\(s_i\)互为相反数的情况。
如果\(u_3\ge 2\), 那么我们可以将上面每个乘积再乘上\(1-\omega\)又得出 \(\prod_{i=1}^t (2\omega+1)\)个结果，同样一半互为共轭。然后所有\(s_i=0\)的场景下，是1和\(1-\omega\)，前者不符合要求需要淘汰。
在得出每个\(u-(-v)\omega\)后，计算\(h=\frac{n}{(u^2+u(-v)+(-v)^2)^2}\)，并且取\(a=|u|,b=|v|\)即可。
northwolves 发表于 2025-3-10 16:06
1亿内的解(原来的计算有问题，重发，？ 暂未确认)：
前15项：
建议这样写。前15项：
a = {9, 441, 441, 8281, 21609, 74529, 74529, 74529, 74529, 2989441, 2989441, 2989441, 2989441, 3651921, 3651921}
a(22)=146482609——x
a(27)=26904969
a(40)=2872852801