1572*8634=13572648

王守恩

northwolves 发表于 2025-11-11 00:33

贪心一点。再看一串数。回头看39#可能会简单些。

1. ParallelDo[Module[{n,d,v},n=7*Ceiling[m/2]+m;d=Mod[Range[n],9];v=Sort[Join[d,d]];
   
2. Do[r=RandomSample[d];If[r[[1]]>2,s=FromDigits[r];pr=s^2;
   
3. If[Sort[IntegerDigits[pr]]==v,Print[{n,ToString@s<>"^2="<>ToString@pr}];
   
4. Break[]]],{3*10^7}]],{m,20}]

复制代码

a(8)。  43165782^2=1863284735671524}
a(9)。  753218046^2=567337424820058116}
a(17)。58147732134665208^2=3381158752404776867151230245683264}
a(18)。682760045832751431^2=466161280185540834700223735832547761}
a(26)。65781081236460247533217854^2=4327150648637782448438756352711008207621155824365316}
a(27)。622536108710412088537647534^2=387551206648302017853442350124737702763046854816281156}
a(35)。
a(36)。
a(44)。
a(45)。
a(53)。
a(54)。

直觉:   1, 都有解？  2, 这串数可以有无限项？

{8, 9, 17, 18, 26, 27, 35, 36, 44, 45, 53, 54, 62, 63, 71, 72, 80, 81, 89, 90, 98, 99, 107, 108, 116, 117, 125, 126, 134, 135, 143, 144, ——有个超级简单的通项公式——Table[7*Ceiling[n/2] + n, {n, 60}]

上面用的9个数码 = 1,2,3,4,5,6,7,8,0。如果9个数码改成 = 1,2,3,4,5,6,7,8,9。{8, 9, 17, 18, 26, 27, 35, 36, 44, 45, 53, 54, 62, 63, 71, 72, 80, 81, 89, 90, 98, 99, 107, 108, ..... 这串数不会变？

王守恩

{8, 9, 17, 18, 26, 27, 35, 36, 44, 45, 53, 54, 62, 63, 71, 72, 80, 81, 89, 90, 98, 99, 107, 108, 116, 117, 125, 126, 134, 135, 143, 144, ——有个超级简单的通项公式——Table[7*Ceiling[n/2] + n, {n, 60}]

A274406——Numbers m such that 9 divides m*(m + 1).——有这串数——有公式——我们的公式还是比她好一些。

王守恩

northwolves 发表于 2025-11-11 00:33

1. ParallelDo[Module[{n,d,v},n=7*Ceiling[m/3]+m;d=Mod[Range[n],10];v=Sort[Join[d,d]];Do[r=RandomSample[d];If[r[[1]]>2,s=FromDigits[r];pr=s^2;
   
2. If[Sort[IntegerDigits[pr]]==v,Print[{n,ToString@s<>"^2="<>ToString@pr}];Break[]]],{3*10^7}]],{m,20}]

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我还是不会用。譬如
上面代码的 "n=7*Ceiling[m/3]+m;d=Mod[Range[n],10]"——改成 "n=7*Ceiling[m/2]+m;d={1,2,3,4,5,6,7,8,9}"。出来是这样了。
{8,345918672^2=119659727638243584}
{9,528714396^2=279538912537644816}
17,528714396^2=279538912537644816}
{18,351987624^2=123895287449165376}
{26,572493816^2=327749169358241856}
{27,572493816^2=327749169358241856}
{35,527394816^2=278145291943673856}
{36,359841267^2=129485737436165289}
{44,487256193^2=237418597616853249}

这8,9,17,18,26,27,...本意是表示数位。不知道怎么改。谢谢！！！

northwolves

王守恩 发表于 2025-11-14 07:06
我还是不会用。譬如
上面代码的 "n=7*Ceiling[m/3]+m;d=Mod[Range[n],10]"——改成 "n=7*Ceiling[m/2]+m; ...

1. d = 1 + Mod[Range[n], 9]

复制代码

王守恩

northwolves 发表于 2025-11-14 07:53

嗨！这么简单就可以了？！！谢谢！！！

{123456789^1=123456789}
{527394816^2=278145291943673856}
{726918453^3=384111297639829428356545677}
{961527834^5=821881685441327565743977956591832631269739424}

其它指数是无解的吗？

northwolves

A358705
Zeroless pandigital numbers whose square has each digit 1 to 9 twice.

王守恩

northwolves 发表于 2025-11-15 09:38
A358705
Zeroless pandigital numbers whose square has each digit 1 to 9 twice.

A358705——这个可以出来。我只是不会排序。
ParallelDo[n = 7 Ceiling[m/2] + m; d = {1, 2, 3, 4, 5, 6, 7, 8, 9}; Do[r = RandomSample[d]; If[r[[1]] > 2, s = FromDigits[r]; p = s^2;
   If[Sort[IntegerDigits[p]] == Sort[Join[d, d]],  Print[{n, ToString@s <> "^2=" <> ToString@p}]; Break[]]], {9*10^9}], {m, 99}]
——这个——OEIS就不一定有了。
ParallelDo[n = 7 Ceiling[m/2] + m; d = {1, 2, 3, 4, 5, 6, 7, 8, 0}; Do[r = RandomSample[d]; If[r[[1]] > 2, s = FromDigits[r]; p = s^2;
   If[Sort[IntegerDigits[p]] == Sort[Join[d, d]],  Print[{n, ToString@s <> "^2=" <> ToString@p}]; Break[]]], {9*10^9}], {m, 99}]
——这个——OEIS就不一定有了。
ParallelDo[n = 7 Ceiling[m/2] + m; d = {1, 2, 3, 4, 5, 6, 7, 8}; Do[r = RandomSample[d]; If[r[[1]] > 2, s = FromDigits[r]; p = s^2;
   If[Sort[IntegerDigits[p]] == Sort[Join[d, d]],  Print[{n, ToString@s <> "^2=" <> ToString@p}]; Break[]]], {9*10^9}], {m, 99}]
——这个没有解——很多想法就有解了。
ParallelDo[n = 7 Ceiling[m/2] + m; d = {1, 2, 3, 4, 5, 6, 7}; Do[r = RandomSample[d]; If[r[[1]] > 2, s = FromDigits[r]; p = s^2;
   If[Sort[IntegerDigits[p]] == Sort[Join[d, d]],  Print[{n, ToString@s <> "^2=" <> ToString@p}]; Break[]]], {9*10^9}], {m, 99}]

王守恩

northwolves 发表于 2025-10-30 18:04
{1,0,{}}
{2,1,{{3,51,153}}}
{3,2,{{3,501,1503},{3,510,1530}}}

这帖子有点乱。整理一下。

3A=B。左边的数码=右边的数码。

3×51=153。
2位数有1个。

3×501=1503。
3×510=1530。
3位数有2个。

3×4128=12384。
3×4281=12843。
3×4515=13545。
3×5001=15003。
3×5010=15030。
3×5100=15300。
3×5145=15435。
3×7125=21375。
3×7251=21753。
4位数有9个。

3×40128=120384。
3×40281=120843。
3×41028=123084。
3×41280=123840。
3×41298=123894。
3×42801=128403。
3×42810=128430。
3×42981=128943。
3×45015=135045。
3×45150=135450。
3×45618=136854。
3×47913=143739。
3×47931=143793。
3×48561=145683。
......
1位数有0个。2位数有1个。3位数有2个。4位数有9个。5位数有32个。6位数有138个。7位数有817个。——0, 1, 2, 9, 32, 138, 817, 5005, 31732, ...——OEIS没有这串数。——详见17#,18#。

第1个数(最小数)是这样一串数——后面的是畅想——还得请 northwolves验算。谢谢 northwolves！！！
{"51", "153"},-2
{"501", "1503"},-3
{"4128", "12384"},-4
{"40128", "120384"},-5
{"350511", "1051533"},-6
{"3460812", "10382436"},-7
{"34051128", "102153384"},-8
{"340051128", "1020153384"},-9
{"3400051128", "10200153384"},
{"34000051128", "102000153384"},
{"340000051128", "1020000153384"},
{"3400000051128", "10200000153384"},
{"34000000051128", "102000000153384"},
{"340000000051128", "1020000000153384"},
{"3400000000051128", "10200000000153384"},
{"34000000000051128", "102000000000153384"},
{"340000000000051128", "1020000000000153384"},
{"3400000000000051128", "10200000000000153384"},
{"34000000000000051128", "102000000000000153384"},
{"340000000000000051128", "1020000000000000153384"},
{"3400000000000000051128", "10200000000000000153384"}

Table[k = 34*10^n + 51128; {k, 3 k}, {n, 6, 20}] // TableForm

最后1个数(最大数)是这样一串数——后面的是畅想——还得请 northwolves验算。谢谢 northwolves！！！
{"51", "153"},-2
{"510", "1530"},-3
{"7251", "21753"},-4
{"97251", "291753"},-5
{"997251", "2991753"},-6
{"9997251", "29991753"},-7
{"99997251", "299991753"},-8
{"999997251", "2999991753"},-9
{"9999997251", "29999991753"},
{"99999997251", "299999991753"},
{"999999997251", "2999999991753"},
{"9999999997251", "29999999991753"},
{"99999999997251", "299999999991753"},
{"999999999997251", "2999999999991753"},
{"9999999999997251", "29999999999991753"},
{"99999999999997251", "299999999999991753"},
{"999999999999997251", "2999999999999991753"},
{"9999999999999997251", "29999999999999991753"},
{"99999999999999997251", "299999999999999991753"}

Table[k = 10^n - 2749; {k, 3 k}, {n, 4, 20}] // TableForm

northwolves

1. f[n_]:=(s=10^n-1;While[s>=10^n-3000,If[DigitCount[3*s]-DigitCount[s]=={0,0,1,0,0,0,0,0,0,0},Return[s]];
   
2. s--]);Do[Print[{n,f[n]}],{n,2,20}]

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{2,51}
{3,510}
{4,7251}
{5,97251}
{6,997251}
{7,9997251}
{8,99997251}
{9,999997251}
{10,9999997251}
{11,99999997251}
{12,999999997251}
{13,9999999997251}
{14,99999999997251}
{15,999999999997251}
{16,9999999999997251}
{17,99999999999997251}
{18,999999999999997251}
{19,9999999999999997251}
{20,99999999999999997251}

王守恩

northwolves 发表于 2025-11-18 15:15
{2,51}
{3,510}
{4,7251}

继续学习。——我还是不会用这些按钮。

1. For[n = 1, n <= 8, n++, s = {}; For[y = 10^n + 2, y < 10^(n + 1), y += 3, x = y/3;
   
2. a = SortBy[Tally[Join[IntegerDigits[x]]], First];  b = SortBy[Tally[IntegerDigits[y]], First];  If[a == b, AppendTo[s, {x, y}]]]; Print[{n, Length@s, s}]]

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3A=B。A的数码=B的数码=没有重复的数码。——不是上面的代码——我不知道怎么改。

3×1035=3105,
3×2475=7425,
3×12375=37125,
3×23751=71253,
3×24750=74250,
3×24876=74628,
3×24975=74925,
3×102375=307125,
3×103428=310284,
3×107235=321705,
3×123507=370521,
3×123750=371250,
3×123876=371628,
3×123975=371925,
3×128034=384102,
3×138456=415368,
3×138546=415638,
3×142857=428571,
3×143856=431568,
3×145386=436158,
3×153846=461538,
3×154386=463158,
3×157284=471852,
3×158427=475281,
3×189657=568971,
3×196587=589761,
3×206793=620379,
3×206856=620568,
3×207693=623079,
3×230679=692037,
3×230769=692307,
3×235071=705213,
3×235107=705321,
3×237501=712503,
3×237510=712530,
3×238761=716283,
3×239751=719253,
3×248760=746280,
3×248976=746928,
3×249750=749250,
3×249876=749628,
3×271584=814752,
3×276489=829467
3×280341=841023,
3×281034=843102,
3×284157=852471,
3×285714=857142,
3×287649=862947,
3×306792=920376,
3×307692=923076,
3×314379=943137,
3×320679=962037,
3×320769=962307,