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[分享] 这是“雷劈数”吗

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发表于 3 天前 | 显示全部楼层 |阅读模式

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(1860-496)^2=1860496,

(12312-1216)^2=123121216,

(139672-21489)^2=13967221489,

(1786590-449956)^2=178659044956,

(16198350-3471076)^2=161983503471076,

(152595156-29065744)^2=15259515629065744,

(1205042682-107298321)^2=1205042682107298321,
毋因群疑而阻独见  毋任己意而废人言
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发表于 3 天前 来自手机 | 显示全部楼层
$10^k>b\ge 10^{k-1},a^2-(2b+10^k)a+b^2-b=0$
所以对每个b解这个方程即可
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 3 天前 | 显示全部楼层
这个数字串——OEIS好像没有?

(10-0)^2=100,

(12-1)^2=121,

(1656-369)^2=1656369,

(1860-496)^2=1860496,

(12312-1216)^2=123121216,

(139672-21489)^2=13967221489,

(1786590-449956)^2=178659044956,

(16198350-3471076)^2=161983503471076,

(152595156-29065744)^2=15259515629065744,

(1205042682-107298321)^2=1205042682107298321,

(1256370006-135490884)^2=1256370006135490884,

(1299640762-159622884)^2=1299640762159622884,
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 3 天前 | 显示全部楼层
A228103
Numbers k whose base-10 digits can be split into two parts, q and r, with k = (q-r)^2.

100, 121, 6084, 10000, 10201, 82369, 132496, 1000000, 1002001, 1162084, 1201216, 1656369, 1860496, 100000000, 100020001, 123121216, 330621489, 10000000000, 10000200001, 13967221489, 113322449956, 1000000000000, 1000002000001, 1786590449956, 7438023471076
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 3 天前 | 显示全部楼层
  1. Do[s=k^2;l=IntegerLength[s];Do[{a,b}={Quotient[s,10^(l-i)],Mod[s,10^(l-i)]};If[k==Abs[a-b],Print[s,"=(",a,"-",b,")^2"];Break[]],{i,l-1}],{k,3,10^5}]
复制代码


100=(10-0)^2
121=(12-1)^2
6084=(6-84)^2
10000=(100-0)^2
10201=(102-1)^2
82369=(82-369)^2
132496=(132-496)^2
1000000=(1000-0)^2
1002001=(1002-1)^2
1162084=(1162-84)^2
1201216=(120-1216)^2
1656369=(1656-369)^2
1860496=(1860-496)^2
100000000=(10000-0)^2
100020001=(10002-1)^2
123121216=(12312-1216)^2
330621489=(3306-21489)^2
10000000000=(100000-0)^2
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 3 天前 | 显示全部楼层
northwolves 发表于 2025-11-19 15:53
A228103
Numbers k whose base-10 digits can be split into two parts, q and r, with k = (q-r)^2.

雷劈数——A006886——Kaprekar numbers: positive numbers n such that n = q+r and n^2 = q*10^m+r, for some m >= 1, q >= 0 and 0 <= r < 10^m, with n != 10^a, a >= 1.

{9, 45, 55, 99, 297, 703, 999, 2223, 2728, 4879, 4950, 5050, 5292, 7272, 7777, 9999, 17344, 22222, 38962, 77778, 82656, 95121, 99999, 142857, 148149, 181819, 187110, 208495, 318682, 329967, 351352, 356643, 390313,
461539, 466830, 499500, 500500, 533170, 538461, 609687, 627615, 643357, 648648, 670033, 681318, 791505, 812890, 818181, 851851, 857143, 961038, 994708, 999999, 4444444, 4927941, 5072059, 5479453, 5555556,
8161912, 9372385, 9999999, 11111112, 13641364, 16590564, 19273023, 19773073, 24752475, 25252525, 30884184, 36363636, 38883889, 44363341, 44525548, 49995000, 50005000, 55474452, 55636659, 61116111,
63636364, 69115816, 74747475, 75247525, 80226927, 80726977, 83409436, 86358636, 88888888, 91838088, 94520547, 99999999, 234567901, 332999667, 432432432, 567567568, 667000333, 765432099, 999999999}

Sort@Flatten@Table[c /. Solve[{(c - 1)/(10^k - 1) == a/c, c > a > 0}, {a, c}, Integers], {k, 6}]——这个公式比A006886的公式还是好一些。——还可以调吗?谢谢!!!——主要是能出来4879, 5292。
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 3 天前 | 显示全部楼层
  1. Flatten[Table[Values@Solve[{k^2==a*10^n+b==(a-b)^2,b<10^n,k<50000},{k,a,b},PositiveIntegers],{n,5}],1]
复制代码


{{11,12,1},{101,102,1},{78,6,84},{287,82,369},{364,132,496},{1001,1002,1},{1078,1162,84},{1287,1656,369},{1364,1860,496},{1096,120,1216},{10001,10002,1},{11096,12312,1216},{18183,3306,21489}}
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 3 天前 | 显示全部楼层
  1. Sort@Flatten@Table[c/. Solve[{(c-1)/(10^k+1)==a/c,c>a>0},{a,c},Integers],{k,10}]
复制代码


{11,78,101,287,364,638,715,924,1001,1096,8906,10001,18183,81819,100001,336634,663368,1000001,2727274,7272728,10000001,19138757,23529412,25974026,76470590,97744361,100000001,120879122,140017878,165991904,237762239,288553552,307692308,333666334,405436669,428571430,440553516,447710186,454545455,473684212,526315790,545454547,552289816,571428572,594563333,666333668,692307694,711446450,762237763,834008098,859982124,879120880,902255641,974025976,980861245,1000000001,1539644505,1980198021,8019801981,8460355497,9559446486,10000000001}
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 前天 08:41 | 显示全部楼层
A228381——Unabridged sub-Kaprekar numbers (A118936, but allowing powers of ten).——这是“雷劈数”吗 ?

{10, 11, 78, 100, 101, 287, 364, 1000, 1001, 1078, 1096, 1287, 1364, 10000, 10001, 11096, 18183, 100000, 100001, 118183, 336634, 1000000, 1000001, 1336634, 2727274, 10000000, 10000001, 12727274, 19138757, 23529412, 25974026,

Sort@Flatten@Table[Abs[a - b] /. Solve[{(a - b)^2 == 10^k a + b, a > 0, 10^k > b >≥ 0}, {a, b}, Integers], {k, 9}]——继续学习。
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 前天 16:28 | 显示全部楼层
王守恩 发表于 2025-11-20 08:41
A228381——Unabridged sub-Kaprekar numbers (A118936, but allowing powers of ten).——这是“雷劈数” ...
  1. ToString[#1]<>"=("<>ToString[#2]<>"-"<>ToString[#3]<>")^2"&@@@Sort[Flatten[Table[{(a-b)^2,a,b}/.Solve[{(a-b)^2==10^k a+b,a>0,10^k>b>=0},{a,b},Integers],{k,8}],1]]//MatrixForm
复制代码


100=(10-0)^2
121=(12-1)^2
6084=(6-84)^2
10000=(100-0)^2
10201=(102-1)^2
82369=(82-369)^2
132496=(132-496)^2
1000000=(1000-0)^2
1002001=(1002-1)^2
1162084=(1162-84)^2
1201216=(120-1216)^2
1656369=(1656-369)^2
1860496=(1860-496)^2
100000000=(10000-0)^2
100020001=(10002-1)^2
123121216=(12312-1216)^2
330621489=(3306-21489)^2
10000000000=(100000-0)^2
10000200001=(100002-1)^2
13967221489=(139672-21489)^2
113322449956=(113322-449956)^2
1000000000000=(1000000-0)^2
1000002000001=(1000002-1)^2
1786590449956=(1786590-449956)^2
7438023471076=(743802-3471076)^2
100000000000000=(10000000-0)^2
100000020000001=(10000002-1)^2
161983503471076=(16198350-3471076)^2
553633229065744=(5536332-29065744)^2
10000000000000000=(100000000-0)^2
10000000200000001=(100000002-1)^2
15259515629065744=(152595156-29065744)^2
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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