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楼主: 王守恩

[求助] OEIS序列A325112的通项公式

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发表于 2023-1-1 13:31:16 | 显示全部楼层
$S_122=3.1416164519106777$
毋因群疑而阻独见  毋任己意而废人言
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 楼主| 发表于 2023-1-2 17:44:40 | 显示全部楼层
northwolves 发表于 2023-1-1 13:29
$s=3.1459691497419673>\pi$

谢谢 northwolves !

1,A325112还没有7楼的通项公式。

2,A325112还没有把这串数(取倒数)加起来,答案就算不是 pi,23/7也是一个简单的得数。

         A261189         整数,使得十进制表示的子序列不能被 5 整除。               
{1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24, 26, 27, 28, 29, 31, 32, 33, 34, 36, 37, 38, 39, 41, 42, 43, 44, 46,
47, 48, 49, 61, 62, 63, 64, 66, 67, 68, 69, 71, 72, 73, 74, 76, 77, 78, 79, 81, 82, 83, 84, 86, 87, 88, 89, 91, 92, 93, 94, 96, 97, 98, 99,
111, 112, 113, 114, 116, 117, 118, 119, 121, 122, 123, 124, 126, 127, 128, 129, 131, 132, 133, 134, 136, 137, 138, 139, 141, 142, 143, 144, 146, 147, 148, 149,
161, 162, 163, 164, 166, 167, 168, 169, 171, 172, 173, 174, 176, 177, 178, 179, 181, 182, 183, 184, 186, 187, 188, 189, 191, 192, 193, 194, 196, 197, 198, 199,
211, 212, 213, 214, 216, 217, 218, 219, 221, 222, 223, 224, 226, 227, 228, 229, 231, 232, 233, 234, 236, 237, 238, 239, 241, 242, 243, 244, 246, 247, 248, 249,
261, 262, 263, 264, 266, 267, 268, 269, 271, 272, 273, 274, 276, 277, 278, 279, 281, 282, 283, 284, 286, 287, 288, 289, 291, 292, 293, 294, 296, 297, 298, 299,
311, 312, 313, 314, 316, 317, 318, 319, 321, 322, 323, 324, 326, 327, 328, 329, 331, 332, 333, 334, 336, 337, 338, 339, 341, 342, 343, 344, 346, 347, 348, 349,
361, 362, 363, 364, 366, 367, 368, 369, 371, 372, 373, 374, 376, 377, 378, 379, 381, 382, 383, 384, 386, 387, 388, 389, 391, 392, 393, 394, 396, 397, 398, 399,
411, 412, 413, 414, 416, 417, 418, 419, 421, 422, 423, 424, 426, 427, 428, 429, 431, 432, 433, 434, 436, 437, 438, 439, 441, 442, 443, 444, 446, 447, 448, 449,
461, 462, 463, 464, 466, 467, 468, 469, 471, 472, 473, 474, 476, 477, 478, 479, 481, 482, 483, 484, 486, 487, 488, 489, 491, 492, 493, 494, 496, 497, 498, 499}

Select[Range@1000, NoneTrue[DeleteCases[FromDigits /@ Rest@Subsequences[IntegerDigits@#], 0], Mod[#, 5] == 0 &] &](* Michael De Vlieger, Mar 31 2019 *)

     把这串数(取倒数)加起来,答案差不多就是 pi^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
发表于 2023-1-2 22:31:17 | 显示全部楼层
王守恩 发表于 2023-1-2 17:44
谢谢 northwolves !

1,A325112还没有7楼的通项公式。

A261189  的生成比较容易,求出第一个n 位的序号,八进制转换即可

点评

没那么复杂。A261189只是去掉0,5,所以每个数位的数字都是12346789,即[(5n-1)/4]  发表于 2023-1-3 10:37
说得轻松,没几十年的功底那出得来(我只会2019,依样画葫芦)。  发表于 2023-1-3 10:01
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 2023-1-2 22:33:06 | 显示全部楼层
  1. from math import log
  2. def a(n):
  3.     k=int(log(7*n+1,8))
  4.     t=[str(1+(5*int(i))//4) for i in ('0'*k+(oct(n-(8**k-1)//7))[2:])[-k:]]
  5.     return(int(''.join(t)))
  6. print([a(n) for n in range(9801,10001)])
复制代码


[23111, 23112, 23113, 23114, 23116, 23117, 23118, 23119, 23121, 23122, 23123, 23124, 23126, 23127, 23128, 23129, 23131, 23132, 23133, 23134, 23136, 23137, 23138, 23139, 23141, 23142, 23143, 23144, 23146, 23147, 23148, 23149, 23161, 23162, 23163, 23164, 23166, 23167, 23168, 23169, 23171, 23172, 23173, 23174, 23176, 23177, 23178, 23179, 23181, 23182, 23183, 23184, 23186, 23187, 23188, 23189, 23191, 23192, 23193, 23194, 23196, 23197, 23198, 23199, 23211, 23212, 23213, 23214, 23216, 23217, 23218, 23219, 23221, 23222, 23223, 23224, 23226, 23227, 23228, 23229, 23231, 23232, 23233, 23234, 23236, 23237, 23238, 23239, 23241, 23242, 23243, 23244, 23246, 23247, 23248, 23249, 23261, 23262, 23263, 23264, 23266, 23267, 23268, 23269, 23271, 23272, 23273, 23274, 23276, 23277, 23278, 23279, 23281, 23282, 23283, 23284, 23286, 23287, 23288, 23289, 23291, 23292, 23293, 23294, 23296, 23297, 23298, 23299, 23311, 23312, 23313, 23314, 23316, 23317, 23318, 23319, 23321, 23322, 23323, 23324, 23326, 23327, 23328, 23329, 23331, 23332, 23333, 23334, 23336, 23337, 23338, 23339, 23341, 23342, 23343, 23344, 23346, 23347, 23348, 23349, 23361, 23362, 23363, 23364, 23366, 23367, 23368, 23369, 23371, 23372, 23373, 23374, 23376, 23377, 23378, 23379, 23381, 23382, 23383, 23384, 23386, 23387, 23388, 23389, 23391, 23392, 23393, 23394, 23396, 23397, 23398, 23399, 23411, 23412, 23413, 23414, 23416, 23417, 23418, 23419]
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 2023-1-3 12:55:55 | 显示全部楼层
northwolves 发表于 2023-1-2 22:31
A261189  的生成比较容易,求出第一个n 位的序号,八进制转换即可

整数,使得十进制表示的子序列不能被 6 整除。这可是OEIS没有的,你还能找出通项来(你的方法我还是不会)?

{1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 17, 19, 20, 21, 22, 23, 25, 27, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 43, 44, 45, 47, 49,
50, 51, 52, 53, 55, 57, 58, 59, 70, 71, 73, 74, 75, 77, 79, 80, 81, 82, 83, 85, 87, 88, 89, 91, 92, 93, 94, 95, 97, 98, 99 100, 101, 103, 104,
105, 107, 109, 110, 111, 113, 115, 117, 119, 131, 133, 134, 135, 137, 139, 140, 141, 143, 145, 147, 149, 151, 152, 153, 155, 157, 158, 159,
170, 171, 173, 175, 177, 179, 191, 193, 194, 195, 197, 199, 200, 201, 202, 203, 205, 207, 208, 209, 211, 213, 214, 215, 217, 219, 220, ......

点评

这个序列暂时找不到简单的计算方法,太烧脑了  发表于 2023-1-4 23:52
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 2023-1-4 00:01:35 | 显示全部楼层
不能被 6 整除的不太好找,有空时试试
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 2023-1-4 14:36:21 | 显示全部楼层
northwolves 发表于 2023-1-2 22:31
A261189  的生成比较容易,求出第一个n 位的序号,八进制转换即可
  1. f[n_] := Module[{s}, k = IntegerPart[Log[8, 7 n + 1]];
  2.   s = FromDigits[
  3.     Floor[5*Take[
  4.          Join[{0}*k, IntegerDigits[n - Floor[(8^k - 1)/7], 8]], -k]/
  5.         4] + 1]]
  6. f[10000]
  7. f[9^999]
复制代码


23419

3289173788763248693278874182979677473297199877814498464377646716236494\
1284884429296677692617479226397191796136869289327829239497986267478933\
1827798781318218733172766279468432967124676644683921442121214977348467\
3184446146323694986824428826386237724678474444826446428481996464642923\
2662164168276612728641948744771728328474823621112372291646644969876946\
4121991821478434829977496148732394481389921923137263182418384997861916\
3769249636929924373779842888968797324893712711646688292483663817747493\
7883791183714624831991718747447994726686186792624346417484633962914287\
9476889987391377747267714494331944282679961997348989198739736171969429\
7249837183792331339127638488673443332233868934686979477936312969479829\
3324611848314894726716668789823377829288298822114687837722687988447263\
4121492186922649617219921276828471136746477964338441361493172489626932\
3337217782126232673362481978183479889693624227731673849726294118432639\
4242611223611324412286622496377297197222384962214142279899432979248179\
4163447764617411998721866743418389244618419464428941793662831631987233\
443181
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 2023-1-4 22:28:23 | 显示全部楼层
王守恩 发表于 2023-1-2 17:44
谢谢 northwolves !

1,A325112还没有7楼的通项公式。

经验证,和值大于10:

a(610989765)=13891339296,S(610989765)=9.869604401097234>pi^2

a(5200530434)=47687346892,S(5200530434)=10.000000000006029

n>5000000000之后,对和值的贡献已经很小很小了:
a(5150000000)=47286641689,S(5150000000)=9.998935429886851
a(5160000000)=47343764889,S(5160000000)=9.999146769985744
a(5170000000)=47411879189,S(5170000000)=9.999357844970831
a(5180000000)=47468993389,S(5180000000)=9.9995686495833
a(5190000000)=47627217689,S(5190000000)=9.99977903880075
a(5200000000)=47684331889,S(5200000000)=9.999988876528164
a(5200530434)=47687346892,S(5200530434)=10.000000000006029
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 2023-1-5 08:48:39 | 显示全部楼层
事不过三,第3次不能有错了。

整数,使得十进制表示的子序列不能被 36 整除。把这串数(取倒数)加起来,答案是 e^5=148.413159
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 2023-1-6 13:00:51 | 显示全部楼层
本帖最后由 northwolves 于 2023-1-6 22:41 编辑

整数,使得十进制表示的非零子序列不能被 n 整除。

把这串数(取倒数)加起来,猜想:和也就是下面这串数。

{3, 5, 8, 11, 14, 18, 21, 25, 29, 33, 37, 41, 45, 49, 54, 58, 62, 67, 71, 76, 81, 85, 90, 95, 100, 105, 109, 114, 119, 124, 129, 134, 140, 145, 150}

把这串数(取倒数)加起来的和

\(=\D\sum_{k=1}^{n-1}\ \frac{1}{k}*\sum_{j=0}^{\infty}\bigg(\frac{n-1}{n}\bigg)^j\)

\(=\D\sum_{k=1}^{n-1}\ \frac{1}{k}*n\)

\(=\D\bigg\lfloor\sum_{k=1}^n\ \frac{n}{k}\bigg\rfloor\)
    2022 年 8 月 10 日

参考:《分母缺数码 0 的调和级数,和不大于 90》谢谢mathe!

参考:《所有 “降序数” 的和是多少?》《最少马步函数》谢谢 hujunhua!
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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