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[转载] 兔子问题

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发表于 2019-2-8 11:42:00 | 显示全部楼层
本帖最后由 dlpg070 于 2019-2-8 16:37 编辑

非常高兴你做了大量修改
19#修改了Axxxxxx 8项,改的好
修改后和我的选择完全一致
王守恩的通项公式正确的项可能增加 7项
有多项原有怀疑(n=2,3,---,7),
经计算数列符合本主题的兔子数,怀疑取消
通项公式成立

   王守恩给出通项公式对应的数列
n  原始 17#   修改后19#    dlpg070     
2   A000045:  A000045 :   A000045       
3   A000073:  A000073 :   A000073       
4   A000078:  A000078 :   A000078       
5   A001591:  A001591 :   A001591       
6   A006261:  A001592*:   A001592       
7   A066178:  A066178 :   A066178       
8   A079262:  A079262 :   A079262
9   A104144:  A104144 :   A104144
10 A008862:  A122265*:   A122265
11 A008863:  A168082*:   A168082
12 A219531:  A168083*:   A168083
13 A133025:  A168084*:   A168084
14 A219676:  A220469*:   A220469       
15 A220051:  A220493*:   A220493
16 A097029:  A249169*:   A249169       
17 ........  A.......
18 ........  A.......
19 ........  A.......
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 2019-2-8 13:33:49 | 显示全部楼层
因帖子交叉,没有来得及及时回应你的验算结果
下面给出相关数列的前19个数,供参考

n-step Fibonacci numbers 的前19个数   (n=2---16,k=0---18)
n                                        k=  0  1  2  3  4  5    6    7     8     9     10     11     12    13      14       15      16        17       18        ---
2    A000045  Fibonacci number   1, 1, 2, 3, 5,  8,  13,  21,  34,   55,  89,   144,   233,  377,   610,    987,   1597,   2584,   4181,  ---
3    A000073  tribonacci number   1, 1, 2, 4, 7, 13, 24,  44,  81,  149, 274, 504,   927,  1705, 3136,  5768,  10609, 19513, 35890, ---
4    A000078  tetranacci number   1, 1, 2, 4, 8, 15, 29,  56, 108, 208, 401, 773,  1490, 2872, 5536, 10671, 20569, 39648, 76424, ---
5    A001591  pentanacci number  1, 1, 2, 4, 8, 16, 31,  61, 120, 236, 464, 912,  1793, 3525, 6930, 13624, 26784, 52656, 103519,---
6    A001592  hexanacci number        1, 1, 2, 4, 8, 16, 32,  63, 125, 248, 492, 976,  1936, 3840, 7617, 15109, 29970, 59448, 117920,---
7    A066178  heptanacci number  1, 1, 2, 4, 8, 16, 32,  64, 127,  253, 504,  1004, 2000, 3984, 7936, 15808, 31489, 62725, 124946,---
8    A079262  Fibonacci 8-step      1, 1, 2, 4, 8, 16, 32, 64, 128, 255, 509, 1016, 2028, 4048, 8080, 16128, 32192, 64256, 128257,---
9    A104144  Fibonacci 9-step      1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1021, 2040, 4076, 8144, 16272, 32512, 64960, 129792,---
10  A122265  Fibonacci 10-step    1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1023, 2045, 4088, 8172, 16336, 32656, 65280, 130496,---
11  A168082  Fibonacci 11-step    1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2047, 4093, 8184, 16364, 32720, 65424, 130816,---
12  A168083  Fibonacci 12-step    1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4095, 8189, 16376, 32748, 65488, 130960,---
13  A168084  Fibonacci 13-step    1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8191, 16381, 32760, 65516, 131024,---
14  A220469  Fibonacci 14-step    1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16383, 32765, 65528, 131052,---
15  A220493  Pentadecanacci       1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32767, 65533, 131064,---
16  A249169  Fibonacci 16-step    1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65535, 131069,---
---
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 2019-2-9 13:20:26 | 显示全部楼层
长寿命兔子数
n=17,18,19 兔子数(寿命为n,先生后死)在OEIS没有录入,没有特殊性,以后可能不会再录入
这里补充给出相应数列,供参考
n=16重录,用作对比
16 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192,16384, 32768, 65535, 131069, 262136, 524268, 1048528, 2097040,---
17 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192,16384, 32768, 65536, 131071, 262141, 524280, 1048556, 2097104 ---
18 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192,16384, 32768, 65536, 131072, 262143, 524285, 1048568, 2097132.---
19 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192,16384, 32768, 65536, 131072, 262144, 524287, 1048573, 2097144.---

点评

该说的差不多都说了,该转话题了,我以此提练习Mathematica  发表于 2019-2-9 17:08
向 dlpg070 致敬!佩服您的执着精神,把互不相干的事件混为一体,是我们的侧重点。  发表于 2019-2-9 13:51
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 2019-2-10 16:28:43 | 显示全部楼层
dlpg070 发表于 2019-2-9 13:20
长寿命兔子数
n=17,18,19 兔子数(寿命为n,先生后死)在OEIS没有录入,没有特殊性,以后可能不会再录入
...

         对 ”兔子数列“,我特别宠爱这一串。 A000045        
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181,
6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229,
832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817,
39088169, 63245986, 102334155, 165580141, 267914296,433494437, 701408733,
  
\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=1\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=2\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=3\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}=5\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{6}}=8\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{7}}=13\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{8}}=21\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{9}}=34\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{10}}=55\)

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 2019-2-10 16:46:37 | 显示全部楼层
本帖最后由 王守恩 于 2019-2-10 19:01 编辑
王守恩 发表于 2019-2-10 16:28
对 ”兔子数列“,我特别宠爱这一串。 A000045        
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 1 ...



      A006355
2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194,
5168, 8362, 13530, 21892, 35422, 57314, 92736, 150050, 242786, 392836,
635622, 1028458, 1664080, 2692538, 4356618, 7049156, 11405774, 18454930,
29860704, 48315634,78176338, 126491972, 20466831, 331160282, 535828592, ....
  
\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{1}}=2\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=4\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=6\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=10\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}=16\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{6}}=26\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{7}}=42\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{8}}=68\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{9}}=110\)

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 2019-2-10 16:56:32 | 显示全部楼层
王守恩 发表于 2019-2-10 16:46
A006355
2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194,
5168, ...

      A022087
4, 8, 12, 20, 32, 52, 84, 136, 220, 356, 576, 932, 1508, 2440,
3948, 6388, 10336, 16724, 27060, 43784, 70844, 114628, 185472
300100,  485572, 785672, 1271244, 2056916, 3328160, 5385076, 8713236,
14098312, 22811548, 36909860, 59721408, 96631268, 156352676,252983944,

  
\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{0}}=4\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{1}}=8\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=12\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=20\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=32\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}=52\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{6}}=84\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{7}}=136\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{8}}=220\)
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 2019-2-10 17:10:02 | 显示全部楼层
本帖最后由 王守恩 于 2019-2-10 19:02 编辑
王守恩 发表于 2019-2-10 16:56
A022087
4, 8, 12, 20, 32, 52, 84, 136, 220, 356, 576, 932, 1508, 2440,
3948, 6388, 10336 ...


      A022091
8, 16, 24, 40, 64, 104, 168, 272, 440, 712, 1152, 1864, 3016,
4880, 7896, 12776, 20672, 33448, 54120, 87568, 141688, 229256,
370944, 600200, 971144, 1571344, 2542488, 4113832, 6656320, 10770152,
17426472, 28196624, 45623096, 73819720, 119442816, 193262536, 312705352,

\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-1}}=8\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{0}}=16\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{1}}=24\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=40\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=64\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=104\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}=168\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{6}}=272\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{7}}=440\)
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 2019-2-10 17:31:32 | 显示全部楼层
本帖最后由 王守恩 于 2019-2-10 19:03 编辑
王守恩 发表于 2019-2-10 17:10
A022091
8, 16, 24, 40, 64, 104, 168, 272, 440, 712, 1152, 1864, 3016,
4880, 7896, 1277 ...


      A022350
16, 32, 48, 80, 128, 208, 336, 544, 880, 1424, 2304, 3728,
6032, 9760, 15792, 25552, 41344, 66896, 108240, 175136,
283376, 458512, 741888, 1200400, 1942288, 3142688, 5084976,
8227664, 13312640, 21540304, 34852944, 56393248, 91246192,....

\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-2}}=16\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-1}}=32\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{0}}=48\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{1}}=80\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=128\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=208\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=336\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}=544\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{6}}=880\)
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 2019-2-10 17:40:12 | 显示全部楼层
本帖最后由 王守恩 于 2019-2-10 17:43 编辑
王守恩 发表于 2019-2-10 17:31
A022350
16, 32, 48, 80, 128, 208, 336, 544, 880, 1424, 2304, 3728,
6032, 9760, 15792, 25 ...


      A022366
32, 64, 96, 160, 256, 416, 672, 1088, 1760, 2848,
4608, 7456, 12064, 19520, 31584, 51104, 82688,
133792, 216480, 350272, 566752, 917024, 1483776,
2400800, 3884576, 6285376, 10169952, 16455328

\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-3}}=32\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-2}}=64\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-1}}=96\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{0}}=160\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{1}}=256\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=416\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=672\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=1088\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}=1760\)
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 2019-2-10 17:53:13 | 显示全部楼层
王守恩 发表于 2019-2-10 17:40
A022366
32, 64, 96, 160, 256, 416, 672, 1088, 1760, 2848,
4608, 7456, 12064, 19520, 3 ...


64, 128, 192, 320, 512, 832, 1344, 2176, 3520, 5696, 9216

\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-4}}=64\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-3}}=128\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-2}}=192\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-1}}=320\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{0}}=512\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{1}}=832\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=1344\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=2176\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=3520\)
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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