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.......
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=012345 6 7 8 9 10 11 12 13 14 15 16 17 18 ---
2 A000045Fibonacci number 1, 1, 2, 3, 5,8,13,21,34, 55,89, 144, 233,377, 610, 987, 1597, 2584, 4181,---
3 A000073tribonacci number 1, 1, 2, 4, 7, 13, 24,44,81,149, 274, 504, 927,1705, 3136,5768,10609, 19513, 35890, ---
4 A000078tetranacci number 1, 1, 2, 4, 8, 15, 29,56, 108, 208, 401, 773,1490, 2872, 5536, 10671, 20569, 39648, 76424, ---
5 A001591pentanacci number1, 1, 2, 4, 8, 16, 31,61, 120, 236, 464, 912,1793, 3525, 6930, 13624, 26784, 52656, 103519,---
6 A001592hexanacci number 1, 1, 2, 4, 8, 16, 32,63, 125, 248, 492, 976,1936, 3840, 7617, 15109, 29970, 59448, 117920,---
7 A066178heptanacci number1, 1, 2, 4, 8, 16, 32,64, 127,253, 504,1004, 2000, 3984, 7936, 15808, 31489, 62725, 124946,---
8 A079262Fibonacci 8-step 1, 1, 2, 4, 8, 16, 32, 64, 128, 255, 509, 1016, 2028, 4048, 8080, 16128, 32192, 64256, 128257,---
9 A104144Fibonacci 9-step 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1021, 2040, 4076, 8144, 16272, 32512, 64960, 129792,---
10A122265Fibonacci 10-step 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1023, 2045, 4088, 8172, 16336, 32656, 65280, 130496,---
11A168082Fibonacci 11-step 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2047, 4093, 8184, 16364, 32720, 65424, 130816,---
12A168083Fibonacci 12-step 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4095, 8189, 16376, 32748, 65488, 130960,---
13A168084Fibonacci 13-step 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8191, 16381, 32760, 65516, 131024,---
14A220469Fibonacci 14-step 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16383, 32765, 65528, 131052,---
15A220493Pentadecanacci 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32767, 65533, 131064,---
16A249169Fibonacci 16-step 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65535, 131069,---
---
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.---
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,
\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=1\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=2\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=3\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}=5\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{6}}=8\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{7}}=13\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{8}}=21\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{9}}=34\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{10}}=55\)
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, ....
\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{1}}=2\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=4\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=6\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=10\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}=16\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{6}}=26\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{7}}=42\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{8}}=68\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{9}}=110\)
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,
\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{0}}=4\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{1}}=8\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=12\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=20\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=32\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}=52\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{6}}=84\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{7}}=136\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{8}}=220\)
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,
\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-1}}=8\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{0}}=16\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{1}}=24\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=40\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=64\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=104\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}=168\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{6}}=272\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{7}}=440\)
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,....
\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-2}}=16\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-1}}=32\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{0}}=48\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{1}}=80\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=128\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=208\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=336\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}=544\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{6}}=880\)
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
\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-3}}=32\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-2}}=64\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-1}}=96\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{0}}=160\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{1}}=256\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=416\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=672\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=1088\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{5}}=1760\)
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
\(\D f(1)=\sum_{k=0}^{1}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-4}}=64\)
\(\D f(2)=\sum_{k=0}^{2}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-3}}=128\)
\(\D f(3)=\sum_{k=0}^{3}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-2}}=192\)
\(\D f(4)=\sum_{k=0}^{4}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{-1}}=320\)
\(\D f(5)=\sum_{k=0}^{5}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{0}}=512\)
\(\D f(6)=\sum_{k=0}^{6}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{1}}=832\)
\(\D f(7)=\sum_{k=0}^{7}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{2}}=1344\)
\(\D f(8)=\sum_{k=0}^{8}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{3}}=2176\)
\(\D f(9)=\sum_{k=0}^{9}\frac{(1+\sqrt{5})^{k}+(1-\sqrt{5})^{k}}{2^{4}}=3520\)