$n=0,a_{n}=0$
$n>0,a_{n}=2*floor(en!)-n!-1$
本帖最后由 王守恩 于 2019-4-24 19:46 编辑
王守恩 发表于 2019-3-28 16:22
8 个数有 40320 道题，应该有普遍的通项公式！各位大侠可有线索？谢谢！
\(a_{1},a_{2},a_{3},a_{4}, ...
82#通项公式。 n 从“1”开始。
\(a(n)=\mod(n-1,8)+\lfloor\frac{n-1}{8}\rfloor+5\lfloor\frac{n-2}{8}\rfloor-6\lfloor\frac{n-3}{8}\rfloor+3\lfloor\frac{n-4}{8}\rfloor-4\lfloor\frac{n-5}{8}\rfloor+\lfloor\frac{n-6}{8}\rfloor-2\lfloor\frac{n-7}{8}\rfloor+3\lfloor\frac{n-8}{8}\rfloor+1\)
本帖最后由 王守恩 于 2019-4-25 07:22 编辑
王守恩 发表于 2019-3-30 03:22
可以有 9 个数 362880 道题的通项公式。
\(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7},a_{8},a_{9} ...
84#的通项公式 。9 个数 362880 道题的通项公式 。 n 从“1”开始。
譬如：\(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7},a_{8},a_{9}=4,7,5,9,2,6,8,1,3\)
\(a(n)=\mod(n-1,9)+10\lfloor\frac{n-1}{9}\rfloor+2\lfloor\frac{n-2}{9}\rfloor-3\lfloor\frac{n-3}{9}\rfloor+3\lfloor\frac{n-4}{9}\rfloor-8\lfloor\frac{n-5}{9}\rfloor+3\lfloor\frac{n-6}{9}\rfloor+\lfloor\frac{n-7}{9}\rfloor-8\lfloor\frac{n-8}{9}\rfloor+\lfloor\frac{n-9}{9}\rfloor-5\)
本帖最后由 王守恩 于 2019-4-25 10:30 编辑
王守恩 发表于 2019-3-30 03:47
可以有 k 个数的通项公式。
\(a_{1},a_{2},a_{3},a_{4},a_{5},\cdots\cdots,a_{k-2},a_{k-1},a_{k}\)...
85#通项公式。可以有 k 个数的通项公式。 n 从“1”开始。
譬如：\(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7},a_{8},a_{9},a_{10}=11,14,18,23,19,14,8,17,9,20\)
\(a(n)=\mod(n-1,10)+10\lfloor\frac{n}{10}\rfloor+\lfloor\frac{n-1}{10}\rfloor+2\lfloor\frac{n-2}{10}\rfloor+3\lfloor\frac{n-3}{10}\rfloor+4\lfloor\frac{n-4}{10}\rfloor-5\lfloor\frac{n-5}{10}\rfloor-6\lfloor\frac{n-6}{10}\rfloor-7\lfloor\frac{n-7}{10}\rfloor+8\lfloor\frac{n-8}{10}\rfloor-9\lfloor\frac{n-9}{10}\rfloor+1\)
本帖最后由 王守恩 于 2019-4-25 10:27 编辑
王守恩 发表于 2019-3-30 03:52
一般地，可以有 k 个任意数的通项公式。这是我个人的想法，更是论坛共同的财富！谢谢各位大侠！
86#通项公式。一般地，可以有 k 个任意数的通项公式。谢谢各位大侠！
譬如：\(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7},a_{8},a_{9},a_{10},a_{11},a_{12},a_{13}=27,18,28,1,82,84,59,04,52,35,36,02,87\)
\(a(n)=\mod(n-1,13)+84\lfloor\frac{n}{13}\rfloor-47\lfloor\frac{n-1}{13}\rfloor-10\lfloor\frac{n-2}{13}\rfloor+9\lfloor\frac{n-3}{13}\rfloor-28\lfloor\frac{n-4}{13}\rfloor\)
\(+80\lfloor\frac{n-5}{13}\rfloor
+\lfloor\frac{n-6}{13}\rfloor-26\lfloor\frac{n-7}{13}\rfloor-56\lfloor\frac{n-8}{13}\rfloor+47\lfloor\frac{n-9}{13}\rfloor-18\lfloor\frac{n-10}{13}\rfloor-35\lfloor\frac{n-12}{13}\rfloor-9\)
n 从“1”开始。
本帖最后由 王守恩 于 2019-4-25 18:30 编辑
王守恩 发表于 2019-3-30 03:52
一般地，可以有 k 个任意数的通项公式。这是我个人的想法，更是论坛共同的财富！谢谢各位大侠！
小结：可以有 k 个任意数的通项公式。谢谢各位大侠！
譬如：\(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7},a_{8},a_{9},a_{10},a_{11},a_{12},a_{13},a_{14},a_{15},a_{16},a_{17}=27,28,17,18,2,3,4,5,90,4,5,2,3,53,60,28,7\)
\(a(n)=\mod(n-1,17)-22\lfloor\frac{n}{17}\rfloor+37\lfloor\frac{n-1}{17}\rfloor-12\lfloor\frac{n-3}{17}\rfloor-17\lfloor\frac{n-5}{17}\rfloor+84\lfloor\frac{n-9}{17}\rfloor
-87\lfloor\frac{n-10}{17}\rfloor
-4\lfloor\frac{n-12}{17}\rfloor+49\lfloor\frac{n-14}{17}\rfloor+6\lfloor\frac{n-15}{17}\rfloor-33\lfloor\frac{n-16}{17}\rfloor+13\)
n 从“1”开始。
northwolves 发表于 2019-4-23 22:10
$n=0,a_{n}=0$
$n>0,a_{n}=2*floor(en!)-n!-1$
这串数可以有通项公式吗？
7, 11, 13, 14, 19, 21, 22, 25, 26, 28, 35, 37, 38, 41, 42, 44, 49, 50,
52, 56, 67, 69, 70, 73, 74, 76, 81, 82, 84, 88, 97, 98, 100, 104, 112,
131, 133, 134, 137, 138, 140, 145, 146, 148, 152, 161, 162, 164, 168,
176, 193, 194, 196, 200, 208, 224, 259, 261, 262, 265, 266, 268, 273,
274, 276, 280, 289, 290, 292, 296, 304, ..........
王守恩 发表于 2019-5-19 09:00
这串数可以有通项公式吗？
7, 11, 13, 14, 19, 21, 22, 25, 26, 28, 35, 37, 38, 41, 42, 44, 49, 50, ...
当然有通项,老问题,见 A014311
dlpg070 发表于 2019-5-19 09:47
当然有通项,老问题,见 A014311
a(n) =a(n)= 2^u + 2^v + 2^w
——
a(01)= 07 = 2^2 + 2^1 + 2^0
——
a(02)= 11 = 2^3 + 2^1 + 2^0
a(03)= 13 = 2^3 + 2^2 + 2^0
a(04)= 14 = 2^3 + 2^2 + 2^1
——
a(05)= 19 = 2^4 + 2^1 + 2^0
a(06)= 21 = 2^4 + 2^2 + 2^0
a(07)= 22 = 2^4 + 2^2 + 2^1
a(08)= 25 = 2^4 + 2^3 + 2^0
a(09)= 26 = 2^4 + 2^3 + 2^1
a(10)= 28 = 2^4 + 2^3 + 2^2
——
a(11)= 35 = 2^5 + 2^1 + 2^0
a(12)= 37 = 2^5 + 2^2 + 2^0
a(13)= 38 = 2^5 + 2^2 + 2^1
a(14)= 41 = 2^5 + 2^3 + 2^0
a(15)= 42 = 2^5 + 2^3 + 2^1
a(16)= 44 = 2^5 + 2^3 + 2^2
a(17)= 49 = 2^5 + 2^4 + 2^0
a(18)= 50 = 2^5 + 2^4 + 2^1
a(19)= 52 = 2^5 + 2^4 + 2^2
a(20)= 56 = 2^5 + 2^4 + 2^3
——
a(21)= 67 = 2^6 + 2^1 + 2^0
a(22)= 69 = 2^6 + 2^2 + 2^0
a(23)= 70 = 2^6 + 2^2 + 2^1
a(24)= 73 = 2^6 + 2^3 + 2^0
a(25)= 74 = 2^6 + 2^3 + 2^1
a(26)= 76 = 2^6 + 2^3 + 2^2
a(27)= 81 = 2^6 + 2^4 + 2^0
a(28)= 82 = 2^6 + 2^4 + 2^1
a(29)= 84 = 2^6 + 2^4 + 2^2
a(30)= 88 = 2^6 + 2^4 + 2^3
a(31)= 97 = 2^6 + 2^5 + 2^0
a(32)= 98 = 2^6 + 2^5 + 2^1
a(33)=100= 2^6 + 2^5 + 2^2
a(34)=104= 2^6 + 2^5 + 2^3
a(35)=112= 2^6 + 2^5 + 2^4
——
a(36)=131= 2^7 + 2^1 + 2^0
本帖最后由 dlpg070 于 2019-5-19 11:32 编辑
王守恩 发表于 2019-5-19 10:37
a(n) =a(n)= 2^u + 2^v + 2^w
——
a(01)= 07 = 2^2 + 2^1 + 2^0
转贴一段代码
Select, (Count, 1] == 3) &]
nn = 8; Flatten[
Table[2^i + 2^j + 2^k, {i, 2, nn}, {j, 1, i - 1}, {k, 0,
j - 1}]] (*T.D.Noe,Nov 05 2013*)
{7, 11, 13, 14, 19, 21, 22, 25, 26, 28, 35, 37, 38, 41, 42, 44, 49, \
50, 52, 56, 67, 69, 70, 73, 74, 76, 81, 82, 84, 88, 97, 98, 100, 104, \
112, 131, 133, 134, 137, 138, 140, 145, 146, 148, 152, 161, 162, 164, \
168, 176, 193, 194, 196, 200, 208, 224, 259, 261, 262, 265, 266, 268, \
273, 274, 276, 280, 289, 290, 292, 296, 304, 321, 322, 324, 328, 336, \
352, 385, 386, 388, 392, 400, 416, 448}