{a(n)}+{b(n)}+{c(n)}能跑遍所有正整数吗？

王守恩

{a(n)}+{b(n)}+{c(n)}= N 吗？为什么？
这里N为自然数集，+表示集合的并，因为三者的两两之交为空。

{a(n)}={1, 4, 07, 09, 12, 15, 16, 19, 22, 25, 27, 30, 32, 34, 37, 40, 43, 45, 47, 50, 52, 55, 58, 60, 63, 65,...}
{b(n)}={2, 5, 08, 11, 14, 18, 20, 23, 26, 29, 33, 36, 38, 41, 44, 48, 51, 54, 56, 59, 62, 66, 69, 72, 75, 77,...}
{c(n)}={3, 6, 10, 13, 17, 21, 24, 28, 31, 35, 39, 42, 46, 49, 53, 57, 61, 64, 67, 71, 74, 79, 82, 85, 89, 92,...}

\(\D a(n)=n+\lfloor n*\sqrt[4]{1/2}\rfloor+\lfloor n*\sqrt[4]{1/4}\rfloor\)

\(\D b(n)=n+\lfloor n*\sqrt[4]{1/2}\rfloor+\lfloor n*\sqrt[4]{2}\rfloor\)

\(\D c(n)=n+\lfloor n*\sqrt[4]{4}\rfloor+\lfloor n*\sqrt[4]{2}\rfloor\)

lsr314

题目出处？

王守恩

lsr314 发表于 2019-10-11 09:10
题目出处？

各位高手！能不能找一个”函数“(我不会)：把 a(n),b(n),c(n)混在一起，
从小到大排列，每个正整数恰好是出现 ”1“ 次(看看有没有漏了的，看看有没有重复的)。
我只是想把所有正整数拆分成若干份，不允许重复，又不想周期循环，大家可有类似参考资料。

lsr314

貌似只要$t>0$且$t^2,t+1/t,t+t^2,1/t+1/t^2$都是无理数，
$a_n=n+[n/t]+[n/t^2],$
$b_n=n+[n/t]+[nt],$
$c_n=n+[nt^2]+[nt],$
$[x]$表示$x$的整数部分，那么每个正整数刚好在$a_n,b_n,c_n$中出现一次。本题中$t=2^(1/4)$.

王守恩

lsr314 发表于 2019-10-11 14:22
貌似只要$t>0$且$t^2,t+1/t,t+t^2,1/t+1/t^2$都是无理数，
$a_n=n+[n/t]+[n/t^2],$
$b_n=n+[n/t]+[nt],$
...

也可以这样：
{a(n)}={1, 4, 7, 10, 13, 16, 19, 22, 24, 27, 30, 33, 36, 39, 42, 45, 47, 49, 52, 55, 58, 61, 64,...}
{b(n)}={2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 46, 50, 53, 56, 59, 62, 65, 68,...}
{c(n)}={3, 6, 9, 12, 15, 18, 21, 25, 28, 31, 34, 37, 40, 43, 48, 51, 54, 57, 60, 63, 66, 69, 73,...}
\(a(n)=n+\lfloor n∗\sqrt\frac{7}{9}\rfloor+\lfloor n∗\sqrt\frac{8}{9}\rfloor\)
\(b(n)=n+\lfloor n∗\sqrt\frac{7}{8}\rfloor+\lfloor n∗\sqrt\frac{9}{8}\rfloor\)
\(c(n)=n+\lfloor n∗\sqrt\frac{8}{7}\rfloor+\lfloor n∗\sqrt\frac{9}{7}\rfloor\)

还可以有吗？
{a(n)}+{b(n)}+{c(n)}+{d(n)}= N
这里N为自然数集，+表示集合的并，因为四者的两两之交为空。

lsr314

如果$t_1,t_2,…,t_k$都是正数，且任意两个数的比值都是无理数，
$a_i(n)=\sum_{j=1}^k[t_j/t_i n],i=1,2,…k.$
那么每个正整数在数列$a_1(n),…a_k(n)$中恰好出现一次。

manthanein

说实话，看了楼主这么多帖子，觉得楼主有拉马努金的潜质

王守恩

{a(n)}+{b(n)}+{c(n)}= N 吗？
{a(n)}+{b(n)}+{c(n)}能跑遍所有正整数吗？
这里N为自然数集，+表示集合的并，因为三者的两两之交为空。

{a(n)}={0, 5, 7, 09, 11, 13, 18, 19, 26, 27, 31, 36, 38, 44, 45, 46, 50, 52, 54, 62, 63, 65, 68,...}
{b(n)}={1, 3, 8, 12, 14, 16, 20, 21, 22, 28, 30, 32, 34, 39, 40, 47, 48, 53, 55, 57, 58, 66, 71,...}
{c(n)}={2, 4, 6, 10, 15, 17, 23, 24, 25, 29, 33, 35, 37, 41, 42, 43, 49, 51, 56, 59, 60, 61, 64,...}

  a：3*(3k+0)，3*c+1，3*b+2,   k=0,1,2,3,...
  b：3*(3k+1)，3*a+1，3*c+2,   k=0,1,2,3,...
  c：3*(3k+2)，3*b+1，3*a+2,   k=0,1,2,3,...

王守恩

[欣赏] 下面的每个数字串(k)都可以跑遍所有自然数！——不重不漏, 又不周期循环。

数字串(0) = {0, 2, 1, 5, 4, 3, 9, 8, 7, 6, 14, 13, 12, 11, 10, 20, 19, 18, 17, 16, 15, 27, 26, 25, 24, 23, 22, 21, 35, 34, 33, 32, 31, 30, 29, 28, 44, 43, 42, 41, 40, 39, 38, 37, 36, 54, 53, 52},
数字串(1) = {1, 0, 5, 4, 3, 2, 11, 10, 9, 8, 7, 6, 19, 18, 17, 16, 15, 14, 13, 12, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 55, 54, 53, 52, 51, 50},
数字串(2) = {2, 1, 0, 8, 7, 6, 5, 4, 3, 17, 16, 15, 14, 13, 12, 11, 10, 9, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 62, 61, 60},
数字串(3) = {3, 2, 1, 0, 11, 10, 9, 8, 7, 6, 5, 4, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 59, 58, 57, 56, 55, 54, 53, 52},
数字串(4) = {4, 3, 2, 1, 0, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32},
数字串(5) = {5, 4, 3, 2, 1, 0, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48},
数字串(6) = {6, 5, 4, 3, 2, 1, 0, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 69, 68, 67, 66, 65, 64},
数字串(7) = {7, 6, 5, 4, 3, 2, 1, 0, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24},
数字串(8) = {8, 7, 6, 5, 4, 3, 2, 1, 0, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33},
数字串(9) = {9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42},
......

A061579——{0, 2, 1, 5, 4, 3, 9, 8, 7, 6, 14, 13, 12, 11, 10, 20, 19, 18, 17, 16, 15, 27, 26, 25, 24, 23, 22, 21, 35, 34, 33, 32, 31, 30, 29, 28, 44, 43, 42, 41, 40, 39, 38, 37, 36, 54, 53, 52},

A061579——有个数字串(0)——可通项公式没我们的好——a(n) = Round[Sqrt[2 n]]^2 - n 。

其它的数字串(k)——OEIS就没有了。

上面的数字串可以有一个统一的通项公式——a(k) = k*Round[Sqrt[(2 n - 1)/k + 1/4]]^2 - n 。—— 精妙在于:  "2"不能改。 "1"不能少。 动"1/4"就漏解。不精妙的就不显臭了。

王守恩

[欣赏] 下面这串数也可以跑遍所有自然数！——不重不漏, 又不周期循环。

OEIS没有这串数。看5个公式之间的差异——拿捏不定。

{1, 0, 3, 2, 6, 5, 4, 9, 8, 7, 13, 12, 11, 10, 17, 16, 15, 14, 22, 21, 20, 19, 18, 27, 26, 25, 24, 23, 33, 32, 31, 30, 29, 28, 39, 38, 37, 36, 35, 34, 46, 45, 44, 43, 42, 41, 40, 53, 52, 51, 50, 49, 48, 47, 61, 60, 59, 58, 57,
56, 55, 54, 69, 68, 67, 66, 65, 64, 63, 62, 78, 77, 76, 75, 74, 73, 72, 71, 70, 87, 86, 85, 84, 83, 82, 81, 80, 79, 97, 96, 95, 94, 93, 92, 91, 90, 89, 88, 107, 106, 105, 104, 103, 102, 101, 100, 99, 98, 118, 117, 116, ...}

M = 20; s = {1}; t = {2}; d = 2; For[i = 2, i <= M, i++, If[i == 2, d = 2, If[OddQ[i], d = d + 1]]; AppendTo[s, s[[i - 1]] + d]; AppendTo[t, If[OddQ[i], t[[i - 1]] + 1, t[[i - 1]]]];]
f = {}; For[i = 1, i <= M, i++, f = Join[f, Range[s[[i]], s[[i]] - t[[i]] + 1, -1]];]  Take[f, 120]

s = 1; t = 2; d = 2; f = {1, 0}; Do[If[i > 2 && OddQ@i, d++]; s += d; t += Boole@OddQ@i; f = Join[f, Range[s, s - t + 1, -1]], {i, 2, 20}]; Take[f, 120]

Flatten[Table[Reverse@Range[Floor[(n + 3)/2]] + Floor[(n^2 + 4 n - 8)/4], {n, 20}]][[;; 120]]

Flatten[Table[Reverse@Range[Floor[(n + 3)/2]] + Floor[(n + 2)^2/4] - 3, {n, 20}]][[;; 120]]

Flatten[Table[Reverse@Range[Ceiling[n/2]] + Floor[(n/2)^2] - 3, {n, 3, 22}]][[;; 120]]