本帖最后由 dlpg070 于 2020-3-10 15:00 编辑
1#的我信手按回文展开得到的数列
northwolves 在10# ,11# 给出了通项公式
23# 给出了 A[n,k] 公式
下面给出我认为有趣的原因,希望专家指点迷津:
1#数列在OEIS中查不到
但是如果在OEIS中查找
给定n的数列 则有许多
例如 n=4: 4,3,3,2,2,1,1
49个结果
对于任意的n(<10),都可以找到,大大意外
它们有什么关系?
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我们看2个典型的数列:
例1:A194543 Triangle T (n, k), n >= 0, 0 <= k <= n,
read by rows :
T (n, k) is the number of partitions of n
into parts p_i such that |p_i - p_j | >= k for i != j.
代码:
- Clear["Global`*"]
- b[n_,i_,k_]:=b[n,i,k]=If[n<0,0,If[n==0,1,
- Sum[b[n-i-j,i+j,k],{j,k,n-i}]]];
- T[n_,k_]:=If[n==0,1,0]+Sum[b[n-i,i,k],{i,1,n}];
- t=Table[Table[T[n,k],{k,0,n}],{n,0,30}];
- (*//Flatten *)
- (*Jean-Fran?ois Alcover,Jan 19 2015,after Alois P.Heinz*)
- Column[t,Left]
复制代码
结果
{1}
{1,1}
{2,1,1}
{[3,2,1,1}
{5,2,2,1,1}
{7,3,2,2,1,1}
{11,4,3,2,2,1,1}
{15,5,3,3,2,2,1,1}
{22,6,4,3,3,2,2,1,1}
{30,8,5,4,3,3,2,2,1,1}
{42,10,6,4,4,3,3,2,2,1,1}
{56,12,7,5,4,4,3,3,2,2,1,1}
{77,15,9,6,5,4,4,3,3,2,2,1,1}
{101,18,10,7,5,5,4,4,3,3,2,2,1,1}
{135,22,12,8,6,5,5,4,4,3,3,2,2,1,1}
{176,27,14,10,7,6,5,5,4,4,3,3,2,2,1,1}
{231,32,17,11,8,6,6,5,5,4,4,3,3,2,2,1,1}
{297,38,19,13,9,7,6,6,5,5,4,4,3,3,2,2,1,1}
{385,46,23,15,11,8,7,6,6,5,5,4,4,3,3,2,2,1,1}
{490,54,26,17,12,9,7,7,6,6,5,5,4,4,3,3,2,2,1,1}
{627,64,31,19,14,10,8,7,7,6,6,5,5,4,4,3,3,2,2,1,1}
{792,76,35,22,16,12,9,8,7,7,6,6,5,5,4,4,3,3,2,2,1,1}
{1002,89,41,25,18,13,10,8,8,7,7,6,6,5,5,4,4,3,3,2,2,1,1}
{1255,104,46,28,20,15,11,9,8,8,7,7,6,6,5,5,4,4,3,3,2,2,1,1}
{1575,122,54,32,23,17,13,10,9,8,8,7,7,6,6,5,5,4,4,3,3,2,2,1,1}
{1958,142,61,36,25,19,14,11,9,9,8,8,7,7,6,6,5,5,4,4,3,3,2,2,1,1}
{2436,165,70,41,28,21,16,12,10,9,9,8,8,7,7,6,6,5,5,4,4,3,3,2,2,1,1}
{3010,192,79,46,31,24,18,14,11,10,9,9,8,8,7,7,6,6,5,5,4,4,3,3,2,2,1,1}
{3718,222,91,52,35,26,20,15,12,10,10,9,9,8,8,7,7,6,6,5,5,4,4,3,3,2,2,1,1}
{4565,256,102,58,38,29,22,17,13,11,10,10,9,9,8,8,7,7,6,6,5,5,4,4,3,3,2,2,1,1}
{5604,296,117,66,43,32,25,19,15,12,11,10,10,9,9,8,8,7,7,6,6,5,5,4,4,3,3,2,2,1,1}
包含1#数列的各个子数列,谁能解其中奥妙?
如 2,1,1
3,2,2,1,1
---
10,9,9,8,8,---,2,2,1,1
---
例1:A063995 Irregular triangle read by rows :
T (n, k), n >= 1, -(n - 1) <= k <= n - 1, =
number of partitions of n with rank k
代码:
- t=Table[Count[(First[#]-Length[#]&/@IntegerPartitions[k]),#]&/@Range[-k+1,k-1],{k,16}];
- Column[t,Center]
- Column[t,Left]
复制代码
结果:
{1}
{1,0,1}
{1,0,1,0,1}
{1,0,1,1,1,0,1}
{1,0,1,1,1,1,1,0,1}
{1,0,1,1,2,1,2,1,1,0,1}
{1,0,1,1,2,1,3,1,2,1,1,0,1}
{1,0,1,1,2,2,3,2,3,2,2,1,1,0,1}
{1,0,1,1,2,2,3,3,4,3,3,2,2,1,1,0,1}
{1,0,1,1,2,2,4,3,5,4,5,3,4,2,2,1,1,0,1}
{1,0,1,1,2,2,4,3,6,5,6,5,6,3,4,2,2,1,1,0,1}
{1,0,1,1,2,2,4,4,6,6,8,7,8,6,6,4,4,2,2,1,1,0,1}
{1,0,1,1,2,2,4,4,6,7,9,8,11,8,9,7,6,4,4,2,2,1,1,0,1}
{1,0,1,1,2,2,4,4,7,7,10,10,13,11,13,10,10,7,7,4,4,2,2,1,1,0,1}
{1,0,1,1,2,2,4,4,7,7,11,11,15,14,16,14,15,11,11,7,7,4,4,2,2,1,1,0,1}
{1,0,1,1,2,2,4,4,7,8,11,12,17,16,20,19,20,16,17,12,11,8,7,4,4,2,2,1,1,0,1}
----------
The partition 5 = 4+1 has largest summand 4 and 2 summands, hence has rank 4-2 = 2.
Triangle begins:
[ 1] 1,
[ 2] 1, 0, 1,
[ 3] 1, 0, 1, 0, 1,
[ 4] 1, 0, 1, 1, 1, 0, 1,
[ 5] 1, 0, 1, 1, 1, 1, 1, 0, 1,
[ 6] 1, 0, 1, 1, 2, 1, 2, 1, 1, 0, 1,
[ 7] 1, 0, 1, 1, 2, 1, 3, 1, 2, 1, 1, 0, 1,
[ 8] 1, 0, 1, 1, 2, 2, 3, 2, 3, 2, 2, 1, 1, 0, 1,
[ 9] 1, 0, 1, 1, 2, 2, 3, 3, 4, 3, 3, 2, 2, 1, 1, 0, 1,
[10] 1, 0, 1, 1, 2, 2, 4, 3, 5, 4, 5, 3, 4, 2, 2, 1, 1, 0, 1,
[11] 1, 0, 1, 1, 2, ...
这个数列输出对称,含有许多 4,3,3,2,2,1,1等
左侧是反向排列1,1,2,2,3,3,4 等
没想到此数列最新修改 Jan 23 2020
没想到 11 22 33 44 或44 33 22 11 有这么奇妙!
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