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楼主: 王守恩

[原创] 数字串的通项公式

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 楼主| 发表于 2023-7-4 14:14:06 | 显示全部楼层
  1, 5, 10, 14, 19, 23, 28, 32, 37, 41, 46, 50, 55, 59, 64, 68, 73,77, 82, 86, 91, 95, 100, 104, 109, 113,
  1, 0, 5, 0, 10, 0, 14, 0, 19, 0, 23, 0, 28, 0, 32, 0, 37, 0, 41, 0, 46, 0, 50, 0, 55, 0, 59, 0, 64, 0, 68, 0, 73,
  1, 0, 0, 5, 0, 0, 10, 0, 0, 14, 0, 0, 19, 0, 0, 23, 0, 0, 28, 0, 0, 32, 0, 0, 37, 0, 0, 41, 0,  0, 46, 0, 0, 50, 0, 0,                 
  1, 0, 0, 0, 5, 0, 0, 0, 10, 0, 0, 0, 14, 0, 0, 0, 19, 0, 0, 0, 23, 0, 0, 0, 28, 0, 0, 0, 32, 0, 0, 0, 37, 0, 0, 0, 41, 0,
   
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 2023-7-4 14:32:40 | 显示全部楼层
王守恩 发表于 2023-7-3 19:34
边长为n(正整数)的正方形ABCD内动点P,三角形ABP三边长为整数,问动点P可能有几个?

得到一串数: 1,3,8, ...

a(8)=50
{{1,8},{2,7},{2,8},{3,6},{3,7},{3,8},{4,5},{4,6},{4,7},{4,8},{5,4},{5,5},{5,6},{5,7},{5,8},{5,9},{6,3},{6,4},{6,5},{6,6},{6,7},{6,8},{6,9},{7,2},{7,3},{7,4},{7,5},{7,6},{7,7},{7,8},{7,9},{7,10},{8,1},{8,2},{8,3},{8,4},{8,5},{8,6},{8,7},{8,8},{8,9},{8,10},{8,11},{9,5},{9,6},{9,7},{9,8},{10,7},{10,8},{11,8}}
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 2023-7-4 14:34:05 | 显示全部楼层
  1. Table[Length@Select[Tuples[Range@(Sqrt[2]*n),{2}],Total@#>n&&#[[1]]^2-n^2<#[[2]]^2<(#[[1]]^2+n^2)&&(#[[1]]^2-#[[2]]^2)^2+5n^4>2(#[[1]]^2+#[[2]]^2)*n^2&],{n,40}]
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{1,3,8,12,21,29,40,50,66,82,99,113,138,160,183,207,239,265,294,324,361,399,434,466,511,558,597,639,692,740,789,837,894,953,1004,1062,1125,1185,1254,1314,1385,1447,1523,1591,1664,1742,1821,1897,1976,2058}
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 2023-7-4 14:57:42 | 显示全部楼层
王守恩 发表于 2023-7-4 14:14
1, 5, 10, 14, 19, 23, 28, 32, 37, 41, 46, 50, 55, 59, 64, 68, 73,77, 82, 86, 91, 95, 100, 104, 109 ...

  1. Array[CoefficientList[Series[(2x^#+1)^2/((x^#-1)^2(x^#+1)),{x,0,29}],x]&,4]
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$\{\{1,5,10,14,19,23,28,32,37,41,46,50,55,59,64,68,73,77,82,86,91,95,100,104,109,113,118,122,127,131\},\{1,0,5,0,10,0,14,0,19,0,23,0,28,0,32,0,37,0,41,0,46,0,50,0,55,0,59,0,64\},\{1,0,0,5,0,0,10,0,0,14,0,0,19,0,0,23,0,0,28,0,0,32,0,0,37,0,0,41\},\{1,0,0,0,5,0,0,0,10,0,0,0,14,0,0,0,19,0,0,0,23,0,0,0,28,0,0,0,32\}\}$
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
 楼主| 发表于 2023-7-4 15:35:03 | 显示全部楼层
northwolves 发表于 2023-7-4 14:57
$\{\{1,5,10,14,19,23,28,32,37,41,46,50,55,59,64,68,73,77,82,86,91,95,100,104,109,113,118,122 ...
  1. Table[CoefficientList[Series[(2 x^n + 1)^2/((x^n + 1) (x^n - 1)^2), {x, 0, 29 n}], x], {n,1, 4}]
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{{1, 5, 10, 14, 19, 23, 28, 32, 37, 41, 46, 50, 55, 59, 64, 68, 73, 77, 82, 86, 91, 95, 100, 104, 109, 113, 118, 122, 127, 131},
{1, 0, 5, 0, 10, 0, 14, 0, 19, 0, 23, 0, 28, 0, 32, 0, 37, 0, 41, 0, 46, 0, 50, 0, 55, 0, 59, 0, 64, 0, 68, 0, 73, 0, 77, 0, 82, 0, 86, 0, 91, 0, 95, 0, 100, 0, 104, 0, 109, 0, 113, 0, 118, 0, 122, 0, 127, 0, 131},
{1, 0, 0, 5, 0, 0, 10, 0, 0, 14, 0, 0, 19, 0, 0, 23, 0, 0, 28, 0, 0, 32, 0, 0, 37, 0, 0, 41, 0, 0, 46, 0, 0, 50, 0, 0, 55, 0, 0, 59, 0, 0, 64, 0, 0,
68, 0, 0, 73, 0, 0, 77, 0, 0, 82, 0, 0, 86, 0, 0, 91, 0, 0, 95, 0, 0, 100, 0, 0, 104, 0, 0, 109, 0, 0, 113, 0, 0, 118, 0, 0, 122, 0, 0, 127, 0, 0, 131},
{1, 0, 0, 0, 5, 0, 0, 0, 10, 0, 0, 0, 14, 0, 0, 0, 19, 0, 0, 0, 23, 0, 0, 0, 28, 0, 0, 0, 32, 0, 0, 0, 37, 0, 0, 0, 41, 0, 0, 0, 46, 0, 0, 0, 50, 0, 0, 0, 55, 0, 0, 0, 59, 0, 0, 0, 64, 0, 0, 0,
68, 0, 0, 0, 73, 0, 0, 0, 77, 0, 0, 0, 82, 0, 0, 0, 86, 0, 0, 0, 91, 0, 0, 0, 95, 0, 0, 0, 100, 0, 0, 0, 104, 0, 0, 0, 109, 0, 0, 0, 113, 0, 0, 0, 118, 0, 0, 0, 122, 0, 0, 0, 127, 0, 0, 0, 131}}

点评

Table & Array功能一样,Array似乎更简单些  发表于 2023-7-4 15:48
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 2023-7-4 18:15:46 | 显示全部楼层
  1. Table[Length@Select[Tuples[Range@(Sqrt[2]*n),{2}],Total@#>n&&#[[1]]^2-n^2<=#[[2]]^2<=(#[[1]]^2+n^2)&&(#[[1]]^2-#[[2]]^2)^2+5n^4>2(#[[1]]^2+#[[2]]^2)*n^2&],{n,100}]
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{1,3,8,14,21,29,40,52,66,82,99,117,138,160,185,209,239,265,294,326,363,399,434,472,511,558,597,641,692,742,789,839,894,953,1006,1066,1125,1185,1254,1318,1385,1449,1523,1593,1668,1742,1821,1903,1976,2058,2146,2234,2313,2399,2502,2592,2677,2771,2872,2973,3064,3170,3279,3383,3488,3592,3705,3820,3931,4045,4166,4282,4399,4521,4644,4768,4900,5028,5155,5285,5426,5560,5685,5829,5975,6115,6254,6396,6555,6693,6842,6994,7155,7306,7455,7619,7784,7946,8103,8265}

评分

参与人数 1威望 +9 金币 +9 贡献 +9 经验 +9 鲜花 +9 收起 理由
王守恩 + 9 + 9 + 9 + 9 + 9 正确!

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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 2023-7-4 18:16:33 | 显示全部楼层
整数的运算比小数还是要快得多

评分

参与人数 1威望 +3 金币 +3 贡献 +3 经验 +3 鲜花 +3 收起 理由
王守恩 + 3 + 3 + 3 + 3 + 3 40就用了1天。

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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 2023-7-4 23:03:40 | 显示全部楼层
用 Subsets还能再稍快些:

  1. Table[Floor[Sqrt[5]*n/2]-Ceiling[(n-1)/2]+2Length@Select[Select[Subsets[Range@(Sqrt[2]*n),{2}],Total@#>n&&#[[2]]^2<=#[[1]]^2+n^2&],(#[[1]]^2-#[[2]]^2)^2+5n^4>2(#[[1]]^2+#[[2]]^2)*n^2&],{n,100}]
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点评

Ceiling[(n-1)/2]=Floor[n/2]  发表于 2023-7-5 10:33
都是有实力之人,我都不知道,在探讨什么问题。  发表于 2023-7-4 23:18
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
 楼主| 发表于 2023-7-5 09:07:51 | 显示全部楼层
边长为n(正整数)的正三角形ABC内动点P,三角形ABP三边长为整数,问动点P可能有几个?

得到一串数(OEIS好像没有): 1,1,2,4,7,11,16,20,21,27,34,42,51,61,72,78,85,97,110,124,...,
  1. Table[Dimensions@Flatten[Table[NSolve[{m/Sin[a]==n/Sin[b]==k/Sin[a+b],\[Pi]/3≥b>0,\[Pi]/3≥a>0},{a,b}],{m,1,k},{n,1,k}]]/2,{k,1,20}]
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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
 楼主| 发表于 2023-7-5 14:16:13 | 显示全部楼层
上底(AB)为n(正整数)的等腰梯形(底角=60,上底=高)内动点P,三角形ABP三边长为整数,问动点P可能有几个?

得到一串数(OEIS好像没有):1,3,10,16,29,37,48,68,84,110,127,147,182,206,245,273,315,347,380,432,...,
  1. Table[Dimensions@Flatten[Table[NSolve[{m/Sin[a]==n/Sin[b]==k/Sin[a+b],(m*n)/(k*k)≤Sin[\[Pi]/2]/Sin[a+b],2\[Pi]/3≥a>0,2\[Pi]/3≥b>0},{a,b}],{m,1,Sqrt[2]k},{n,1,Sqrt[2]k}]]/2,{k,1,20}]
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算式没问题,就是太慢了。
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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