数字串的通项公式

northwolves

王守恩 发表于 2023-5-7 16:18
若 n={2, 8, 10, 12, 18, 20, 28, 30, 32, 40, 42, 48, 50, 60, 68, 70, 72, 78, 80, 88, 90, 98, 102, 10 ...

$a<=b<=c$
$3/2c^4>=a*b*c (a + b + c)/2>=3/2a^4$
$3/2c^4>=n^2>=3/2a^4$

王守恩

northwolves 发表于 2023-5-12 21:14
$a=3/2a^4$
$3/2c^4>=n^2>=3/2a^4$

求助，是这么一道题:
\(\cot^{-1}{(n)}=\cot^{-1}(a)+\cot^{-1}(b),\ \ \ n=0,1,2,3,4,5,6...\)   a,b=正整数。
\(a(0)=1: \cot^{-1}{(0)}=\cot^{-1}{(1)}+\cot^{-1}{(1)}\)
\(a(1)=1: \cot^{-1}{(1)}=\cot^{-1}{(2)}+\cot^{-1}{(3)}\)
\(a(2)=1: \cot^{-1}{(2)}=\cot^{-1}{(3)}+\cot^{-1}{(7)}\)
\(a(3)=2: \cot^{-1}{(3)}=\cot^{-1}{(4)}+\cot^{-1}{(13)},\cot^{-1}{(3)}=\cot^{-1}{(5)}+\cot^{-1}{(8)}\)
\(a(4)=1: \cot^{-1}{(4)}=\cot^{-1}{(5)}+\cot^{-1}{(21)}\)
\(a(5)=2: \cot^{-1}{(5)}=\cot^{-1}{(6)}+\cot^{-1}{(31)},\cot^{-1}{(5)}=\cot^{-1}{(7)}+\cot^{-1}{(18)}\)
\(a(6)=1: \cot^{-1}{(6)}=\cot^{-1}{(7)}+\cot^{-1}{(43)}\)
\(a(7)=3: \cot^{-1}{(7)}=\cot^{-1}{(8)}+\cot^{-1}{(57)},\cot^{-1}{(7)}=\cot^{-1}{(9)}+\cot^{-1}{(32)},\cot^{-1}{(7)}=\cot^{-1}{(12)}+\cot^{-1}{(17)}\)
......
得到这样一串数:
{1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 2, 2, 4, 1, 2, 1, 4, 3, 2, 1, ....

我没有“方法”让这串数出来，更不说通项公式(OEIS好像没有类似的)，

nyy

王守恩 发表于 2023-5-15 08:45
求助，是这么一道题:
\(\cot^{-1}{(n)}=\cot^{-1}(a)+\cot^{-1}(b),\ \ \ n=0,1,2,3,4,5,6...\)   a,b= ...

1. Clear["Global`*"];(*清除所有变量*)
   
2. xx={2, 3, 5, 8, 37, 45, 82, 127, 590, 717, 1307, 2024, 9403, 11427, 20830, 32257, 149858, 182115};
   
3. (*求解n的表达式,n是变量*)
   
4. xxx=FindSequenceFunction[xx,n]//FullSimplify
   

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代码写出来了，但是求解结果不很满意

\[\text{DifferenceRoot}[\{\unicode{f818},\unicode{f80d}\}\unicode{f4a1}\{\unicode{f818}(\unicode{f80d})-16 \unicode{f818}(\unicode{f80d}+4)+\unicode{f818}(\unicode{f80d}+8)=0,\unicode{f818}(1)=2,\unicode{f818}(2)=3,\unicode{f818}(3)=5,\unicode{f818}(4)=8,\unicode{f818}(5)=37,\unicode{f818}(6)=45,\unicode{f818}(7)=82,\unicode{f818}(8)=127\}][n]\]

nyy

nyy 发表于 2023-5-15 11:44
代码写出来了，但是求解结果不很满意

\[\text{DifferenceRoot}[\{%unicode{f818},%unicode{f80d}\ ...

根据这个函数，得到第991项是
5100132119933373636843402476364806819280548191237026976059321345334242\
7577827429784536226030517275209585619623022177931044357655197292424313\
7742664165914475199049989177625135111605189266770303036353154862941307\
1959317278922470849479129463925931004707265703540489243297723616696472\
774399983636860222

northwolves

王守恩 发表于 2023-5-15 08:45
求助，是这么一道题:
\(\cot^{-1}{(n)}=\cot^{-1}(a)+\cot^{-1}(b),\ \ \ n=0,1,2,3,4,5,6...\)   a,b= ...

1. Table[Length@
   
2.   Solve[a*b == 1 + n*(a + b) && a >= b > n, {a, b}, Integers], {n, 0,
   
3.   99}]

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{1,1,1,2,1,2,1,3,2,2,1,2,2,4,1,2,1,4,3,2,1,4,2,4,1,2,1,4,2,2,2,4,3,4,2,2,1,4,3,2,1,3,2,6,2,2,2,8,2,2,2,2,2,4,1,4,1,8,2,2,2,2,2,4,2,2,1,4,4,2,3,2,4,8,1,4,2,4,2,2,2,4,3,8,1,2,2,4,2,4,1,4,2,6,1,2,2,4,4,6}

王守恩

northwolves 发表于 2023-5-15 16:43
{1,1,1,2,1,2,1,3,2,2,1,2,2,4,1,2,1,4,3,2,1,4,2,4,1,2,1,4,2,2,2,4,3,4,2,2,1,4,3,2,1,3,2,6,2,2,2 ...

\(a(1)=007: 007^2+1=0002*05^2\)  因子数有6个{1,2,5,10,25,50}
\(a(2)=018: 018^2+1=0013*05^2\)  因子数有6个{1,5,13,25,65,325}
\(a(3)=032: 032^2+1=0041*05^2\)  因子数有6个{1,5,25,41,205,1025}
\(a(4)=038: 038^2+1=0005*17^2\)  因子数有6个{1,5,17,85,289,1445}
\(a(5)=041: 041^2+1=0002*29^2\)  因子数有6个{1,2,29,58,841,1682}
\(a(6)=070: 070^2+1=0029*13^2\)  因子数有6个{1,13,29,169,377,4901}
\(a(7)=082: 082^2+1=0269*05^2\)  因子数有6个{1,5,25,269,1345,6725}
\(a(8)=118: 118^2+1=0557*05^2\)  因子数有6个{1,5,25,557,2785,13925}
\(a(9)=168: 168^2+1=1129*05^2\)  因子数有6个{1,5,25,1129,5645,28225}
............
\(a(24)=?\)

northwolves

王守恩 发表于 2023-5-15 18:34
\(a(1)=007: 007^2+1=0002*05^2\)  因子数有6个{1,2,5,10,25,50}
\(a(2)=018: 018^2+1=0013*05^2\)  因 ...

1. Take[Select[Range@1000, Length@Divisors[#^2 + 1] == 6 &], {24}]

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776

northwolves

1. Select[Range@10000, Length@Divisors[#^2 + 1] == 6 &]
   

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{7,18,32,38,41,70,82,118,168,218,232,282,332,368,468,482,500,518,532,540,618,732,768,776,800,832,868,944,1018,1082,1132,1194,1218,1332,1368,1418,1468,1482,1486,1518,1582,1618,1632,1710,1718,1732,1744,1760,1882,1982,2032,2118,2296,2518,2532,2564,2632,2668,2768,2774,2968,2984,3018,3032,3218,3282,3332,3368,3382,3430,3450,3506,3618,3868,4030,4032,4060,4084,4126,4224,4468,4632,4732,4782,5082,5132,5140,5168,5240,5332,5518,5532,5604,5632,5668,5676,5718,5732,5744,5782,5816,5846,5882,5968,5982,6132,6168,6268,6320,6332,6346,6382,6468,6582,6632,6690,6718,6732,6760,6868,6968,6974,6982,7032,7082,7218,7232,7268,7366,7368,7476,7482,7582,7618,7718,7844,7882,7982,8032,8082,8118,8119,8124,8268,8382,8418,8482,8520,8582,8618,8782,8882,8918,8968,9032,9132,9196,9286,9332,9418,9534,9632,9700,9718,9868,9918,9968}

王守恩

northwolves 发表于 2023-5-15 21:33
{7,18,32,38,41,70,82,118,168,218,232,282,332,368,468,482,500,518,532,540,618,732,768,776,800,832 ...

  太好了！  谢谢 northwolves！     A049532       
7,18,32,38,41,43,57,68,70,82,93,99,107,117,118,132,143,157,168,182,193,207,218,232,239,243,251,257,268,282,293,307,318,327,332,

1. n = 1; Reap[Do[While[SquareFreeQ[n^2 + 1], n++]; Sow[n]; n++, {c, 100}]][[2, 1]]

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这通项也可以调吗?

northwolves

王守恩 发表于 2023-5-16 05:42
太好了！  谢谢 northwolves！     A049532       
7,18,32,38,41,43,57,68,70,82,93,99,107,117,118,132,14 ...

1. Select[Range@1000, ! SquareFreeQ[#^2 + 1] &]

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{7,18,32,38,41,43,57,68,70,82,93,99,107,117,118,132,143,157,168,182,193,207,218,232,239,243,251,257,268,282,293,307,318,327,332,343,357,368,378,382,393,407,408,418,432,437,443,457,468,482,493,500,507,515,518,532,540,543,557,568,577,582,593,606,607,616,618,632,643,657,668,682,693,707,718,732,743,746,757,768,775,776,782,793,800,807,818,829,832,843,857,868,882,893,905,907,915,918,932,943,944,957,968,982,993}