数字串的通项公式

王守恩

R(n)表示正整数n除以2,3,4,5,6,7,8,9,10的余数之和,
若要求R(n)=R(n-1),  则n是这样一串数:
14, 98, 154, 182, 238, 266, 322, 406, 434, 518, 574, 602, 658, 686, 742, 826, 854, 938, 994, 1022, 1078, 1106, 1162,
1246, 1274, 1358, 1414, 1442, 1498, 1526, 1582, 1666, 1694, 1778, 1834, 1862, 1918, 1946, 2002, 2086, 2114, 2198,
2254, 2282, 2338, 2366, 2422, 2506, 2534, 2618, 2674, 2702, 2758, 2786, 2842, 2926, 2954, 3038, 3094, 3122, ......

1. Table[14(4n-3-2Floor[(n-1)/8]+(1+(-1)^Floor[(n-2)/2])(-1)^Floor[(n-2)/4]),{n,1,60}]

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northwolves

王守恩 发表于 2023-9-25 18:46
R(n)表示正整数n除以2,3,4,5,6,7,8,9,10的余数之和,
若要求R(n)=R(n-1),  则n是这样一串数:
14, 98, 154, 1 ...

1. Select[Range@4000,Sum[Mod[#,k]-Mod[#-1,k],{k,10}]==0&]

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王守恩

northwolves 发表于 2023-9-25 19:54

太好了！不能就这样溜走！！
1,R(n)表示正整数n除以2,3,4,5,6,7,8,9,10的余数之和,
若要求R(n)=R(n-1),  则n是这样一串数:
Select[Range@4000,Sum[Mod[#,k]-Mod[#-1,k],{k,10}]==0&]
2,R(n)表示正整数n除以2,3,4,5,6,7,8,9,10的余数之和,
若要求R(n)=R(n-a),  则n是这样一串数:
Select[Range@4000,Sum[Mod[#,k]-Mod[#-a,k],{k,10}]==0&]
3,R(n)表示正整数n除以2,3,4,5,6,7,8,9,10的余数之和,
若要求R(n)=R(n-1)+b,  则n是这样一串数:
Select[Range@4000,Sum[Mod[#,k]-Mod[#-1,k],{k,10}]==b&]
4,R(n)表示正整数n除以c,(c+1),...,(d-1),d的余数之和,
若要求R(n)=R(n-1),  则n是这样一串数:
Select[Range@4000,Sum[Mod[#,k]-Mod[#-1,k],{k,c,d}]==0&]
......
这样的通项也可以有！
1,5,11,13,17,19,23,25,29,31,37,41,43,47,53,55,59,61,65,67,71,73,79,83,85,89, 95, 97, 101,103,107,109,113,115,121,125,127,131,137,139,143,145,149,151,155,157,163,167,169,173,179,181,185,187,191,193,197,199,205,209,211,215,221,223,...
1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,59,61,67,71,73,77,79,83,89,91,97,101,103,107,109,113,119,121,127,131,133,137,139,143,149,151,157,161,163,167,169,173,179,181,187,191,193,197,199,203,209,211,217,221,223,227,229,233,239,...
......

王守恩

1:|pi-3/1|=0.141692
2:|pi-16/5|=0.0584073
3:|pi-22/7|=0.00126448
4:|pi-201/64|=0.000967653
5:|pi-333/106|=0.0000832196
6:|pi-355/113|=2.66764*10^-7
7:|pi-355/113|=2.66764*10^-7
8:|pi-75948/24175|= 9.92981*10^-8
9:|pi-100798/32085|=9.05184*10^-9
......
我们专门把分母拉出来,是这样一串数:
1, 5, 7, 64, 106, 113, 113, 24175, 32085, 33102, 99532, 265381, 1360120,
1725033, 18610450, 25510582, 78256779, 340262731,811528438, ..........
这串数可是在OEIS找不到的。可有好的通项公式？谢谢！

northwolves

王守恩 发表于 2023-9-28 16:48
1:|pi-3/1|=0.141692
2:|pi-16/5|=0.0584073
3:|pi-22/7|=0.00126448

1. p[n_]:=Min@Denominator@Select[Table[Rationalize[Pi+k*10^(-n-1),10^-n],{k,-5,5}],Abs[Pi-#]<10^-n&];Table[p[n],{n,0,40}]

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{1,5,7,64,106,113,113,24175,32085,33102,99532,265381,1360120,1725033,18610450,25510582,78256779,340262731,811528438,1963319607,6701487259,6701487259,413528890451,554260122890,1142027682075,2851718461558,2851718461558,41633749241295,91822653867264,136308121570117,1543874804974140,1952799169684491,9627687726852338,21208174623389167,115668560843798173,136876735467187340,842468587426513207,842468587426513207,49842523393631466553,79328923953559428798,84383735478118508040}

王守恩

northwolves 发表于 2023-9-28 19:46
{1,5,7,64,106,113,113,24175,32085,33102,99532,265381,1360120,1725033,18610450,25510582,78256779, ...

我们这串数(pi的误差分数)还是比A002486(pi的连分数)误差分布均匀些。
1, 5, 7, 64, 106, 113, 113, 24175, 32085, 33102, 99532, 265381, 1360120, 1725033, 18610450, 25510582, 78256779, 340262731, 811528438, 1963319607, 6701487259, 6701487259, 413528890451, 554260122890, 1142027682075,......
1, 7, 106, 113, 33102, 33215, 66317, 99532, 265381, 364913, 1360120, 1725033, 25510582, 52746197, 78256779, 131002976, 340262731, 811528438, 1963319607, 4738167652, 6701487259, 567663097408, 1142027682075,......

王守恩

再来一串! OEIS没有的。e 的误差分数比 e 的连分数(A007677)误差分布均匀些。

1. Table[Min@Denominator@Select[Table[Rationalize[E+k/10^(n+1),10^-n],{k,-5,5}],Abs[E-#]<10^-n&],{n,0,40}]

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1, 3, 7, 32, 71, 394, 1001, 5541, 8544, 18089, 154257, 398959, 398959, 4597073, 10391023, 10391023, 140478290, 312129649, 312129649, 4843205071, 10622799089, 10622799089, 175432249793, 403978495031, 403978495031, ......
1, 3, 4, 7, 32, 39, 71, 465, 536, 1001, 8544, 9545, 18089, 190435, 208524, 398959, 4996032, 5394991, 10391023, 150869313, 161260336, 312129649, 5155334720, 5467464369, 10622799089, 196677847971, 207300647060, 403978495031, ......
这两者摆在一起，连分数还是有缺陷的：连分数有重叠，有空白。误差分数就没有这些缺陷。

王守恩

A033485    a(n)=a(n-1)+a(floor(n/2)), a(1)=1.
{1, 2, 3, 5, 7, 10, 13, 18, 23, 30, 37, 47, 57, 70, 83, 101, 119, 142, 165, 195, 225, 262, 299, 346, 393, 450, ...

1. a[1]=1;a[n_]:=a[2n]=Sum[a[k],{k,1,n}];Table[a[2n],{n,1,26}]

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A033485好像不是这样的。还可以有其他表达方式吗？

northwolves

王守恩 发表于 2023-10-4 17:00
A033485    a(n)=a(n-1)+a(floor(n/2)), a(1)=1.
{1, 2, 3, 5, 7, 10, 13, 18, 23, 30, 37, 47, 57, 70, 83 ...

1. a[n_] := If[n < 4, n, a[Floor[n/2]] + a[n - 1]]; Table[a[n], {n, 36}]

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{1,2,3,5,7,10,13,18,23,30,37,47,57,70,83,101,119,142,165,195,225,262,299,346,393,450,507,577,647,730,813,914,1015,1134,1253,1395}

王守恩

northwolves 发表于 2023-10-4 18:04
{1,2,3,5,7,10,13,18,23,30,37,47,57,70,83,101,119,142,165,195,225,262,299,346,393,450,507,577,647 ...

第2串:{1,2,3,5,7,10,13,18,23,30,37,47,57,70,83,101,119,142,165,195,225,262,299,346,393,450,507,577,647,730,813,914,1015,1134,1253,1395,

1. a[n_]:=If[n<2,n,a[Floor[(n+0)/2]]+a[n-1]];Table[a[n],{n,36}]

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第3串:a(1)=1, a(n) = a(n-1) + a(floor((n+1)/3))A089649

1. a[n_]:=If[n<2,n,a[Floor[(n+1)/3]]+a[n-1]];Table[a[n],{n,36}]

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第4串:a(1)=1, a(n) = a(n-1) + a(floor((n+1)/4))A089651

1. a[n_]:=If[n<2,n,a[Floor[(n+2)/4]]+a[n-1]];Table[a[n],{n,36}]

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第5串:a(1)=1, a(n) = a(n-1) + a(floor((n+1)/5))

1. a[n_]:=If[n<2,n,a[Floor[n+3)/5]]+a[n-1]];Table[a[n],{n,36}]

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第6串:a(1)=1, a(n) = a(n-1) + a(floor((n+1)/6))

1. a[n_]:=If[n<2,n,a[Floor[(n+4)/6]]+a[n-1]];Table[a[n],{n,36}]

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第7串:a(1)=1, a(n) = a(n-1) + a(floor((n+1)/7))

1. a[n_]:=If[n<2,n,a[Floor[(n+5)/7]]+a[n-1]];Table[a[n],{n,36}]

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  OEISH还没有第5,6,7,...串数。