数字串的通项公式

王守恩

northwolves 发表于 2024-12-20 15:54
这两个是等价的
$\sum _{i=1}^{k+1} \lfloor \frac{n-i}{k-1}\rfloor -\sum _{i=1}^{n-k+1} \lfloor \frac{ ...

这些按钮我还是用不了。OEIS没有这串数。

周长相同的梯形。梯形={上底, 左腰, 右腰, 下底}=4个不同的整数=左腰<右腰。
a(1)=0
a(2)=0
a(3)=0
a(4)=0
a(5)=0
a(6)=0
a(7)=0
a(8)=0
a(9)=0,
a(10)=1=1,2,3,4,
a(11)=3=1,2,3,5=2,1,3,5=3,1,2,5,
a(12)=1=1,2,4,5,
a(13)=6=1,2,4,6=1,3,4,5=1,3,5,4=1,4,5,3=2,1,4,6=4,1,2,6,
a(14)=3=1,2,5,6=1,3,4,6=2,3,4,5,
a(15)=12=1,2,5,7=1,3,4,7=1,3,5,6=1,3,6,5=1,5,6,3=2,1,5,7=2,3,4,6=3,1,47=3,2,4,6=4,1,3,7=4,2,3,6=5,1,2,7,
a(16)=9=1,2,6,7=1,3,5,7=1,4,5,6=1,4,6,5=1,5,6,4=2,3,4,7=2,3,5,6=3,2,4,7=4,2,3,7,
a(17)=21=1268=1358=1367=1376=1457=1475=1574=1673=2168=2348=2357=2456=2465=2564=3158=3248=3257=4238=5138=5237=6128,
a(18)=12=1278=1368=1458=1467=1476=1674=2358=2367=2457=3258=3456=5238,
a(19)=33,
a(20)=24,
a(21)=48,
a(22)=34,
a(23)=69,
a(24)=52,
a(25)=
a(26)=
a(27)=
a(28)=

王守恩

已知梯形周长,  用4条整数边={上底,左腰,右腰,下底}={a,b,c,d},  来表示梯形最大面积。可以有公式(A)。

1. 公式(A)。Table[Maximize[{((d + a)Sqrt[(b + c - (d - a)) ((d - a) + b - c) (c + (d - a) - b) (b + c + (d - a))])/(4 (d - a)), a+b+c+d==n, 0<a<d, 0<b≤c, c-b<d-a<b+c},{a,b,c,d},Integers],{n,7,35}]

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a(07)={1,2,2,2}
a(08)={1,2,2,3}
a(09)={2,2,2,3}
a(10)={1,3,3,3}
a(11)={2,3,3,3}
a(12)={2,3,3,4}
a(13)={3,3,3,4}
a(14)={2,4,4,4}
a(15)={3,4,4,4}
a(16)={3,4,4,5}
a(17)={4,4,4,5}
a(18)={3,5,5,5}
a(19)={4,5,5,5}
a(20)={4,5,5,6}
a(21)={3,5,5,6}
a(22)={4,6,6,6}
a(23)={5,6,6,6}
a(24)={5,6,6,7}
a(25)={4,6,6,7}
a(26)={5,7,7,7}
a(27)={6,7,7,7}
a(28)={6,7,7,7}
a(29)={7,7,7,8}
公式(A)速度慢了。换成公式(B),答案没有变。

1. 公式(B)。Table[((n - 2 Floor[(n + 2)/4]) Sqrt[(2 Floor[(n + 4)/4] - n) (n - 4 Floor[(n + 2)/4] - 2 Floor[(n + 4)/4])])/4, {n, 7, 350}]

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对公式(B)取整数。得到公式(C)。

1. 公式(C)。Table[Round[((n - 2 Floor[(n + 2)/4]) Sqrt[(2 Floor[(n + 4)/4] - n) (n - 4 Floor[(n + 2)/4] - 2 Floor[(n + 4)/4])])/4], {n, 7, 350}]

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3, 3, 5, 6, 7, 8, 10, 12, 14, 15, 18, 20, 22, 24, 27, 30, 33, 35, 39, 42, 45, 48, 52, 56, 60, 63, 68, 72, 76, 80, 85, 90, 95, 99, 105, 110, 115, 120, 126, 132, 138, 143, 150, 156, 162, 168, 175, 182, 189, 195,
203, 210, 217, 224, 232, 240, 248, 255, 264, 272, 280, 288, 297, 306, 315, 323, 333, 342, 351, 360, 370, 380, 390, 399, 410, 420, 430, 440, 451, 462, 473, 483, 495, 506, 517, 528, 540, 552, 564, 575, 588,
600, 612, 624, 637, 650, 663, 675, 689, 702, 715, 728, 742, 756, 770, 783, 798, 812, 826, 840, 855, 870, 885, 899, 915, 930, 945, 960, 976, 992, 1008, 1023, 1040, 1056, 1072, 1088, 1105, 1122, 1139, 1155,
1173, 1190, 1207, 1224, 1242, 1260, 1278, 1295, 1314, 1332, 1350, 1368, 1387, 1406, 1425, 1443, 1463, 1482, 1501, 1520, 1540, 1560, 1580, 1599, 1620, 1640, 1660, 1680, 1701, 1722, 1743, 1763, 1785, 1806}
单独把第1, 3, 5, 7, 9, ...项取出来。
3, 5, 7, 10, 14, 18, 22, 27, 33, 39, 45, 52, 60, 68, 76, 85, 95, 105, 115, 126, 138, 150, 162, 175, 189, 203, 217, 232, 248, 264, 280, 297, 315, 333, 351, 370, 390, 410, 430, 451, 473, 495, 517, 540, 564, 588,
612, 637, 663, 689, 715, 742, 770, 798, 826, 855, 885, 915, 945, 976, 1008, 1040, 1072, 1105, 1139, 1173, 1207, 1242, 1278, 1314, 1350, 1387, 1425, 1463, 1501, 1540, 1580, 1620, 1660, 1701, 1743, 1785,
可以有公式(D)。

1. 公式(D)。Table[Floor[(n/(1 + Power[E, (n)^-1]))^2], {n, 4, 100}]

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注意：公式(D)可是出现了大名鼎鼎的常数" e " !

northwolves

王守恩 发表于 2024-12-23 09:42
这些按钮我还是用不了。OEIS没有这串数。

周长相同的梯形。梯形={上底, 左腰, 右腰, 下底}=4个不同的整 ...

1. h[r_]:=Module[{a=r[[1]],b=r[[2]],c=r[[3]],d=r[[4]]},CountDistinct@r==4&&d<a+b+c&&b^2>((d-a)/2-(c^2-b^2)/(2(d-a)))^2];
   
2. v=Table[Table[{x,n-x},{x,n/2}],{n,100}];
   
3. y[n_]:=Module[{c=0},For[k=2,k<n-1,k++,z=Flatten[Table[{x[[1]],y[[1]],y[[2]],x[[2]]},{x,v[[k]]},{y,v[[n-k]]}],1];q=Select[z,h@#==True&];c+=Length@q];c];
   
4. Table[y[n],{n,100}]

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{0,0,0,0,0,0,0,0,0,1,3,1,6,3,12,9,21,12,33,24,48,34,69,52,93,69,123,99,159,123,201,165,249,202,306,256,369,306,441,378,522,441,612,531,711,613,822,721,942,822,1074,954,1218,1074,1374,1230,1542,1375,1725,1555,1920,1725,2130,1935,2355,2130,2595,2370,2850,2596,3123,2866,3411,3123,3717,3429,4041,3717,4383,4059,4743,4384,5124,4762,5523,5124,5943,5544,6384,5943,6846,6405,7329,6847,7836,7351,8364,7836,8916,8388}

northwolves

northwolves 发表于 2024-12-28 06:16
{0,0,0,0,0,0,0,0,0,1,3,1,6,3,12,9,21,12,33,24,48,34,69,52,93,69,123,99,159,123,201,165,249,202,3 ...

研究发现，奇数项 $a_n=lfloor \frac{n+1}{6}\rfloor +\frac{n^3-12 n^2+17 n+12+18 i^{n+1}}{96} $

王守恩

northwolves 发表于 2024-12-29 09:43
研究发现，奇数项 $a_n=lfloor \frac{n+1}{6}\rfloor +\frac{n^3-12 n^2+17 n+12+18 i^{n+1}}{96} $ ...

00001, 00003, 00001, 00006,
00003, 00012, 00009, 00021,
00012, 00033, 00024, 00048,
00034, 00069, 00052, 00093,
00069, 00123, 00099, 00159,
00123, 00201, 00165, 00249,
00202, 00306, 00256, 00369,
00306, 00441, 00378, 00522,
00441, 00612, 00531, 00711,
00613, 00822, 00721, 00942,
00822, 01074, 00954, 01218,
01074, 01374, 01230, 01542,
01375, 01725, 01555, 01920,
01725, 02130, 01935, 02355,
02130, 02595, 02370, 02850,
02596, 03123, 02866, 03411,
03123, 03717, 03429, 04041,
03717, 04383, 04059, 04743,
04384, 05124, 04762, 05523,
05124, 05943, 05544, 06384,
05943, 06846, 06405, 07329,
06847, 07836, 07351, 08364,
07836, 08916, 08388, 09492,
08916, 10092, 09516, 10716,
10093, 11367, 10741, 12042,
11367, 12744, 12069, 13473,
12744, 14229, 13500, 15012,
14230, 15825, 15040, 16665,
15825, 17535, 16695, 18435,
17535, 19365, 18465, 20325,
19366, 21318, 20356, 22341,
21318, 23397, 22374, 24486,
23397, 25608, 24519, 26763,
25609, 27954, 26797, 29178,
27954, 30438, 29214, 31734,

王守恩

分成4列还是可以有。

{1, 3, 12, 34, 69, 123, 202, 306, 441, 613, 822, 1074, 1375, 1725, 2130, 2596, 3123, 3717, 4384, 5124, 5943, 6847, 7836, 8916, 10093, 11367, 12744, 14230, 15825, 17535, 19366, 21318, 23397, 25609, 27954, 30438}
Table[Round[(4 n^3 - 3 n^2 - 3 n + 4 - 2 Cos[(2 n \[Pi])/3] + Sin[(2 n \[Pi])/3])/6], {n, 36}]

{3, 12, 33, 69, 123, 201, 306, 441, 612, 822, 1074, 1374, 1725, 2130, 2595, 3123, 3717, 4383, 5124, 5943, 6846, 7836, 8916, 10092, 11367, 12744, 14229, 15825, 17535, 19365, 21318, 23397, 25608, 27954, 30438, 33066}
Table[Round[(4 n^3 + 9 n^2 + 3 n)/6 ], {n, 36}]

{1, 9, 24, 52, 99, 165, 256, 378, 531, 721, 954, 1230, 1555, 1935, 2370, 2866, 3429, 4059, 4762, 5544, 6405, 7351, 8388, 9516, 10741, 12069, 13500, 15040, 16695, 18465, 20356, 22374, 24519, 26797, 29214, 31770}
Table[Round[(4 n^3 + 3 n^2 + 3 n - 7 Sin[(2 n \[Pi])/3])/6], {n, 36}]

{6, 21, 48, 93, 159, 249, 369, 522, 711, 942, 1218, 1542, 1920, 2355, 2850, 3411, 4041, 4743, 5523, 6384, 7329, 8364, 9492, 10716, 12042, 13473, 15012, 16665, 18435, 20325, 22341, 24486, 26763, 29178, 31734, 34434}
Table[Round[(4 n^3 + 15 n^2 + 15 n + 2)/6], {n, 36}]

王守恩

northwolves 发表于 2024-12-29 09:43
研究发现，奇数项 $a_n=lfloor \frac{n+1}{6}\rfloor +\frac{n^3-12 n^2+17 n+12+18 i^{n+1}}{96} $ ...

每个信封装信A,信B两封信,  信封,信A,信B都有编号1～N,  要求每个信封与信有不同的编号,  有多少种装法?

每个信封装信A,信B两封信,  信封,信A,信B都有编号1～N,  要求每个信封与信=3个不同的编号,  有多少种装法?

王守恩

A033294——Squares which when written backwards remain square (final 0's excluded).

1, 4, 9, 121, 144, 169, 441, 484, 676, 961, 1089, 9801, 10201, 10404, 10609, 12321, 12544, 12769, 14641, 14884, 40401, 40804, 44521, 44944, 48841, 69696, 90601, 94249, 96721, 698896, 1002001, 1004004, 1006009, 1022121, 1024144,

1. Select[Range[400]^2, Mod[#, 10] != 0 && IntegerQ[Sqrt[IntegerReverse[#]]] &]

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短一点了。怎样可以不出现：441, 961, 9801, 40401, 90601, 44521, 96721, 48841, 4004001, 9006001, 1212201, 4414201, ......

northwolves

王守恩 发表于 2025-1-6 07:56
A033294——Squares which when written backwards remain square (final 0's excluded).

1, 4, 9, 121, 1 ...

1. Select[Range[400]^2,
   
2. Mod[#, 10] > 0 && IntegerReverse[#] >= # &&
   
3.    IntegerQ[Sqrt[IntegerReverse[#]]] &]

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{1,4,9,121,144,169,484,676,1089,10201,10404,10609,12321,12544,12769,14641,14884,40804,44944,69696,94249}

王守恩

A008904——a(n) is the final nonzero digit of n!.

1, 1, 2, 6, 4, 2, 2, 4, 2, 8, 8, 8, 6, 8, 2, 8, 8, 6, 8, 2, 4, 4, 8, 4, 6, 4, 4, 8, 4, 6, 8, 8, 6, 8, 2, 2, 2, 4, 2, 8, 2, 2, 4, 2, 8, 6, 6, 2, 6, 4, 2, 2, 4, 2, 8, 4, 4, 8, 4, 6, 6, 6, 2, 6, 4, 6, 6, 2, 6, 4, 8, 8, 6, 8, 2, 4, 4, 8, 4, 6, 8, 8, 6, 8, 2, 2, 2, 4, 2, 8, 2, 2, 4, 2,

Table[Block[{d = IntegerDigits[n!]}, While[d[[-1]] == 0, d = Most@d]; d[[-1]]], {n, 0, 25}]