王守恩 发表于 2023-9-4 16:31
本来是这样一道题: f(n)表示是由 0,1,2,3 组成的长度为 n 的所有序列中连续两个 0 出现的次数总和。
f(n)=0 ...
$f(n,k)=(n - 1)*(k+1)^{n-2}$
(1),f(n) 表示由 0,1,2,3,...k 组成的长度为 n 的所有序列中连续两个 0 出现的次数总和，
(1),Table, x], {k, 2, 5}]
{0, 1, 4, 12, 32, 80, 192, 448, 1024, 2304, 5120, 11264, 24576, 53248, 114688,245760,524288,
{0, 1, 6, 27, 108, 405, 1458, 5103, 17496, 59049, 196830, 649539, 2125764, 6908733, 22320522,
{0, 1, 8, 48, 256, 1280, 6144, 28672, 131072, 589824, 2621440, 11534336,50331648,218103808,
{0, 1, 10, 75, 500, 3125, 18750, 109375, 625000,3515625,19531250,107421875,585937500,
(2),f(n) 表示由 0,1,2,3,...k 组成的长度为 n 的所有序列中连续两个 0 出现的序列数量，
(2),Table,x],{k,2,5}]
{0, 0, 1, 3, 8, 19, 43, 94, 201, 423, 880, 1815, 3719, 7582, 15397, 31171, 62952, 126891, 255379,
{0, 0, 1, 5, 21, 79, 281, 963, 3217, 10547, 34089, 108955, 345137, 1085331, 3392377, 10549739,
{0, 0, 1, 7, 40, 205, 991, 4612, 20905, 92935, 407056, 1762117, 7556095, 32148940, 135892321,
{0, 0, 1, 9, 65, 421, 2569, 15085,86241,483429,2669305,14564061,78699089,421880725,
(1)-(2)=(3), 有关(3)这些数字串还可以有通项公式吗？
(1),f(n) 表示由 0,1,2,3,...k 组成的长度为 n 的所有序列中连续两个 0 出现的次数总和，
(1),Table, x], {k, 2, 5}]
{0, 1, 4, 12, 32, 80, 192, 448, 1024, 2304, 5120, 11264, 24576, 53248, 114688,245760,524288,
{0, 1, 6, 27, 108, 405, 1458, 5103, 17496, 59049, 196830, 649539, 2125764, 6908733, 22320522,
{0, 1, 8, 48, 256, 1280, 6144, 28672, 131072, 589824, 2621440, 11534336,50331648,218103808,
{0, 1, 10, 75, 500, 3125, 18750, 109375, 625000,3515625,19531250,107421875,585937500,
(2),f(n) 表示由 0,1,2,3,...k 组成的长度为 n 的所有序列中连续两个 0 出现的序列数量，
(2),Table,x],{k,2,5}]
{0, 0, 1, 3, 8, 19, 43, 94, 201, 423, 880, 1815, 3719, 7582, 15397, 31171, 62952, 126891, 255379,
{0, 0, 1, 5, 21, 79, 281, 963, 3217, 10547, 34089, 108955, 345137, 1085331, 3392377, 10549739,
{0, 0, 1, 7, 40, 205, 991, 4612, 20905, 92935, 407056, 1762117, 7556095, 32148940, 135892321,
{0, 0, 1, 9, 65, 421, 2569, 15085,86241,483429,2669305,14564061,78699089,421880725,
(1)-(2)=(3), 好像没有特别好的方法？至少在OEIS是没有这些数字串的。
Table,x],{k,2,5}]
{0, 1, 3, 9, 24, 61, 149, 354, 823, 1881, 4240,9449,20857,45666,99291,214589,461336,
{0, 1, 5, 22, 87, 326, 1177, 4140, 14279, 48502, 162741, 540584, 1780627, 5823402, 18928145,
{0, 1, 7, 41, 216, 1075, 5153, 24060, 110167, 496889, 2214384, 9772219, 42775553,185954868,
{0, 1, 9, 66, 435,2704,16181,94290,538759,3032196,16861945,92857814,507238411,
边长和面积都是整数的三角形，面积一定是6的倍数。
a(01)=1, 01*6={5,4,3}
a(02)=2, 02*6={6,5,5}={8,5,5}
a(03)=0, 03*6
a(04)=2, 04*6={10,8,6}={15,13,4}
a(05)=1, 05*6={13,12,5}
a(06)=2, 06*6={17,10,9}={26,25,3}
a(07)=1, 07*6={20,15,7}
a(08)=2, 08*6={12,10,10}={16,10,10}
a(09)=1, 09*6={15,12,9}
a(10)=4, 10*6={13,13,10}={17,15,8}={24,13,13}={29,25,6}
a(11)=1, 11*6={20,13,11}
a(12)=1, 12*6={30,29,5}
a(13)=0, 13*6
a(14)=4, 14*6={15,14,13}={21,17,10}={25,24,7}={35,29,8}
......
得到这样一串数:1,2,0,2,1,1,2,2,1,4,1,1,0,4,1,2,0,2,1,4,......速度太慢了！
Table/4!==n,40>a≥b≥c>a-b},{a,b,c},Integers],{n,1,20}]
王守恩 发表于 2023-9-8 16:00
边长和面积都是整数的三角形，面积一定是6的倍数。
a(01)=1, 01*6={5,4,3}
a(02)=2, 02*6={6,5,5}={8,5,5}
Select/4},{c,2,40},{b,c,40},{a,b,40}]],#>0&&IntegerQ[#]&]/6],#[]<21&]
{{1,1},{2,2},{4,2},{5,1},{6,2},{7,1},{8,2},{9,1},{10,4},{11,1},{12,1},{14,4},{15,2},{16,2},{18,2},{19,1},{20,4}}
northwolves 发表于 2023-9-10 21:54
{{1,1},{2,2},{4,2},{5,1},{6,2},{7,1},{8,2},{9,1},{10,4},{11,1},{12,1},{14,4},{15,2},{16,2},{18,2 ...
Select/4!},{c,2,40},{b,c,40},{a,b,b+c}]],#>0&&IntegerQ[#]&]],#[]<21&]
假如我们想写成: 1, 2, 0, 2, 1, 2, 1, 2, 1, 4, 1, 1, 0, 4, 2, 2, 0, 2, 1, 4,......还有方法吗？
假如我们想写成: 1, 2, 0, 2, 1, 2, 1, 2, 1, 4, 1, 1, 0, 4, 2, 2, 0, 2, 1, 4,......还有方法吗？
嗨！还真是有的,A051585:1, 2, 0, 2, 1, 2, 1, 2, 1, 4, 1, 1, 0, 4, 2, 2, 0, 2, 1, 4, 3, 1, 0, 2, 1, 2, 0, 4, 0, 2, 0, 2, 1, 1, 6, 3, 0, 0, 1, 5, 0,......
a(n)=sum(z=sqrtint(sqrtint(192*n^2)-1)+1, sqrtint(9*(64*n^2+5)\20), sum(y=z\2+1, z, my(t=(y*z)^2-(12*n)^2, x); if(issquare(t, &t), (issquare(y^2+z^2-2*t, &x) && x<=y) + (t && issquare(y^2+z^2+2*t, &x) && x<=y), 0)))
有公式，可惜我的计算软件出不来。
x=Select/24},{c,2,40},{b,c,40},{a,b,b+c}]],#<21&];Array&,20]
$S_{\triangleABC}=6n=\frac{1}{2}bcsinA$
$(12n)^2=(bc)^2sin^2A=(bc)^2(1-cos^2A)=b^2c^2-(\frac{b^2+c^2-a^2}{2})^2$
$a^2=b^2+c^2 \pm2\sqrt{b^2c^2-(12n)^2}$
1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1,
2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2,
3, 2, 2, 3, 2, 2, 3, 3, 2, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3,
4, 3, 3, 4, 3, 3, 4, 4, 3, 4, 4, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4,
......