f(n) 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，求 f(9)

wintex

Treenewbee

\[ f(n)=round ([4^n-(\frac{1}{2}+\frac{5\sqrt{21}}{42}) (\frac{\sqrt{21}+3}{2} )^n)\]

Treenewbee

前50项：

{0,1,7,40,205,991,4612,20905,92935,407056,1762117,7556095,32148940,135892321,571232647,2389810360,9956870845,41335010911,171055514452,705891052825,2905717608775,11934337612576,48918212175157,200149835407615,817572886925980,3334643143711441,13582547998754887,55255173054769480,224527561670055085,911405798212401631,3696030455799081892,14975230266641297545,60625468185748526215,245248839430879022896,991409899144720853797,4005124120906152455935,16170193680870031232620,65248319888198196279361,263145006638683263390727,1060737843306558702429400,4273880004739383191136925,17212779363752454855405151,69295681383934030838450932,278868195356893523876867065,1121862882677818931327149255,4511678243925482434336830016,18138563419095284371891062037,72902485146203821518280173055,292930186324463402068899692860,1176726176926266008355083548081}

Treenewbee

1. a[n] = 7 a[n - 1] - 9 a[n - 2] - 12 a[n - 3]
   
2. 
   
3. a[1] = 0, a[2] = 1, a[3] = 7

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王守恩

1. Table[CoefficientList[Series[x^2/((1-xk)(1-(k-1)x(x+1))),{x,0,16}],x],{k,2,7}]

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{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,
{0, 0, 1, 11, 96, 751, 5531, 39186, 270241,1827071,12166176,80043931,521516711,3370600266,
{0, 0, 1, 13, 133, 1219, 10513,  87199,  703921,  5570263,  43409905,  334234615,  2548342369,
{0, 0, 1, 15, 176, 1849, 18271, 173608, 1605297, 14549487,129860704,1145089065,9998390207,

Treenewbee

To  lihp2020
别个 示例a【3】=8？
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若000算两次，那就是a(3)=8,得重新推导

Treenewbee

\[a(n)=(n - 1)*2^{2n -4 }\]

Treenewbee

\[a(9)=2^{17}\],故共有18个正因子

Treenewbee

前50项：
{0,1,8,48,256,1280,6144,28672,131072,589824,2621440,11534336,50331648,218103808,939524096,4026531840,17179869184,73014444032,309237645312,1305670057984,5497558138880,23089744183296,96757023244288,404620279021568,1688849860263936,7036874417766400,29273397577908224,121597189939003392,504403158265495552,2089670227099910144,8646911284551352320,35740566642812256256,147573952589676412928,608742554432415203328,2508757194024499019776,10330176681277348904960,42501298345826806923264,174727559866176872906752,717799705396186072481792,2946756685310658613346304,12089258196146291747061760,49565958604199796162953216,203099537695257701350637568,831740963894864872197849088,3404335108034795755972591616,13926825441960528092615147520,56945241807127492645359714304,232732727385651491681035354112,950737950171172051122527404032,3882179963198952542083653566464}

王守恩

主帖还是容易些。谢谢 Treenewbee！

1. Table[CoefficientList[Series[x/(1 - k x - x)^2, {x, 0, 21}], x], {k, 1, 7}]

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{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,
{0, 1, 12, 108, 864, 6480, 46656,326592,2239488,15116544,100776960,665127936,4353564672,