例如:A(x)=1+x+x^2/2!-x^3/3!+2*x^4/4!+6*x^5/5!-264*x^6/6!+5370*x^7/7!+。。。
使A(x)=A'(x*A(x
A(x/A’(x))=A'(x)=1+x-x^2/2!+2*x^3/3!+6*x^4/4!-264*x^5/5!+。。。
为了说明a(n)=[x^n/n!]a'(x)^(n+1)/(n+1),创建一个x^k/k!的系数表!,k> =0,在A'(x)^n中如下:
A'^1:[1,1,-1,2,6,-264,5370,-937501315706,…];
A'^2:[1,2,0,-2,34,-508,7472,-100392,774076,…];
A'^3:[1,3,3,-6,48,-522,6036,-54030,-435618,…];
A'^4:[1,4,8,-4,36,-336,2832,3672,-1469680,…];
A'^5:[1,5,15,10,10,-100,-130,44490,-1964390,…];
A'^6:[1,6,24,42,6,36,-1680,59520,-1938564,…];
A'^7:[1,7,35,98,84,42,-1848,54978,-1605394,…];
A'^8:[1,8,48,184,328,128,-1504,42960,-1194368,…];
A'^9:[1,9,63,306,846,864,-1278,32202,-843750,…]。。。
然后上表中的对角线生成此序列:
[1/1, 2/2, 3/3, -4/4, 10/5, 36/6, -1848/7, 42960/8, -843750/9, ...].
期末剩余模量之和2^n。
对于k>=(8*n-6),对于n>1,给定a(k)==0(mod 2^n),那么考虑n>=1的所有模2^n项的剩余之和是很有趣的。
设b(n)=Sum_{k>=0}a(k)(mod 2^n)for n>=1,则序列{b(n,n)}开始:
[4, 16, 40, 144, 432, 1008, 3184, 6384, 15600, 33520, 75504, 159472, 356080, 798448, 1797872, 3895024, 8089328, 16609008, 37842672, 76639984, 166817520, ...].
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