Fred> The term-by-term sum and product of such LFSR sequences are also LFSR sequences, but not in general the quotient. BIll Gosper recently asked for an algorithm to detect the quotient of LFSR sequences. I have been waiting for sombody like Neil Sloane or Simon Plouffe to wade in, but in vain ... maybe there's a trade secret to protect? Or maybe they just can't think of anything intelligent to say about it. Like me. WFL I actually have an application for this problem--determining if and how two continued fraction formulæ are related. Fooling around with special cases, two heuristic methods seem to work. Looking at arithmetic subsequences of numerators and denominators separately should eventually score, although there might be multivalued ambiguity in filling in the missing values. Alternatively, taking repeated ratios of a long sample seems to smooth out sporadic cancellations, at the expense of large numbers of large numbers. Neither method suggests when to give up. Was anybody surprised by Fibonacci[6*n]*Fibonacci[n]/Fibonacci[3*n]/Fibonacci[2*n] - Fibonacci[3*n]/Fibonacci[n] ? NeilB found a short Mma proof. --rwg On Fri, Jul 5, 2013 at 6:13 AM, Bill Gosper <billgosper@gmail.com> wrote:
is sporadically impressive: In[127]:= FindSequenceFunction[Table[2^n - 1/n, {n, 11}], n]
Out[127]= (-1 + 2^n n)/n
In[133]:= FindSequenceFunction[Table[3^n/n - 2^n, {n, 11}], n]
Out[133]= FindSequenceFunction[{1, 1/2, 1, 17/4, 83/5, 115/2, 1291/7, 4513/8, 1675, 48809/10, 154619/11}, n]
In[130]:= FindSequenceFunction[{1, 2, 4, 8, 15}, n]
Out[130]= 1/6 (8 n - 3 n^2 + n^3)
In[131]:= Table[2^n/%, {n, 11}]
Out[131]= {2, 2, 2, 2, 32/15, 32/13, 64/21, 4, 512/93, 512/65, 128/11}
In[132]:= FindSequenceFunction[%, n]
Out[132]= (3 2^(1 + n))/(n (8 - 3 n + n^2))
In[156]:= FindSequenceFunction[Table[1/(3^n - n), {n, 12}], n]
Out[156]= FindSequenceFunction[{1/2, 1/7, 1/24, 1/77, 1/238, 1/723, 1/ 2180, 1/6553, 1/19674, 1/59039, 1/177136, 1/531429}, n]
Is there known a generalization of reciprocal differences, say, that can detect rational_function(n,2^n,(-1)^n,Fib(n),...)? --rwg I have one that can detect Fibonacci[6*n]*Fibonacci[n]/Fibonacci[3*n]/Fibonacci[2*n] - Fibonacci[3*n]/Fibonacci[n] ;-)