Oops, replace "Now, adding that 2 Cos[Pi n] to it and evaluating at the integers .." by "Now, adding that 2 Cos[Pi n] to it and evaluating at the half-integers .." I just skipped over the fact their sum is exactly zero *at* the integers ... Wouter. -----Original Message----- From: Wouter Meeussen Sent: Saturday, July 13, 2013 4:09 PM To: math-fun Subject: Re: [math-fun] FindSequenceFunction If you plot Fibonacci[6*n]*Fibonacci[n]/Fibonacci[3*n]/Fibonacci[2*n] - Fibonacci[3*n]/Fibonacci[n] then it jumps out that it swiftly approaches -2*Cos[Pi n]. Now, adding that 2 Cos[Pi n] to it and evaluating at the integers gives Table[ c[n] f[n - 1/2]/f[3 n - 3/2] - f[3 n - 3/2]/f[n - 1/2], {n, 5}] with c[n] a set of integers. Which seq is hiding there? {2, 17, 122, 842, 5777, 39602, 271442, 1860497, 12752042, 87403802, 599074577, 4106118242} use FindSequenceFunction: c[n_Integer]:= (-2 + ((7 - 3*Sqrt[5])/2)^n*(3 + Sqrt[5]) - (-3 + Sqrt[5])*((7 + 3*Sqrt[5])/2)^n)/2 does the job. The Sqrt[5]-ish stuff hints at a cool Fibo-format. How to get there from here? Wouter. -----Original Message----- From: Fred lunnon Sent: Saturday, July 13, 2013 9:42 AM To: math-fun Subject: Re: [math-fun] FindSequenceFunction << 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. >> I suppose it is a little unexpected that the answer is not exactly zero ... << Here's van der Poorten's paper which Rumely just refers to as "to be published": http://maths.mq.edu.au/~alf/www-centre/alfpapers/a077.pdf Victor >> Coming back to detecting quotients, it seems from the handy 2-page summary for unenlightened monoglots in the above that (for example) the quotient of integer LFSR sequences _is_ another LFSR sequence if it comprises only integers --- or more generally (I think) if the denominators have finitely many distinct prime factors. But that doesn't help detect a quotient sequence where that is not the case. Back to square one, wiser but sadder. Fred Lunnon On 7/11/13, Bill Gosper <billgosper@gmail.com> wrote:
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] ;-)
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun