[seqfan] Re: a(n-1+a(n))=a(n)+a(n-1)
Andrew N W Hone
A.N.W.Hone at kent.ac.uk
Sat Aug 21 11:34:20 CEST 2010
Viewing this as a functional equation, one can look for power series solutions
a(t) = c_0 + c_1 t + c_2 t^2 + ...
At leading order, c_0 comes out as a factor, so one can set c_0=0. If c_0=0, then
at the next order one gets a relation that includes c_1-c_1^2 plus a sum of terms that is linear
in c_2,c_3,...
I used Maple to calculate the terms, substituting in a truncated series, and found that the linear relation one gets
for c_2,c_3,... includes all the coefficients up to the point of truncation; this suggests that for the full series
there should be an infinite linear combination of all these coefficients, but I have not tried to find a non-trivial solution.
The simple solution is to just set c_2=c_3=...=0, leaving the equation c_1-c_1^2=0, so the only solution
(apart from a(t)=0) that arises this way is a(t)=t, with c_1=1.
Even if there are other solutions of the functional equation, then it is not clear that they would give an integer sequence for integer
values of t.
All the best,
Andy
________________________________________
From: seqfan-bounces at list.seqfan.eu [seqfan-bounces at list.seqfan.eu] On Behalf Of Alex M [timeroot.alex at gmail.com]
Sent: 20 August 2010 23:38
To: Sequence Fanatics Discussion list
Subject: [seqfan] a(n-1+a(n))=a(n)+a(n-1)
Recently, I was considering a rather strange recursive function:
a(t-1+a(t))=a(t)+a(t-1), a(1)=1
If we also say that a(2)=2, then we quickly get a simple linear function, so
I took
a(2)=3.
Now, not all terms are defined from the recursion; we'll take the smallest
positive integer term that lead to no contradictions. For example:
a(1)=1
a(2)=3
Therefore a(4)=4 if we plug t=2 into the above recursive equation. Next, we
must figure out a(3). a(3) cannot equal 1, because (taking t=3) it would
imply a(3)=a(3)+a(2) -> 1=1+3. a(3) also cannot equal 2, because (taking
t=3, again) would imply a(2+a(3))=a(3)+a(2) -> a(4)=2+3 -> a(4)=5. But we
already know that a(4) *must* equal 4. But, if we choose a(3)=3, then we
will have no contradictions.
I've started to compute this sequence, and it's rather confusing at times. I
must really check my work, but I get the sequence:
1,3,3,4,6,3,7,9,3,10,12,12,10,2,12,16,2,18,13,5
where a(20)=5. As an example of some the interesting problems that arise:
let us say a(12)=3. This produces no clear contradictions; Taking t=12 gives
us a(11+3)=a(12)+a(11) -> a(14)=3+12=15. a(14) is not determined by any
other recurrence relations. It is only much later, when computing a(22), do
we realize that taking t=11 gives a(10+a(11))=a(11)+a(10) ->a(22)=22, but at
the same time taking t=13 gives a(12+a(13))=a(13)+a(12) -> a(22)=13, which
is contradiction! a(13), a(10), and a(11) are determined previously, so we
know that a(12) must be 12 in order to avoid contradictions; then we have to
go back an change anything we've found in the meantime.
This sequence generally seems to grow linearly, which makes sense
considering that the initial conditions a(1)=1 a(2)=2 is satisfied exactly
by a(x)=x, although this solution isn't unique.
Comments?
~6 out of 5 statisticians say that the number of statistics that either make
no sense or use ridiculous timescales at all has dropped over 164% in the
last 5.62474396842 years.
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