# [seqfan] a(n-1+a(n))=a(n)+a(n-1)

Alex M timeroot.alex at gmail.com
Sat Aug 21 00:38:57 CEST 2010

```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

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

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.