[seqfan] Re: Offsets for Recurrence Sequences

Richard Mathar mathar at strw.leidenuniv.nl
Thu Apr 23 21:29:22 CEST 2009

skd> Creighton Kenneth Dement creighton.k.dement at mail.uni-oldenburg.de
skd> Thu Apr 23 17:52:31 CEST 2009
skd>     * Previous message: [seqfan] Re: Offsets for Recurrence Sequences
skd> I may be the only one, but I think it would be nice to see an example of
skd> what is meant by "the" proper notation in the case of setting up relations
skd> between sequences with different offsets.

a)  Consider A159557 which starts 4,6,26 and has been assigned an offset 3.
By definition, this means A159557(3)=4, A159557(4)=6, A159557(5)=26.
Consider in addition A005843, which starts 0,2,4,6,8 and has been given
an offset 0. This means A005843(0)=0, A005843(1)=2, A005843(2)=4.
--there is no doubt left about this in the OEIS FAQ.

If you want to express the fact that the 4 and the 6 are two consecutive
numbers in both sequences, that is (obviously, because the
equal sign is just what is the equal sign in the mathematical literature)


and is summarized as either
A159557(n)=A005843(n-1), 3<=n<=4
or by substitution n->n+1 in all parts of the formula,
A159557(n+1)=A005843(n), 2<=n<=3
A159557(n+1)=A005843(n), 1<n<4.

Annotations like "started as 0,2,4..." are superfluous, because the index
range of validity can be attached to any formula, and because the offset is
part of the database--for anyone to read. A very sober business, indeed.

b) Another example is  the line
   a(n+1) = A098301(n+1) + A055793(n+2)
in A103974, expanding to the claim
a(1) = A098301(1) + A055793(2) at n=0
a(2) = A098301(2) + A055793(3) at n=1 
a(3) = A098301(3) + A055793(4) at n=2 etc
which can be checked by explicit insertion of the values
1 = 1 + 1 at n=0 (wrong...)
5 = 16 + 4 at n=1 (wrong...)
and is wrong, calling for a correction: If we want to express the alignment

1= 1 + 0, -> A103974(1)= A055793(2) + A098301(0)
5= 4 + 1, -> A103974(2)= A055793(3) + A098301(1)
65= 49 + 16, -> A103974(3)= A055793(4) + A098301(2)
the correct formula is
A103974(n)= A055793(n+1) + A098301(n-1), n>=1
A103974(n)= A098301(n-1) + A055793(n+1), n>=1
A103974(n)= A055793(n+1) + A098301(n-1), n>0
A103974(n)= A098301(n-1) + A055793(n+1), n>0
A103974(n+1)= A055793(n+2) + A098301(n), n>-1
(Of course A103974 would be replaced by "a" in the actual A103974 entry.)
We conclude: *the* proper notation does not exist, and a lot of academic
freedom is left to pick a correct formula. Unfortunately, a lot more of
freedom is left to pick an incorrect formula.

c) A very nice exercise is  (just submitted, so needs to be checked)

%F A159833 a(n)= A008488(n+1)-2 = 4-15*A000292(n+1)+6*A000332(n+4)+20*A000217(n+1)-15*(n+1).


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