[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>
skd> * Previous message: [seqfan] Re: Offsets for Recurrence Sequences
skd>
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)
4=A159557(3)=A005843(2).
6=A159557(4)=A005843(3).
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
or
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
or
A103974(n)= A098301(n-1) + A055793(n+1), n>=1
or
A103974(n)= A055793(n+1) + A098301(n-1), n>0
or
A103974(n)= A098301(n-1) + A055793(n+1), n>0
or
A103974(n+1)= A055793(n+2) + A098301(n), n>-1
or
...
(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).
Richard
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