[seqfan] Re: An equivalence for integer sequences
Jaume Oliver i Lafont
joliverlafont at gmail.com
Sun Mar 1 19:58:00 CET 2009
Neil said "More generally, for a strictly positive sequence with offset O,
we can assign the value Sum_{n >= O} 1/a(n)"
So two seqs are equivalent iff there are integers O_a, O_b such that
Sum_{n >= O_a} 1/a(n) = Sum_{n >= O_b} 1/b(n)
This classifies n*(-1)^(n+1) and (n+1)*(-1)^n into the same log(2)
class, while avoiding some zero divisions.
If the relation has to be also independent from other features, the
equality becomes
c_a*( Sum_{n >= O_a} 1/a(n) )^p_a = c_b*( Sum_{n >= O_b} 1/b(n) )^p_b
with c_a, c_b, p_a, p_b either positive or negative integers.
In this way, classes become broader and are still meaningful. For
instance, sequences whose reciprocals sum up to 2/Pi, Pi^2/6 or
1/sqrt(2Pi) belong to the same class.
Jaume
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