Re: Inverting the floor function (sort of)

David W. Cantrell wrote:
Rainer Rosenthal <r.rosenthal@xxxxxx> wrote:

The sub-sequence A097075(2n+1) is related to the
sequence A053141(n) as I happened to find out:
A097075(2n+1) = floor(A053141(n)*sqrt(2)+1/2) (2)

There is no explicit formula for A053141 in the OEIS.

There is; you just overlooked it. In 2002, Bruce Corrigan gave a(n) =
(1/8)*(-4 + (2 + sqrt(2))*(3 + 2*sqrt(2))^n + (2 - sqrt(2))*(3 - 2*sqrt(2))^n).

With your help and the help from Robert Israel regarding
Maple computations I have a clearer view now. The great OEIS
showed me: A097075(2n+1) = A001652(n). And one comment in
http://www.research.att.com/~njas/sequences/A001652
explicitely mentions the connection to A053141:

=======================================================================
Solution to a(a+1)=2b(b+1) in natural numbers including 0;
a=a(n), b=b(n)= A053141(n)

- The solution of a special case of a binomial problem of H. Finner and
K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).
=======================================================================

From a(a+1)=2b(b+1) it follows immediately that (a/b)^2 ----> 2 or
A001652(n)/A053141(n) ----> sqrt(2).

So finally this nice geometric problem of mine (rift bendings in rotated
grids) shows up as related to some binomial problem, whatever that
means. I'll ask Finner&Strassburger, but other comments are very welcome.

Best regards
Rainer Rosenthal
r.rosenthal@xxxxxx
.

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