[seqfan] A024810 floor(2^n/pi)

[Kevin Ryde]

I was contemplating Clark Kimberling's comment in A024810 about
equivalence to floor(2^n/pi).  Is there an easy argument for that
(which might appear there!) ?

The closest I got was

    x = 2^(n+1)/pi                      = 1/pi bit-shifted upwards
    A024810(n) = floor(cotan(1/x))      the %N definition

    cotan(1/x) = x - (1/3)/x - (1/45)/x^3 - (2/945)/x^5 - ...
    so  floor(x - (1/3)/x - (1/45)/x^3 - (2/945)/x^5 - ...)
    vs  floor(x)

-(1/3)/x is about -1/2^n so it could only push x down to a smaller
integer part if x had a run of about n+1 many 0-bits immediately below
the binary point (a very small fractional part).

This would mean bits of 1/pi (A127266) having a run of about n many
0-bits at bit position n.  The initial bits of 1/pi look like the 1s are
too common, so no such run.  But is it so?