# lost definition of A005598 : speculation

Rainer Rosenthal r.rosenthal at web.de
Fri Mar 9 08:23:57 CET 2007

Jonathan Post wrote:
> A005598  Number of straight binary strings of length n.
> (Formerly M1097)
> 2, 4, 8, 14, 24, 36, 54, 76, 104, 136

Taking half of these terms you get

1,2,4,7,12,18,27,38,52,68

which is the beginning of
http://www.research.att.com/~njas/sequences/A049703

When will there be a first discrepancy? How to
compute more terms of A005598?

Best regards,
Rainer Rosenthal
r.rosenthal at web.de

There seems to be limited interest in this subject, but I will post it
anyway so as to share this with the few who are interested.

I have been interested in the transform of a sequence of reals, {c(k)},
to another sequence of reals, {r(k)}, defined as follows:

(And we can perform the transform on integer sequences, of course,
getting a sequence of +- rationals as output.)

First, {c(k)} must be such that c(n) does not equal c(n+1), and c(n) does
not equal c(n+2), for every positive integer n.
(I am assuming for our purposes that the first terms of {c(k)} is c(1).)

So we want the {r(k)} where c(n) = the continued fraction (of rational
terms) [r(1);r(2),...,r(n)], for every positive integer n.

So, for example:

If r(n) = A095159(n)/A095175(n), then, for every positive integer n, the
continued fraction (of rational terms) [r(1);r(2),...r(n)] = H(n) =

For instance:

The 4th harmonic number, 25/12 = 1 + 1/(1/2 +1/(-5/4 +9/28)).

And the 5th harmonic number, 137/60 = 1 + 1/(1/2 +1/(-5/4 +1/(28/9
-64/81))).

In general, this is how I calculate {r(k)}:

r(1) = c(1).

r(2) = 1/(c(2)-c(1)).

r(3) = (c(3)-c(1))*(c(2)-c(1))/(c(2)-c(3)).

And, for n >= 4,

r(n) =
-(c(n)-c(n-2))*(c(n-1)-c(n-3))/((c(n)-c(n-1))*(c(n-2)-c(n-3))*r(n-1)).

Here are the first few terms of the r sequences based on the c-sequences
of a couple popular integer sequences:
(I may have made an error or two, even though I double checked these.)

c(n) = nth prime:

{r(k)}: 2, 1, -3/2, 4, -3/4, 12, -3/16,...

c(n) = F(n) = the nth Fibonacci number:

{r(k)}: 1, 1, -2, 3/2, -10/3, 6/5, -65/18, 378/325,...

c(n) = n(n+1)/2, the nth triangular number:

{r(k)}: 1, 1/2, -10/3, 21/16, -16/5, 165/128,...

c(n) = x^n, where x is not 0, -1, or 1:

r(1)=x. r(2) = 1/(x(x-1)).
For n >= 2, r(2n-1) = 1 -x^2;  r(2n) = (1+x)/(x(x-1)).

c(n) = n:

r(1)=1. r(2)=1.
For n >= 3, r(n) = (-1)^n *2.

Thanks,
Leroy Quet

I wrote in part:

>c(n) = F(n) = the nth Fibonacci number:
>
>{r(k)}: 1, 1, -2, 3/2, -10/3, 6/5, -65/18, 378/325,...
>

Man, I hate finding a small error just after posting. But I must reply to
my original post to point out that c(n) is really F(n+1), the (n+1)th
Fibonacci number.

Sorry, thanks again,
Leroy Quet