Non-recursive series solution needed
Randall Rathbun
Can anyone find a direct, non-recursive algorithm for the series
listed below? This will help crack a companion series, both of
which create Heron triangles with two rational medians.
T Series
level term
0 1
1 1
2 2
3 3
4 5
5 11
6 37
7 83
8 274
9 1217
10 6161
11 22833
12 165713
13 1249441
14 9434290
15 68570323
16 1013908933
17 11548470571
18 142844426789
19 2279343327171
20 57760865728994
21 979023970244321
22 etc....
T(i-1)*T(i-4)+T(i-2)*T(i-3)
T(i) = ---------------------------
T(i-5)
T(0) = 1 T(1) = 1 T(2) = 2 T(3) = 3 T(4) = 5
Thanks for any help! Please email at ran...@coyote.csusm.edu or
ra...@defcen.gov.au
NOTE:
The smallest Heron triangle with two rational medians is 73 51 26
with area 420 and medians 35/2 97/2.
- Randall
[Mod. note: This either is or is close to a Somos sequence,
described by David Gale in the Intelligencer. - Greg]
Kevin Brown
RR>
RR> T(i-1)*T(i-4)+T(i-2)*T(i-3)
RR> T(i) = ---------------------------
RR> T(i-5)
RR>
RR> T(0) = 1 T(1) = 1 T(2) = 2 T(3) = 3 T(4) = 5
VB = Vincent Broman
VB> While I haven't checked the initial conditions on this, here are
VB> suggestions for semi-general solutions.
VB> T(n) = a*r^n + b*s^n
VB> solves the general equation for any a and b... Otherwise, I'd
VB> also try solutions like a*r^n*sin(c*n) + b*s^n*cos(c*n).
The magnitude of T(n) seems to increase faster than geometrically, so I
think it can't be expressed in either of those forms. If you define
the ratio sequence s(n) = T(n+2)/T(n) then you have s3/s0 = s2/s1 + 1,
which gives an interesting sequence that oscillates somewhat but seems
to grow geometrically. If we define
s(n+1) T(n+3) T(n)
q(n) = -------- = --------------
s(n) T(n+1) T(n+2)
then the values of q(n) seem to exhibit a rough seven-step cycle
with the approximate asymptotic values
0.84988
1.67051
1.88098
0.91686
1.11147
2.07966
1.30297
(These seven values also seem to oscillate slightly.) Notice that
T(2k+1)
-------- = q(0) q(2) q(4) ... q(2(k-1))
T(2k)
so the ratio of consecutive T values increases indefinitely. (The
ratios T(2k)/T(2k-1) are given by 2 q(1) q(3)...q(2k-3).) The geometric
mean of the q values seems to be around 4/3, so a very rough estimate
of T(m)/T(m-1) would be (4/3)^(m/2). Then an even rougher estimate of
T(m) would be (4/3)^(m(m+1)/4). Let's see, for m=20 this gives about
(1.3)10^13, which is at least in the ballpark with the true value
of (5.7)10^13. With some refinements this could probably be improved.
David Savitt
<kevi...@delphi.com> wrote:
>
> T(i-1)*T(i-4)+T(i-2)*T(i-3)
> T(i) = ---------------------------
> T(i-5)
>
> T(0) = 1 T(1) = 1 T(2) = 2 T(3) = 3 T(4) = 5
Nothing nontrivial of the form T(i)=r^i will satisfy the recursion
relation--inserting r^i for T(i), the RHS becomes 2*r^i. However, if c is
any root of x^12-x^4-1=0 and a,b are any constants, T(i)=c^(i^2+a*i+b) does
satisfy the recurrence.
>If you define
>the ratio sequence s(n) = T(n+2)/T(n) then you have s3/s0 = s2/s1 + 1.
>If we define q(n)=s(n+1)/s(n)
>then the values of q(n) seem to exhibit a rough seven-step cycle
>with the approximate asymptotic values
<snip>
> The geometric mean of the q values seems to be around 4/3.
The q(i) satisfy the recurrence relation q(i-1)q(i+1)=1+1/q(i), so if
the q(i) were to tend to a limit k, k would satsify k^2=1+1/k, or
k^3-k-1=0. One such value of k is around 1.32..., which I suppose is
where the 4/3 comes from.
> Then an even rougher estimate of T(m) would be (4/3)^(m(m+1)/4).
Replacing 4/3 by the k above, k^(1/4) is the c introduced before, i.e.
the rough estimate of T(m) is c^(m^2+m).
Also, notice that the q(i) will satisfy the (terminating) continued fraction
1/q(i)=q(i-2)/(1+q(i-3)/(1+q(i-4)/(...+q(2)/(1+q(1)/(3/2))...)
-Dave
--
------------------------------------------------------------------------------
Dave Savitt | 3rd year mathematics | AKA Little Dave | Go Canucks!
dsa...@unixg.ubc.ca | University of B.C. | AKA Goliath | Go Grizzlies!
--- "I am not a crook" - Richard Nixon ----- "I am not your cook" - my mom ---
Kevin Brown
RR>
RR> T(i-1)*T(i-4)+T(i-2)*T(i-3)
RR> T(i) = ---------------------------
RR> T(i-5)
RR>
RR> T(0) = 1 T(1) = 1 T(2) = 2 T(3) = 3 T(4) = 5
VB = Vincent Broman
VB> While I haven't checked the initial conditions on this, here are
VB> suggestions for semi-general solutions.
VB> T(n) = a*r^n + b*s^n
VB> solves the general equation for any a and b... Otherwise, I'd
VB> also try solutions like a*r^n*sin(c*n) + b*s^n*cos(c*n).
The magnitude of T(n) seems to increase faster than geometrically, so I
think it can't be expressed in either of those forms. If you define
the ratio sequence s(n) = T(n+2)/T(n) then you have s3/s0 = s2/s1 + 1,
which gives an interesting sequence that oscillates somewhat but seems
to grow geometrically. If we define
s(n+1) T(n+3) T(n)
q(n) = -------- = --------------
s(n) T(n+1) T(n+2)
then the values of q(n) seem to exhibit a rough seven-step cycle
with the approximate asymptotic values
0.84988
Peter L. Montgomery
>RR>
>RR> T(i-1)*T(i-4)+T(i-2)*T(i-3)
>RR> T(i) = ---------------------------
>RR> T(i-5)
>RR>
>RR> T(0) = 1 T(1) = 1 T(2) = 2 T(3) = 3 T(4) = 5
>
I didn't find an explit formula for T(n), but I have
several observations (long);
. If we try to extend the sequence backwards, we get
T[1] = T[-2] = T[-3] = 1, T[-4] = 2, T[-5] = 3.
In general T[-2 - n] = T[n].
. Define s[n] = T[n-1]. Then
s[-2] = s[-1] = s[0] = s[1] = s[2] = 1,
s[n] * s[n-5] = s[n-1] * s[n-4] + s[n-2] * s[n-3],
s[-n] = s[n].
The last condition (symmetry) suggests it may be simpler
to find a general formula for s[n] than for T[n].
. The s sequence appears to satisfy
s[n]^2 + s[n-2] * s[n+2] = 3 s[n-1] * s[n+1] (n odd)
s[n]^2 + s[n-2] * s[n+2] = 2 s[n-1] * s[n+1] (n even)
(For the T sequence, reverse "odd" and "even".)
If we find a general formula for s[n], we will need to
distinguish odd and even n. Of course this can be hidden,
if we reference (-1)^n or cos(pi * n).
. Numerical data suggest the sequence is periodic modulo primes
and prime powers. For example, the pattern modulo 16 seems to be
n mod 12 0 1 2 3 4 5 6 7 8 9 10 11
s[n] mod 16 1 1 1 2 3 5 11 5 3 2 1 1
There seem to be many patterns like
s[n+10] == 3 * 3^n * s[n] (mod 5) and s[5] == 0 (mod 5)
s[n+12] == - (-1)^n * s[n] (mod 7)
s[n+12] == 8 * 8^n * s[n] (mod 11) and s[6] == 0 (mod 11)
s[n+14] == 25 * 16^n * s[n] (mod 37) and s[7] == 0 (mod 37)
. The value of s[n] seems to be very close to r^(n^2), where
r ~= 1.07425451486467 .
The constant r^4 ~= 1.3317685368462 compares to (4/3) suggested
by Kevin Brown and to 1.3247... (root of x^3 = x + 1) suggested
by David Savitt.
. When n is even, s[n+1] * s[n-1] - s[n]^2 is a perfect square.
When n is odd, this is the product of the neighboring
square roots (up to sign):
n s[n] s[n+1]*s[n-1] - s[n]^2 u[n] v[n] = u[n]/s[n]
0 1 0 0 0
1 1 0 1 1
2 1 1 -1 -1
3 2 -1 -8 -4
4 3 1 57 19
5 5 8 = 1 * 8 455 91
6 11 64 = 8 * 8 -22352 -2032
7 37 -456 = -8 * 57 -47797 -1291
8 83 3249 = 57 * 57 69739671 840237
9 274 25935 = 57 * 455 -3385862936 -12357164
10 1217 207025 = 455 * 455 -1747973613295 -1436297135
The u[n] sequence has the square roots, with the signs
chosen to preserve the pattern in the third column.
Some likely (but unproved) identities are
s[2n]^2 = u[n]^2 - u[n-1] * u[n+1]
s[2n-1] * s[2n+1] = 2 u[n]^2 - u[n-1] * u[n+1]
s[2n-2] * s[2n+2] = 3 u[n]^2 - u[n-1] * u[n+1]
s[2n-3] * s[2n+3] = 5 u[n]^2 - 4 u[n-1] * u[n+1]
s[2n-4] * s[2n+4] = 11 u[n]^2 - 9 u[n-1] * u[n+1]
s[2n-5] * s[2n+5] = 74 u[n]^2 - 25 u[n-1] * u[n+1]
s[2n+1]^2 = -4 u[n] * u[n+1] + u[n-1] * u[n+2]
s[2n ] * s[2n+2] = -3 u[n] * u[n+1] + u[n-1] * u[n+2]
s[2n-1] * s[2n+3] = -5 u[n] * u[n+1] + 2 u[n-1] * u[n+2]
s[2n-2] * s[2n+4] = -11 u[n] * u[n+1] + 3 u[n-1] * u[n+2]
s[2n-3] * s[2n+5] = -37 u[n] * u[n+1] + 10 u[n+1] * u[n+2]
If these patterns are correct, then the coefficients can be found by
(separately) setting n = 0 and n = 1, since u[0] = 0.
s[2n+j] * s[2n-j] = s[j+2] * s[j-2] * u[n]^2
- s[j]^2 * u[n-1] * u[n+1]
s[2n+1+j] * s[2n+1-j] = - s[j+3] * s[j-3] * u[n] * u[n+1]
+ s[j+1] * s[j-1] * u[n-1] * u[n+2]
Now setting n = 2 gives
s[j+4] * s[j-4] = s[j+2] * s[j-2] + 8 s[j]^2
s[j+5] * s[j-5] = -8 s[j+3] * s[j-3] + 57 s[j+1] * s[j-1]
These recurrences reference only even subscripts
(or only odd subscripts) in s, and (if correct)
may be easier to manipulate.
Also, s[2n-1] * s[2n+3] - s[2n+1]^2 = u[n-1] * u[n+2] - u[n] * u[n+1]
appears to be a perfect square, but I see no pattern
for its square root. For n = 0, 1, 2, 3, 4,
it is 1, 1, 7^2, 1, 391^2, respectively.
I do note -391 = 3249 - 3640 = u[4]^2 + u[3] * u[5].
If this conjecture is correct, then s[n] will always
be a sum of two squares when n is odd.
s[1] = 1 = 1^2 + 0^2
s[3] = 2 = 1^2 + 1^2
s[5] = 5 = 2^2 + 1^2
s[7] = 37 = 6^2 + 1^2
s[9] = 274 = 15^2 + 7^2
s[11] = 6161 = 56^2 + 55^2 = 65^2 + 44^2
s[13] = 165713 = 407^2 + 8^2
s[15] = 9434290 = 3071^2 + 57^2 (and other ways)
s[17] = 1013908933 = 22742^2 + 22287^2 (and other ways)
s[2n+1] = (s[n]*s[n+1])^2 + (another square)
The apparent identity u[n]^2 - s[2n]^2 = u[n-1] * u[n+1]
may let us write u[n] as a product of two other sequences,
by looking at expressions like GCD(u[n] + s[2n], u[n-1]).
I have not investigated this.
If we define v[n] = u[n]/s[n], then v[n] appears to be an integer
(see the earlier observation about patterns modulo 5, 11, 37).
The v[n] sequence seems to be periodic modulo small primes,
but with long periods:
Mod 2, period 3: 0, 1, 1
Mod 3, period 16: 0, 1, 2, 2, 1, 1, 2, 2,
0, 1, 1, 2, 2, 1, 1, 2
Mod 4, period 12: 0, 1, 3,
0, 3, 3,
0, 1, 1,
0, 1, 3
v[n+3] == (-1)^n v[n] mod 4
Mod 5, period 40: 0, 1, 4, 1, 4, 1, 3, 4, 2, 1,
0, 1, 3, 4, 2, 1, 1, 1, 1, 1,
0, 4, 4, 4, 4, 4, 3, 1, 2, 4,
0, 4, 3, 1, 2, 4, 1, 4, 1, 4
v[n+10] == 2^(n-1) v[n] (mod 5)
Mod 7, period 20: 0, 1, 6, 3, 5,
0, 5, 4, 6, 6,
0, 1, 1, 3, 2,
0, 2, 4, 1, 6
Mod 11, period 120
Mod 13, period 40: 0, 1, 12, 9, 6,
0, 9, 9, 8, 12,
0, 12, 5, 9, 4,
0, 7, 9, 1, 1,
0, 12, 12, 4, 6,
0, 9, 4, 8, 1,
0, 1, 5, 4, 4,
0, 7, 4, 1, 12
v[5n+6] == +- 4^(n+1) v[5n+1] (mod 13)
Mod 17, period 288
Mod 19, period 16: 0, 1, 18, 15,
0, 15, 1, 1,
0, 18, 18, 4,
0, 4, 1, 18
In order to continue these periodic patterns to negative n,
we should define v[-n] = -v[n] and u[-n] = -u[n]
(i.e., use the other square root there).
--
Peter L. Montgomery pmon...@cwi.nl San Rafael, California
Mathematically gifted, unemployed, U.S. citizen. Interested in computer
architecture, program optimization, computer arithmetic, cryptography,
compilers, computational mathematics. 17 years industrial experience.
Kevin Brown
RR>
RR> T(i-1)*T(i-4)+T(i-2)*T(i-3)
RR> T(i) = ---------------------------
RR> T(i-5)
RR>
RR> T(0) = 1 T(1) = 1 T(2) = 2 T(3) = 3 T(4) = 5
Define the sequence s(n) as follows
T(n) T(n+3)
s(n) = --------------
T(n+1) T(n+2)
The first several values of s(n) are tabulated below (read by rows):
1.50000 0.83333 1.46667 2.01818 1.01966 0.98144 1.97999
1.53352 0.83440 1.43362 2.03445 1.04040 0.96398 1.95826
1.56711 0.83652 1.40094 2.04873 1.06222 0.94762 1.93489
1.60066 0.83971 1.36874 2.06095 1.08509 0.93237 1.91000
1.63407 0.84396 1.33709 2.07106 1.10901 0.91823 1.88370
1.66720 0.84929 1.30606 2.07899 1.13395 0.90519 1.85612
1.69994 0.85569 1.27573 2.08472 1.15987 0.89324 1.82737
1.73215 0.86316 1.24615 2.08821 1.18675 0.88240 1.79757
1.76373 0.87172 1.21740 2.08946 1.21455 0.87264 1.76686
1.79452 0.88137 1.18951 2.08844 1.24322 0.86398 1.73536
1.82441 0.89210 1.16254 2.08517 1.27271 0.85639 1.70320
1.85328 0.90393 1.13651 2.07967 1.30297 0.84988 1.67051
etc
Remarkably, each of these seven columns is just a phase-shifted version
of a single function f(m). The function f appears to be a very pure
harmonic function that oscillates between the values of 0.833 and 2.089
with period near 124.
The function f(m) seems to match very closely to the form
f(m) = A - B sqrt( C - sin(m/k) )
Essentially it's a sine wave, but the peaks are a bit narrower than
the valleys.
It's more convenient to shift the indicies by taking the initial values
T(0)=T(1)=T(2)=T(3)=1 and T(4)=2. Then the first few values of s(n)
are
1 2 3/2 5/6 22/15 111/55 415/407 etc
The values of the T sequence with even indicies can be expressed in terms
of the s(n) values as
T(4) = [(1)(2)]
T(6) = [(1)(2)]^2 [(3/2)(5/6)]
T(8) = [(1)(2)]^3 [(3/2)(5/6)]^2 [(22/15)(111/55)]
etc
and with odd indicies
T(5) = [(2)(3/2)]
T(7) = [(2)(3/2)]^2 [(5/6)(22/15)]
T(9) = [(2)(3/2)]^3 [(5/6)(22/15)]^2 [(111/55)(415/407)]
etc
(Each of the bracketed terms reduces to the form T(k)T(k+4)/(T(k+2)^2),
which can be used to prove that all the T values are integers.)
In general, we have
m-1
T(2m) = PROD [s(2k)s(2k+1)]^(m-1-k)
k=0
m-1
T(2m+1) = PROD [s(2k+1)s(2k+2)]^(m-1-k)
k=0
Substituting for each s(n) the appropriate expression in terms of
the period function f(n), we have a product of many terms of the form
A + B sqrt(C-sin(jn)). Expanding this product will result in a series,
and it might be possible to collect powers of the trig function and sum
the series. This would be more promising if there was a simpler trig
expression for f(n) than the one I have postulated.
Noam Elkies
> T(i-1)*T(i-4)+T(i-2)*T(i-3)
> T(i) = ---------------------------
> T(i-5)
>
> T(0) = 1 T(1) = 1 T(2) = 2 T(3) = 3 T(4) = 5
Two versions of a closed form are given below, together
with consequences for the asymptotics of the T(i).
First, some preliminary observations:
We can extend the functional equation to negative i, finding the
sequence ..., 11, 5, 3, 2, 1, 1, 1, 1, 1, 2, 3, 5, 11, ...
the symmetry suggests renumbering the T's so that the central 1
is the zeroth.
Numerical experiments show that the functional equation has analogs
expressing T(i+k+1)T(i-k) as a linear combination of T(i)T(i+1) and
T(i-1)T(i+2) for all k. We might hope for a similar equation
expressing T(i+k)T(i-k) in terms of T(i)^2 and T(i+1)T(i-1). We don't
quite find that; for instance if k=2 then T(i)^2 + T(i-2)T(i+2)
is either twice or three times T(i-1)T(i+1) depending on the
parity of i. But we can fix this, without losing either the symmetry
or the T(i+k+1)T(i-k) identity, by replacing every other T(i) by r*T(i)
for a suitable constant r. Indeed let r be the 4th root of 2/3,
and define
c(-i) = c(i) = T(i-1) if i is even, r*T(i-1) if i is odd,
so for i=0,1,2,3,... c(i) is 1,r,1,2r,3,2r,5,11r,... . These
satisfy the recurrence
c(i-2) c(i+2) = sqrt(6) * c(i)^2 - c(i-1) c(i+1).
This looks like a Somos recurrence on sequences of elliptic theta
functions (presumably Michael Somos hasn't seen this thread yet,
or he would have remarked on this some time back). Indeed let
q = 0.02208942811097933557356088...
z = 0.1141942041600238048921321...
b = 0.9576898995913810138013844...
u = 0.7889128685374661530379575...
then
c(i) = b * u^(n^2) * [sum from n=-infty to + infty of (q^(n^2)*z^(i*n)].
(These values of q,z,b,u were obtained numerically from the condition
that the formula for c(i) hold for the initial values i=0,1,2,3.)
It follows that c(i) = f(i)*C^(i^2) where
C = u * exp(log^2(z) / 4 log(q)) = 1.074254514864672463110626...
and f(i) is a quasiperiodic function with the single irrational period
2 log(q) / log(z) = 3.51420405240870236831...
which is bounded away from zero; this gives the asymptotic behavior
of c(i), and thus also of T(i) though the T's also have a period of 2
due to the factors of r. The continued fraction of the period starts
3, 1, 1, 17, 9, 1, 15, 2, 7, 2, 7, 1, 1, ... and yields the good
rational approximations 7/2 (whence the apparent period-7 behavior
noted by Kevin Brown <kevi...@delphi.com>), 123/35 (explaining the
oscillation "with period near 124" observed by Kevin), and 1237/352.
The numbers T(i-1) can also be obtained "arithmetically" from the
elliptic curve C/q^(2Z) associated to our theta functions. Wasting
some more time on gp, I found that we're dealing with the elliptic curve
E: y^2 + xy + y = x^3 - 2x (#102-A1(E) in Cremona's table)
which has [a rational 2-torsion point at (0,0) and] a rational point
P: (x,y) = (2,2)
of infinite order. For i=0,1,2,3,... the x-coordinate of the i-th
multiple of P on E in lowest terms is
1/0 [sic], 2/1, 1/1, 8/1, 9/1, 50/49, 121/64, 2738, 6889/3249, ... ,
with numerators
1, 2, 1, 2*2^2, 3^2, 2*5^2, 11^2, 2*37^2, 83^2, ... !
Indeed the numerator of i*P is always T(i-1)^2 or 2*T(i-1)^2
according as i is even or odd, and the canonical height of P on E is
log(C) = 0.0716269464704400357381372...
Note that 7P is an integral point with large (x,y), so very close
to the origin (inf,inf) of E(R); this reflects the fact that z^7
is close to a power of q^2 (namely q^4), which is also what was
responsible for the nearly 7-periodic behavior of T(i).
--Noam D. Elkies (elk...@ramanujan.harvard.edu)
Dept. of Mathematics, Harvard University
Noam Elkies
>[....]
>The numbers T(i-1) can also be obtained "arithmetically" from the
>elliptic curve C/q^(2Z) associated to our theta functions. Wasting
>some more time on gp, I found that we're dealing with the elliptic curve
>E: y^2 + xy + y = x^3 - 2x (#102-A1(E) in Cremona's table)
>which has [a rational 2-torsion point at (0,0) and] a rational point
>P: (x,y) = (2,2)
>[...]
As Dave Rusin (ru...@math.niu.edu) points out, I transposed
two coefficients of the elliptic curve; the curve with
the above equation does not even contain the point (2,2).
The correct equation is
E: y^2 + xy = x^3 + x^2 - 2x.
This was obtained by 1) computing the j-invariant j(E)=j(q^2)
as a real number, 2) using its continued fraction to recognize
j(E) as the rational number 11^6/612, 3) computing the x-coordinate
of the point z on the curve C*/q^(2Z), which determines the correct
quadratic twist, and 4) reducing to standard minimal form, and
5) looking up the resulting curve in Cremona's tables.
Kevin Brown
> T(i-1)*T(i-4)+T(i-2)*T(i-3)
> T(i) = ---------------------------
> T(i-5)
>
> T(0) = 1 T(1) = 1 T(2) = 2 T(3) = 3 T(4) = 5
This sequence (formed by the convolution of successive terms) reminds me
of the sequence of coefficients for the power series solution of the
equation
x x'' + a (x')^2 = b (1)
Among the solutions of this equation (with appropriate choices of a,b)
are exp(t), sin(t), cos(t), (A+Bt)^n, A+Bt+Ct^2, and sqrt(A+Bt+Ct^2).
This last function represents the separation between any two objects
in unaccelerated motion. Other solutions include the cycloid relation
for (non-rotating) gravitational free-fall, and the radial distance
of a mass from a central point about which it revolves with constant
angular velocity and radial freedom.
The power series solution of equation (1) can be written
x(t) = c[0] + c[1] t + c[2] t^2 + c[3] t^3 + ...
where the coefficients c[i] satisfy the convolutions
n / b if n = 2
SUM A(k,n-k) c[k] c[n-k] = (
k=0 \ 0 if n > 2
with
A(k,j) = (a-1) j (k-j) + k(k-1)/2
Any choice of c[0], c[1], c[2], and c[3], with c[1]c[2] not zero,
determines the values of a,b and therefore all the remaining
coefficients. There are many interesting things about these sequences
of c[k] values. Focusing on just the sequences with |c[k]| = 1,
k=0,1,2,3, there are obviously 16 possible choices, but only 8 up to
a simple sign change. These 8 can be arranged as four groups of 2:
k I II III IV
--- -------- ---------- --------- ---------
0 1 1 -1 1 1 1 1 1
1 1 -1 1 1 1 -1 1 -1
2 1 1 1 -1 -1 -1 1 1
3 -1 1 1 1 -1 1 1 -1
4 1/2 1/2 3/2 -3/2 0 0 1 1
5 1/2 -1/2 5/2 5/2 4/5 -4/5 1 -1
6 -3/2 -3/2 9/2 -9/2 2/5 2/5 1 1
7 3/2 -3/2 19/2 19/2 -2/5 2/5 1 -1
8 3/8 3/8 133/8 -133/8 -1/2 -1/2 1 1
9 -29/8 29/8 267/8 267/8 1/30 -1/30 1 -1
etc
Clearly the coefficients in each group differ only in sign. The
coefficients in groups I and II diverge, and those in group IV are
all units. Only the group III sequences converge. Interestingly,
these coefficients are given very closely by
c[k-1] = 2 exp(uk) sin(wk)
for k>2, where
u = -0.145370157...
/ 1.877672951... for III(a)
w = (
\ 1.263919649... for III(b)
Notice that the two possible values of w sum to 3.1415926...
The integer numerators and denominators of these c[k] sequences also
have many interesting properties. For example, primes p congruent to
+1 (mod 4) first appear in the denominator at c[p], whereas primes
congruent to -1 (mod 4) first appear at c[p^2]. The sequence of
numerators is much less regular
1 -1 -1 1 0 -4 2 2 -1 -1 59 -9 -1 233 8 -934 49 .. etc
Incidentally, the value of b in the ubiquitous equation (1) is
essentially just a constant of integration, and the underlying
relation is the derivitive
x x'' + q x' x'' = 0
where q=3 for unaccelerated separations and q=2 for (non-rotating)
gravitational separations. Isolating q and differentiating again
leads to the basic relation, free of arbitrary constants,
x x' x'' x'''' - x x' (x'')^2 - x (x'')^2 x''' + (x')^2 x'' x''' = 0
Dividing by x x' x'' x''' gives the nice form
x'''' x''' x'' x'
------ - ----- - ---- + ----- = 0
x''' x'' x' x