A formula for 35 hexachords

Happy to report that I have found a formula for 35 hexachords.
(Sequences A006840 and A045629) A006840 is the same as A045611 for
n=0 to n=7. n=6 is hexachords in 12-ET.

Sequence A006840 reduces for complementary Z-related sets for C(2n,n)-
reduced sets.

Sequence A045611 also reduces for non-complementary Z-related sets in
C(2n,n)-reduced sets. (Equals the distinct count of interval vectors
for the set)

A045629:

1/2n sum{ d divides n } (phi(n/d)*C(2d-1,d-1) + phi(2n/d)*2^(d-1))

If n is odd, 1/2 (A045629 + 1/2 C(n-1,(n-1)/2) + 2^(n-2)); if n is
even,
1/2 (A045629 + 1/2 C(n,n/2) + 2^(n-2))

I assume phi is the Euler totient function. I am not sure that the
sequence A045611 can even be expressed with a formula. However I am
happy that one can express 35 hexachords in 12-ET and 85 septachords
in 14-ET. After that (16-ET, 18-ET...) the sequences diverge.

All of these sequences can be found on the "Encyclopaedia of Integer
Sequences"

Attn: Jon Wild and Paul Erlich especially, if you are reading this

Paul Hj

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
>
> Happy to report that I have found a formula for 35 hexachords.
> (Sequences A006840 and A045629) A006840 is the same as A045611 for
> n=0 to n=7. n=6 is hexachords in 12-ET.
>
> Sequence A006840 reduces for complementary Z-related sets for C
(2n,n)-
> reduced sets.
>
> Sequence A045611 also reduces for non-complementary Z-related sets
in
> C(2n,n)-reduced sets. (Equals the distinct count of interval vectors
> for the set)
>
> A045629:
>
> 1/2n sum{ d divides n } (phi(n/d)*C(2d-1,d-1) + phi(2n/d)*2^(d-1))
>
> If n is odd, 1/2 (A045629 + 1/2 C(n-1,(n-1)/2) + 2^(n-2)); if n is
> even,
> 1/2 (A045629 + 1/2 C(n,n/2) + 2^(n-2))
>
***This is the formula for A006840, which uses A045629, which reduces
for complementation but not direction. There are 44 sets in C(12,6)-
reduced***
>
> I assume phi is the Euler totient function. I am not sure that the
> sequence A045611 can even be expressed with a formula. However I am
> happy that one can express 35 hexachords in 12-ET and 85
septachords
> in 14-ET. After that (16-ET, 18-ET...) the sequences diverge.
>
> All of these sequences can be found on the "Encyclopaedia of
Integer
> Sequences"
>
> Attn: Jon Wild and Paul Erlich especially, if you are reading this
>
> Paul Hj

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
> >
> > Happy to report that I have found a formula for 35 hexachords.
> > (Sequences A006840 and A045629) A006840 is the same as A045611
for
> > n=0 to n=7. n=6 is hexachords in 12-ET.
> >
> > Sequence A006840 reduces for complementary Z-related sets for C
> (2n,n)-
> > reduced sets.
> >
> > Sequence A045611 also reduces for non-complementary Z-related
sets
> in
> > C(2n,n)-reduced sets. (Equals the distinct count of interval
vectors
> > for the set)
> >
> > A045629:
> >
> > 1/2n sum{ d divides n } (phi(n/d)*C(2d-1,d-1) + phi(2n/d)*2^(d-1))
> >
> > If n is odd, 1/2 (A045629 + 1/2 C(n-1,(n-1)/2) + 2^(n-2)); if n
is
> > even,
> > 1/2 (A045629 + 1/2 C(n,n/2) + 2^(n-2))
> >
> ***This is the formula for A006840, which uses A045629, which
reduces
> for complementation but not direction. There are 44 sets in C(12,6)-
> reduced***
> >
> > I assume phi is the Euler totient function. I am not sure that
the
> > sequence A045611 can even be expressed with a formula. However I
am
> > happy that one can express 35 hexachords in 12-ET and 85
> septachords
> > in 14-ET. After that (16-ET, 18-ET...) the sequences diverge.
> >
> > All of these sequences can be found on the "Encyclopaedia of
> Integer
> > Sequences"
> >
35 is determined by:

1/2(44 + 1/2(C{6,3})+2^4) where 44 is determined by:

SUM:
1/12 (phi(6/6)*C{11,5}+phi(12/6)*2^5
1/12 (phi(6/3)*C{5,2}+phi(12/3)*2^2
1/12 (phi(6/2)*C{3,1}+phi(12/2)*2^1
1/12 (phi(6/1)*C{1,0}+phi(12/1)*2^0

=41 1/6 + 9/6 + 5/6 + 3/6 = 44

phi is the Euler Totient function, the number of numbers less
than a number which are relatively prime to that number, for example:
phi(12) is 4 because 1,5,7,11 are relatively prime to 12.

This stuff is great. Another "coincidence" is that when 44 is
expanded out to include all transpositions you obtain 462, which is C
{11,5}. I tested this for other sets and it holds true. Can anyone
tell me why these formulas WORK?

Paul