g[m_] := PadRight[#, m] & /@
Table[Mod[n, k, 1], {n, 3, m + 2}, {k, 2, n - 1}]
Please see attached Mathematica notebook and corresponding pdf for details.
- George
With f the original matrix, the differences of the first column of the
product of f and g seems to be A069735 (to 88 terms), a multiplicative
function.
In[45]:= fg100 = f[100].g[100];
Differences@First@Transpose[fg100]
Out[49]= {3, 2, 5, 2, 6, 2, 7, 3, 6, 2, 10, 2, 6, 4, 9, 2, 9, 2, 10, 4,
6, 2, 14, 3, 6, 4, 10, 2, 12, 2, 11, 4, 6, 4, 15, 2, 6, 4, 14, 2, 12, 2,
10, 6, 6, 2, 18, 3, 9, 4, 10, 2, 12, 4, 14, 4, 6, 2, 20, 2, 6, 6, 13, 4,
12, 2, 10, 4, 12, 2, 21, 2, 6, 6, 10, 4, 12, 2, 18, 5, 6, 2, 20, 4, 6,
4, 14, 2, 18, 4, 10, 4, 6, 4, 22, 2, 9, 6, 15}
Following a reference in A069735 leads to
A051731 Triangle read by rows: T(n,k)=1 if k divides n, T(n,k)=0
otherwise. There Paul Barry suggests calling this a Mobius matrix.
Perhaps that helps lead back to the Mobius function again.
1 0 0 0 0 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0 0 0 0 0
1 0 1 0 0 0 0 0 0 0 0 0
1 1 0 1 0 0 0 0 0 0 0 0
1 0 0 0 1 0 0 0 0 0 0 0
1 1 1 0 0 1 0 0 0 0 0 0
1 0 0 0 0 0 1 0 0 0 0 0
1 1 0 1 0 0 0 1 0 0 0 0
1 0 1 0 0 0 0 0 1 0 0 0
1 1 0 0 1 0 0 0 0 1 0 0
1 0 0 0 0 0 0 0 0 0 1 0
1 1 1 1 0 1 0 0 0 0 0 1
The first column of the inverse of f is A092673 a(n) = moebius(n)-
moebius(n/2) where moebius(n) is zero if n is not an integer.
(In the definitions, I would drop the first column, which is rather dull
both for the Mobius matrix and f.)
Let i be the matrix with all ones on and below the main diagonal and f
(the matrix of this thread). Then it seems that the sequence of the row
sums of i.f is A032741 a(0) = 0; for n > 0, a(n) = number of proper
divisors of n.
The sequence of partial sums of A03274 seems to be A002541
Sum_{k=1..n-1} floor((n-k)/k). The conjecture is that a(n) = A002541(n)
- A002541(n-1). To relate these two sequences in this way is probably
easier to prove than to find in the literature. So far I could do neither.
A032741 and A002541 are not cross-referenced yet--I couldn't login to OEIS.
Proof
short version: floor((n-k)/k) - floor((n-k-1)/k) only increases by 1
when k | n.
long version:
If k | (n-k), say (n-k)/k = m, then (n-k-1)/k = m - 1/k, and
floor((n-k)/k) - floor((n-k-1)/k) = m - (m-1) = 1.
If k does not divide n-k, (n-k)/k = m + j/k and (n-k-1)/k = m + (j-1)/k,
where m is an integer and j is one of 1, 2, 3, ..., k-1. Then
floor((n-k)/k) - floor((n-k-1)/k) = m - m = 0.
Enrique, I played a little with your idea. See the pdf.
Exactly what you did, Maximilian! Looking for a, b, and c(a,b)
| Omega(n) > a and omega(n) = b
| if and only if
| card { 0<k<n | Moebius(n,k+1) <> Moebius(n,k)} = c(a,b).
and yes, for variations like
| For odd n, Omega(n) = omega(n) = m
In fact when I stumbled over A123712/A178212 I was looking for
Omega(n) > 3 and omega(n) >= 3 (not in OEIS)
although I did not knew this at that time.
> (500 values computed using the original definition in terms of the
> matrix inverse.)
My good old Maple, now in retirement, would say: 'object too large'
and quit ;-))
I will be offline for the next couple of days:
By Roberta F. (Own work) [CC-BY-SA-3.0
(www.creativecommons.org/licenses/by-sa/3.0)]
via Wikimedia Commons from Wikimedia Commons