Help with theory regarding new sequences.

[Ed Jeffery]

Hello,

I was originally invited to this forum by Peter Luschny regarding another topic, but the following seems more important to me now.

I hit a snag with some new sequences, so I thought I'd try asking for help with theory here. The problems have to do with the equation

(1)  n = T(M) - T(M-k),

where T(r) = A000217(r) is the r-th triangular number (r in {0,1,...}), n, M, k are natural numbers, k <= M, and M is the LEAST integer for which the equality holds.

To find M in (1):

Conjecture. If D(n) is the set of all integral divisors of n, and B_n is the set of all INTEGRAL solutions of

     d + ((n/d - 1)/2  (d in D(n)),

then

(2)  M = min(b : b in B_n).

The problems are the following regarding the minimal such M in (1):

  (i) Under what conditions is it true that the associated k does not divide n?
 (ii) Does (i) occur for only finitely many n?
(iii) Are the n of (i) and (ii) necessarily triangular numbers?
 (iv) Can anyone determine a generalized expression for k?
  (v) Can anyone prove (2) for the minimality of M?

For (iv), I admit the mistaken assumption that k always divides n (which is the reason for these questions). Note that this is all related to A209260.

Finally, as for the new sequences I referred to, one has been submitted (the sequence of least such M of (1)) but not approved yet, and the other (the sequence of k associated with n and M in (1)) I am waiting for approval of the pending one before submitting it: the OEIS editors frown on making references to sequences which have not yet been approved, and I agree with that policy.

Thanks for any help.

Ed Jeffery

[susanne.wienand]

Hello Ed,

concerning (i), I mean that n is not divisible by k, if k is an even number:

n = T(M) - T(M-k) = (M*(M+1))/2 - (M-k)*(M-k+1)/2

n = k * (2M + 1 - k)/2 = k * A/2

If k is even, then A is not divisible by 2 and n/k is not integral.

Regards, Susanne

[Ed Jeffery]

Hello Susanne,

Thank you very much for responding. You are of course correct, and
your answer had not occurred to me. What I should have asked in (i)
was "Under what conditions _on n_ is it true...?" However, your
reasoning then answers (ii) and (iii) as well. So that leaves (iv) and
(v).

Since first posting this thread, I found the following:

The answer to (v) should follow from the maximality of k (for which
proof is sketched in A109814).

For the divisor d (of n) such that M is minimal, it seems that the
following four cases are exhaustive:

1. If n is odd and composite, then
d = max(p : p | n, p <= sqrt(n), p is a prime).

2. If n is a power of 2, then d = n.

3. If n is even and not a power of 2, then
d = max(m : m | n, m = 2*j < n, for some integer j>2).

4. If n is an odd prime, then d = 1.

Cases 2 and 4 are known to be true. Are the cases 1 and 3 correct, and/
or have I missed something?.

At any rate, I intend to submit to OEIS the corresponding sequence of
divisors d as soon as the sequence of M's (A212652) is approved.

Finally, in A109814 it is stated (and I am paraphrasing) that for each
odd divisor d of n there is a unique corresponding j = min(d,2n/d),
and that k (corresponding to minimal M in eq. (1)) is the largest
among those j's; but this is not very illuminating and does not really
answer (iv).

Ed Jeffery

On Feb 18, 2:08 am, "susanne.wienand" <susanne.wien...@gmail.com>
wrote:

[Ed Jeffery]

Sorry to reply to my own post, but the cases for the conditions on d
were wrong. There are more cases, so I'll try listing them again (and
rearranged) as follows, in which n, v, d, j, p are all natural
numbers.

1. If n is an odd prime, then d = 1.

2. If n is odd and composite, then

d = max(p : p | n, p <= sqrt(n), p is a prime).

3. If n is equal to a power of 2, then d = n.

4. If n = 2^j*v, v>1, with v odd and j>1, then

d = max(2^j : 2^j | n).

5. If n = 2*v with v odd and composite, then

d = 2*p, where p is the least odd prime such that p | n.

6. If n = 2*p for some odd prime p, then d = 2.

If this list of cases is exhaustive, then is there an analog for k?

Ed Jeffery

[susanne.wienand]

Hello Ed,

Hopefully this is what you are looking for:

k^2 - k(2M+1) + 2n = 0

k1 = 1/2 * [(2M+1) - sqrt( (2M+1)^2 - 8n )]

k2 = 1/2 * [(2M+1) + sqrt( (2M+1)^2 - 8n) ]  (bigger than M)

M   n   k1   k2
---------------
10 10    1   20
"  19    2   19
"  27    3   18
"  34    4   17
"  40    5   16
"  45    6   15
"  49    7   14
"  52    8   13
"  54    9   12
"  55   10   11
9   9    1   18
"  17    2   17

Regards, Susanne