Conjectures 111-113 from "100 Conjectures from the OEIS". Correction.

[David Wilson]

>>  a(n) = 3 or -3 iff n in 3*A036556 [...]
>>
>> then would mean that A065359(n) = 3 or -3 <==> n is an odious number 
>> which is multiple of 3.

No, A065356((2^(12k+6)-1)/3) = 6k+3.

The truth is

a(n) = odd multiple of 3 iff n in 3*A036556.





This is a two-part post, both parts related to the same main idea.

Say we have the partition of the first n positive integers into the two 

We can get the sum S =
product a(k)'s  + product b(k)'s.

1) So, what is the largest prime, if any, S possible for each n?
2) What is the least prime, if any, S possible for each n?

3) What n' s have no prime S's?

If we define the product over the empty set to be 1, so that S for n=1 is 
2, then I get (by hand) the max primes of:
2,3,7,11,43,149,1013

I get the min primes of:
2,3,5,11,23,149,179

I may have very likely made a mistake in calculating these sequences' 
given terms.


One interesting, yet very easily seen, fact that narrows the searches:

If we want S to be a prime, then the positive integers <= n, except 1 and 
the primes > n/2, must all be together in either {a(k)} or {b(k)}.

A simple example shows why this is true.

9 and 7 must be on the same side of the sum S if n is >= 14. Proof: 9 
must be on the same side as 6, or else 3 divides S. 14 must be on the 

But if n is < 2p, where p is prime, then no other multiple of p occurs 
among the integers 1 to n.
So p perhaps can be on either side of the sum if S is prime.


===

The rest of this message is a post I have made to sci.math.
Instead of trying to get primes, in this variation we try to get integers 
with as many divisors as possible.

In his response, J K Haugland calculated these terms for the max-score 

2, 2, 2, 4, 4, 12, 20, 16, 24, 64, 96, 144 

Anyone able to calculate more terms?

---

I wonder what the sequence is of maximal scores (for n integers)
of the game defined below.

-
First, for a given n, each player partitions the sequence of
the first n positive integers into the two sets {a(k)} and {b(k)}.

A player's score is the number of positive divisors of the sum:

product a(k)'s + product b(k)'s.

So, for example, if n = 6, a player may have the sum of the products:

1*3*4*5 + 2*6 = 72.

Since 72 has 12 positive divisors, the player gets 12 points.
-

Allowing for empty sets (so a(1) can be defined), I get that
the maximum scores sequence begins:
2, 2, 2, 4,...

Thanks,
Leroy Quet