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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

647.0. "Recursion defined on relation that doesn't hold" by CLT::GILBERT (eager like a child) Thu Jan 15 1987 21:22

Take a fixed prime p.  Let a  = 0 and, for n >= 1, define a  as the least
			    0				   n
non-negative integer such that the relation

	a  = a    = a     = ... = a
	 n    n-d    n-2d          n-(p-1)d 

does *not* hold for any positive integer d < n/(p-1).  Prove or disprove
that for all k >= 0 and 0 <= r <= p-1,

	a     = [ (pa + r)/(p - 1) ],  where [x] is the integer part of x.
	 pk+r        k

Note that this is true for p = 2, since a  = n.
					 n


This was posed by James Propp and Mike Filaseta.  We noters should be able
to prove the recurrence when p = 3, and possibly also p = 4 and 5.
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647.1CLT::GILBERTeager like a childMon Jan 19 1987 23:1334
There is an apparent typo in the original problem.  I assume it should read:

... the relation ... does *not* hold for any positive integer d <= n/(p-1).

The first few values for p=3 are:

	a(0) = 0	(by definition)
	a(1) = 0	(0 < d <= 1/2 is empty)
	a(2) = 1	(if a(2) were 0 then a(2) = a(1) = a(0) would hold,
			 with d=1)
	a(3) = 0	(a(3) = a(2) = a(1) doesn't hold)
	a(4) = 0	(a(4) = a(3) = a(2) doesn't hold)
	a(5) = 1	(if it were 0, then a(5) = a(4) = a(3) would hold;
			 a(5) = a(4) = a(3) doesn't hold;
			 a(5) = a(3) = a(1) doesn't hold)
	a(6) = 1	(if it were 0, then a(6) = a(3) = a(0) would hold)
			 a(6) = a(5) = a(4) doesn't hold;
			 a(6) = a(4) = a(2) doesn't hold;
			 a(6) = a(3) = a(0) doesn't hold)
	a(7) = 2	(if it were 0, then a(7) = a(4) = a(1) would hold;
			 if it were 1, then a(7) = a(6) = a(5) would hold)
			 a(7) = a(6) = a(5) doesn't hold;
			 a(7) = a(5) = a(3) doesn't hold;
			 a(7) = a(4) = a(1) doesn't hold)
	a(8) = 2	(if it were 0, then a(8) = a(4) = a(0) would hold;
			 if it were 1, then a(8) = a(5) = a(2) would hold)
			 a(8) = a(7) = a(6) doesn't hold;
			 a(8) = a(6) = a(4) doesn't hold;
			 a(8) = a(5) = a(2) doesn't hold;
			 a(8) = a(4) = a(0) doesn't hold)
	a(9) = 0	(a(9) = a(8) = a(7) doesn't hold;
			 a(9) = a(7) = a(5) doesn't hold;
			 a(9) = a(6) = a(3) doesn't hold; and
			 a(9) = a(5) = a(1) doesn't hold)
647.2CLT::GILBERTeager like a childSun Mar 22 1987 23:0213
The conjecture is false.

I'll counter that for p >=5, a[p^3-p^2+p-2] = 1, and that this is
the first a[n] for which the conjecture is false.  This has been
verified for p <= 29.


Since a[r] = 0 for r < p-1, and a[p-1] = 1, we use the conjectured equation:
a[p*k+r] = [ (p*a[k] + r)/(p - 1) ] to see that:

	a[p*(p-1)+0] = [ (p*a[p-1]+0)/(p-1) ] = [ p/(p-1) ] = 1
and
	a[p*(p^2-p)+p-2] = [ (p*a[p^2-p]+p-2)/(p-1) ] = [ (2*p-2)/(p-1) ] = 2.