Problems & Puzzles:
DPS in AP
Frederick Schneider proposes the
following interesting puzzle:
Find n such there exists an
a.p. of length L between some set of
'divisor pair sums'. A 'divisor pair sum' (DPS) for a number n
as d + n/d where d | n. (d should be <= sqrt(n) as well).
So far, there are minimum
values known for L=1 to 5: 1, 4, 72, 1008, 36400
72's solution: 3+24, 4+18,
8+9. Difference=5 (Note: 72 = 3*24 = 4*18 = 8*9)
1008's solution: 16+63=79, 18+56=74, 21+48=69, 28+36=64.
Difference = 5
36400's solution: 91+400=491, 100+364=464, 112+325=437,
130+280=410,175+208=383. Difference = 27
you find any numbers with dps in a.p of length > 5. Minimal
solutions are preferred but all are welcome. If this is
impossible, please explain why.
Note. Fred wrote that the
solution by Jaroslaw Wroblesky to the
has inspired this puzzle. As a matter of fact, JW has demonstrated
for each number a such
that a=d or n/d in the DPS a.p of length l, if (a*n-1) is prime
for all 2*l numbers, then you have a set of l numbers in
arithmetic progression with equal sigma.
Giovanni Resta wrote (22-III-08):
I usually do not send negative result, but sometimes a negative
result is better than nothing...
It seems that this puzzle is quite difficult. I've searched up to
10^9. I found several other values n whose DPS have 5 values in
arithmetic progression, but no longer progressions. I have also
checked, to no avail, some larger numbers, i.e. the Highly composite
numbers (OEIS A002182) up to 4488062423933088000 in the hope that
their large number of divisors could lead to a longer AP.