Problems & Puzzles: Puzzles

Puzzle 413. 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 is defined
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

Q1. Can 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 CYF #35 has inspired this puzzle. As a matter of fact, JW has demonstrated that:

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.


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