Problems & Puzzles: Puzzles

Puzzle 97. Consecutive numbers with the same "sum of prime factors" (SPF)

If n = product (pi^ai) then let's define SPF(n) = sum(pi*ai)

Example n = 12 = 2^2*3^1, then SPF(12) = 2*2+3*1= 7

In this puzzle we ask you to find K consecutive numbers such that all of them share the same SPF.

I have found the least consecutive numbers of this type for K=2 and K=3:

For K=2, SPF(5, 6)=5
For K=3, SPF(417162, 417163, 417164)=533


1. Find the least consecutive numbers with this property, for K = 4, 5 & 6.
2. Do you think that exist a sequence for any K value?


Jud McCranie wrote (24/06/2000):

"I checked up to 4,292,000,000 and I didn't even find another solution for k=3, and of course none for k>3".

Later the same day, he wrote to my question "...Can you devise if exist any theoretical reason for this negative result?" the following:

"Roughly. There are a lot of them for k=2. Maybe the chance of getting 3 of them in a row is very small. The SOPF of n can be as large as n. Assuming that SOPF(n) is distributed randomly (it isn't, but let's work with that) then the chance that SOPF(n+1)=SOPF(n) is approximately 1/n. The change that SOPF(n+2)=SOPF(n) also is about 1/n^2. The sum of this for n from 1 to infinity is pi^2/6, so we expect about 1.6 solutions for k=3. The probability isn't exactly 1/n^2, but at any rate, we expect a small number of solutions for k=3."

Subquestion: the rough Jud's estimation starts from a very crude estimation of the chance of SOPF(n)=SOP(n+1) as 1/n. Can someone improve this estimation in order to improve the estimation of the chance of SOPF(n)=SOPF(n+1)=SOP(n+2)=...





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