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A frugal sequence (Posted on 2012-09-14) Difficulty: 3 of 5
Find the longest(?) string of consecutive frugal numbers.

Def: A frugal number is a natural number that has more digits than the number of digits in its prime factorization (including exponents). For example, 128=2^7 and 29282=2*11^4

See The Solution Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

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Hints/Tips possible approach | Comment 3 of 13 |

A prime cannot be frugal, so a prime or a compound number which is power-free, cannot be in the sequence.

Consider any 3 numbers a-1, a, a+1, One of these is divisible by 2. One of these is divisible by 3. Powers of 2 and 3 cannot be adjacent, except for 3^2=2^3+1 which is not frugal. So at least one of the numbers is a power of 2 times something else, or at least one of them is a power of 3 times something else.

Since the base powers, 2^a, 3^b, tend to spread out as they get bigger, to have them close translates to 'other factors', and to 'have a string of them close' translates to 'lots of other powered factors'.

It's also worth bearing in mind that 3^5=243 is the smallest 'frugal' power of 3, while 2^7=128 is the smallest 'frugal' power of 2. Then, e.g. 2^7*a=3^5*b-1, with {a = 243n+112, b = 128n+59, giving 14336,14337, etc } and 2^7*a=3^5*b+1, with {a = 243n+131, b = 128n+69 giving 16767,16768, etc}. 

It is at once obvious that strings of any length can be constructed using this method; the problem is finding minimum values for the starting points of strings of a given length.

Similarly, it seems that saint-like levels of frugality are possible, if one is prepared to choose large strings of frugal powers!

Edited on September 14, 2012, 12:34 pm
  Posted by broll on 2012-09-14 12:09:13

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