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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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Some Thoughts re(2): interesting side topic -- economical numbers -- transcription error compounded | Comment 10 of 13 |
(In reply to re: interesting side topic -- economical numbers by Dej Mar)

For speed, I used QB64 in the list of purported economical numbers. When that prints to the screen, the screen cannot be copied onto the clipboard, the way it can from QB 4.5's Command-prompt. So I transcribed the first number of the 9-economical-number sequence by hand.

But I took the last three digits from the first number of the previous sequence. I then factored by hand blindly without counting digits. The 9-number sequence of economical numbers actually begins at 10,990,399.

10990399 = 7 * 31 * 50647
10990400 = 2^6 * 5^2 * 6869
10990401 = 3 * 79^2 * 587 (frugal)
10990402 = 2 * 59 * 93139
10990403 = 10990403 (prime)
10990404 = 2^2 * 3^6 * 3769
10990405 = 5 * 271 * 8111
10990406 = 2 * 7^3 * 37 * 433
10990407 = 3 * 3663469

So the prime pages' implication does not hold; so perhaps they weren't really trying to imply that the 9-sequence they found was the lowest.

Correction made for typo in previous version of 10990400 = 2^6 * 5^2 * 6869.

Edited on September 15, 2012, 3:28 pm
  Posted by Charlie on 2012-09-15 10:33:20

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