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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 interesting side topic -- economical numbers | Comment 8 of 13 |

If one settles for economical, rather than frugal, numbers, a sequence of 8 starts at 1,169,312:

  1169312 = 2^5 * 36541
  1169313 = 3*47*8293
  1169314 = 2*73*8009
  1169315 = 5*7*33409
  1169316 = 2^2 * 3^6 * 401
  1169317 = 19*61543
  1169318 = 2*584659
  1169319 = 3*389773
 
 and of 9 begins at 10,990,312:
 
 10990312 = 2^3 * 1373789
 10990313 = 17*139*4651
 10990314 = 2 * 3^2 * 157 * 3889
 10990315 = 5*7*101*3109
 10990316 = 2^2 * 2747579
 10990317 = 3*13*281803
 10990318 = 2*5495159
 10990319 = 10990319 (prime)
 10990320 = 2^4 * 3 * 5 * 11 * 23 * 181

As noted in follow-up comments, the above list is based on a transcription error. The real 9-member sequence does indeed have 8-digit members, but the correct sequence appears in my follow-up reply.

Edited on September 15, 2012, 10:35 am
  Posted by Charlie on 2012-09-15 01:35:11

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