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 A frugal sequence (Posted on 2012-09-14)
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

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 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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