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A165256
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Numbers whose number of distinct prime factors equals the number of digits in the number.
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4
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2, 3, 4, 5, 7, 8, 9, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 33, 34, 35, 36, 38, 39, 40, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58, 62, 63, 65, 68, 69, 72, 74, 75, 76, 77, 80, 82, 85, 86, 87, 88, 91, 92, 93, 94, 95, 96, 98, 99, 102, 105, 110, 114, 120, 126, 130, 132
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OFFSET
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1,1
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COMMENTS
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Is this sequence finite? If the answer is yes, is there any estimate for the number of terms of this sequence? - Parthasarathy Nambi, Nov 16 2009
This sequence is finite since there are only finitely many primes less than 10. - Charles R Greathouse IV, Feb 04 2013
Specifically, all terms have <= 10 digits since primorial(k) = A002110(k) has > k digits for k > 10. - Michael S. Branicky, Apr 13 2023
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LINKS
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EXAMPLE
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The number of distinct prime factors of 4 is 1, which is the same as the number of digits in 4, so 4 is in the sequence.
The number of distinct prime factors of 21 is 2, which is the same as the number of digits in 21, so 21 is in the sequence.
However, 25 is NOT in the sequence because the number of distinct prime factors of 25 is 1, which does not match the number of digits in 25.
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MAPLE
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omega := proc(n) nops(numtheory[factorset](n)) ; end: A055642 := proc(n) max(1, ilog10(n)+1) ; end: A165256 := proc(n) option remember; local a; if n = 1 then 2; else for a from procname(n-1)+1 do if A055642(a) = omega(a) then RETURN(a) ; fi; od: fi; end: seq(A165256(n), n=1..120) ; # R. J. Mathar, Sep 17 2009
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MATHEMATICA
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Select[Range[200], IntegerLength[#] == Length[FactorInteger[#]] &] (* Harvey P. Dale, Mar 20 2011 *)
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PROG
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(Python) # see link for alternate version producing full sequence instantly
from sympy import primefactors
def ok(n): return len(primefactors(n)) == len(str(n))
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CROSSREFS
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KEYWORD
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base,nonn,fini,full
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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