A109425 Numbers k such that tau(k)/omega(k) is an integer, where tau(k) = number of divisors of k and omega(k) = number of distinct prime factors of k.
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 79
Offset: 1
Keywords
Examples
The number 12 is in the sequence because tau(12) = 6 (1,2,3,4,6,12) and omega(12) = 2 (2,3) and so tau(12)/omega(12) = 3. The number 36 is not in the sequence because tau(36) = 9 (1,2,3,4,6,9,12,18,36) and omega(36) = 2 (2,3) and so tau(36)/omega(36) = 9/2.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
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Maple
with(numtheory): a:=proc(n) if type(tau(n)/nops(factorset(n)), integer)=true then n else fi end: seq(a(n),n=2..90);
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Mathematica
f[n_] := DivisorSigma[0, n]/Length[FactorInteger[n]]; Select[ Range[2, 80], IntegerQ[ f[ # ]] &] (* Robert G. Wilson v, Jun 30 2005 *) Select[Range[2,80],IntegerQ[DivisorSigma[0,#]/PrimeNu[#]]&] (* Harvey P. Dale, Sep 29 2024 *)
Comments