A072084 In prime factorization of n replace all primes with their numbers of 1's in binary representation.
1, 1, 2, 1, 2, 2, 3, 1, 4, 2, 3, 2, 3, 3, 4, 1, 2, 4, 3, 2, 6, 3, 4, 2, 4, 3, 8, 3, 4, 4, 5, 1, 6, 2, 6, 4, 3, 3, 6, 2, 3, 6, 4, 3, 8, 4, 5, 2, 9, 4, 4, 3, 4, 8, 6, 3, 6, 4, 5, 4, 5, 5, 12, 1, 6, 6, 3, 2, 8, 6, 4, 4, 3, 3, 8, 3, 9, 6, 5, 2, 16, 3, 4, 6, 4, 4, 8, 3, 4, 8, 9, 4, 10, 5, 6
Offset: 1
Examples
a(30) = a(2*3*5) = a(2)*a(3)*a(5) = 1*2*2 = 4,
as a(2)=a('10')=1, a(3)=a('11')= 2 and a(5)=a('101')=2.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Programs
-
Haskell
a072084 = product . map a000120 . a027746_row -- Reinhard Zumkeller, Feb 10 2013
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Maple
A072084 := proc(n) local a,c; readlib(ifactors): a := n -> add(i,i=convert(n, base, 2)); mul(a(c[1])^c[2],c=ifactors(n)[2]) end: # Peter Luschny, Jan 16 2010
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Mathematica
a[n_] := Times @@ Power @@@ (FactorInteger[n] /. {p_Integer, e_} :> {DigitCount[p, 2, 1], e}); Array[a, 100] (* Jean-François Alcover, Feb 09 2018 *) -
PARI
a(n)=my(f=factor(n)); f[,1]=apply(hammingweight, f[,1]); factorback(f) \\ Charles R Greathouse IV, Aug 06 2015
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Sage
A072084 = lambda n: prod(p.digits(base=2).count(1)**m for p,m in factor(n)) # D. S. McNeil, Jan 17 2011
Formula
Completely multiplicative with a(p) = number of 1's in binary representation of prime p.
Multiplicative with a(p^e) = A000120(p)^e
Comments