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A136566
a(n) = sum of the exponents occurring only once each in the prime-factorization of n.
4
0, 1, 1, 2, 1, 0, 1, 3, 2, 0, 1, 3, 1, 0, 0, 4, 1, 3, 1, 3, 0, 0, 1, 4, 2, 0, 3, 3, 1, 0, 1, 5, 0, 0, 0, 0, 1, 0, 0, 4, 1, 0, 1, 3, 3, 0, 1, 5, 2, 3, 0, 3, 1, 4, 0, 4, 0, 0, 1, 2, 1, 0, 3, 6, 0, 0, 1, 3, 0, 0, 1, 5, 1, 0, 3, 3, 0, 0, 1, 5, 4, 0, 1, 2, 0, 0, 0, 4, 1, 2, 0, 3, 0, 0, 0, 6, 1, 3, 3, 0, 1, 0, 1, 4, 0
OFFSET
1,4
LINKS
Diana Mecum and Michael De Vlieger, Table of n, a(n) for n = 1..10000 (first 1000 terms from Diana Mecum)
EXAMPLE
4200 = 2^3 * 3^1 * 5^2 * 7^1. The exponents of the prime factorization are therefore 3,1,2,1. The exponents occurring exactly once are 2 and 3. So a(4200) = 2+3 = 5.
MATHEMATICA
Table[Total@ Flatten@ Select[Split[Sort[FactorInteger[n][[All, -1]]]], Length@ # == 1 &] - Boole[n == 1], {n, 105}] (* Michael De Vlieger, Sep 21 2017 *)
PROG
(PARI) a(n) = my(f=factor(n)[, 2]); sum(k=1, #f, f[k]*(#select(x->(x==f[k]), f) == 1)); \\ Michel Marcus, Sep 22 2017
CROSSREFS
Sequence in context: A329615 A272894 A268387 * A330235 A048983 A301505
KEYWORD
nonn
AUTHOR
Leroy Quet, Jan 07 2008
EXTENSIONS
More terms from Diana L. Mecum, Jul 17 2008
STATUS
approved