login
A136567
a(n) is the number of exponents occurring only once each in the prime factorization of n.
5
0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 0, 0, 1, 1, 2, 1, 2, 0, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 2, 1, 0, 1, 2, 2, 0, 1, 2, 1, 2, 0, 2, 1, 2, 0, 2, 0, 0, 1, 1, 1, 0, 2, 1, 0, 0, 1, 2, 0, 0, 1, 2, 1, 0, 2, 2, 0, 0, 1, 2, 1, 0, 1, 1, 0, 0, 0, 2, 1, 1, 0, 2, 0, 0, 0, 2, 1, 2, 2, 0, 1, 0, 1, 2, 0
OFFSET
1,12
COMMENTS
Records are in A006939: 1, 2, 12, 360, 75600, ..., . - Robert G. Wilson v, Jan 20 2008
FORMULA
a(n) = A056169(A181819(n)). - Antti Karttunen, Jul 24 2017
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.
MATHEMATICA
f[n_] := Block[{fi = Sort[Last /@ FactorInteger@n]}, Count[ Count[fi, # ] & /@ Union@fi, 1]]; f[1] = 0; Array[f, 105] (* Robert G. Wilson v, Jan 20 2008 *)
Table[Boole[n != 1] Count[Split@ Sort[FactorInteger[n][[All, -1]]], _?(Length@ # == 1 &)], {n, 105}] (* Michael De Vlieger, Jul 24 2017 *)
PROG
(PARI) A136567(n) = { my(exps=(factor(n)[, 2]), m=prod(i=1, length(exps), prime(exps[i])), f=factor(m)[, 2]); sum(i=1, #f, f[i]==1); }; \\ Antti Karttunen, Jul 24 2017
(Scheme) (define (A136567 n) (A056169 (A181819 n))) ;; Antti Karttunen, Jul 24 2017
CROSSREFS
For a(n)=0 see A130092 plus the term 1; for a(n)=1 see A000961.
Sequence in context: A264893 A340653 A334744 * A336569 A324904 A109708
KEYWORD
nonn
AUTHOR
Leroy Quet, Jan 07 2008
EXTENSIONS
More terms from Robert G. Wilson v, Jan 20 2008
STATUS
approved