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%I #19 Sep 28 2023 04:26:09
%S 0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,
%T 0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,
%U 0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1
%N Number of odd exponents larger than one in the canonical prime factorization of n.
%H Antti Karttunen, <a href="/A295662/b295662.txt">Table of n, a(n) for n = 1..65537</a>
%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>.
%F Additive with a(p) = 0, a(p^e) = A000035(e) if e > 1.
%F a(1) = 0; and for n > 1, if A067029(n) = 1, a(n) = a(A028234(n)), otherwise A000035(A067029(n)) + a(A028234(n)).
%F a(n) = A162642(n) - A056169(n).
%F a(n) <= A295659(n).
%F a(n) = 0 iff A295663(n) = 0, and when A295663(n) > 0, a(n) <= A295663(n).
%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} (1/(p^2*(p+1)) = 0.122017493776862257491... . - _Amiram Eldar_, Sep 28 2023
%e For n = 24 = 2^3 * 3^1 there are two odd exponents, but only the other is larger than 1, thus a(24) = 1.
%e For n = 216 = 2^3 * 3^3 there are two odd exponents larger than 1, thus a(216) = 2.
%t Array[Count[FactorInteger[#][[All, -1]], _?(And[OddQ@ #, # > 1] &)] &, 105] (* _Michael De Vlieger_, Nov 28 2017 *)
%o (Scheme, with memoization-macro definec)
%o (definec (A295662 n) (if (= 1 n) 0 (+ (if (= 1 (A067029 n)) 0 (A000035 (A067029 n))) (A295662 (A028234 n)))))
%o (PARI) a(n) = vecsum(apply(x -> x%2 - (x==1), factor(n)[, 2])); \\ _Amiram Eldar_, Sep 28 2023
%Y Cf. A056169, A162642, A295659, A295663, A295664.
%Y Cf. A295661 (positions of nonzero terms).
%K nonn,easy
%O 1,216
%A _Antti Karttunen_, Nov 28 2017
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