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%I #24 Nov 01 2020 21:42:32
%S 0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,
%T 0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,0,0,
%U 0,0,0,1,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1
%N Number of exponents larger than 2 in the prime factorization of n.
%H Antti Karttunen, <a href="/A295659/b295659.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^e) = 1 if e>2, 0 otherwise.
%F a(n) = 0 iff A212793(n) = 1.
%F a(n) = A001221(A053150(n)).
%F a(n) = A056170(A003557(n)).
%F a(n) >= A295662(n) = A162642(n) - A056169(n).
%F a(n) = A295883(n) + A295884(n).
%F Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = Sum_{p prime} 1/p^3 = 0.174762... (A085541). - _Amiram Eldar_, Nov 01 2020
%e For n = 120 = 2^3 * 3^1 * 5^1 there is only one exponent larger than 2, thus a(120) = 1.
%e For n = 216 = 2^3 * 3^3 there are two exponents larger than 2, thus a(216) = 2.
%t Array[Count[FactorInteger[#][[All, -1]], _?(# > 2 &)] &, 105] (* _Michael De Vlieger_, Nov 28 2017 *)
%o (Scheme, with memoization-macro definec) (definec (A295659 n) (if (= 1 n) 0 (+ (if (> (A067029 n) 2) 1 0) (A295659 (A028234 n)))))
%o (PARI) a(n) = { my(v = factor(n)[, 2], i=0); for(x=1, length(v), if(v[x]>2, i++)); i; } \\ _Iain Fox_, Nov 29 2017
%Y Cf. A001221, A003557, A053150, A056169, A056170, A085541, A162642, A295657, A295662, A295883, A295884.
%Y Cf. A004709 (positions of zeros), A046099 (of nonzeros), A212793.
%K nonn
%O 1,216
%A _Antti Karttunen_, Nov 28 2017
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