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a(n) is the greatest prime dividing the n-th cubefree number, for n >= 2; a(1)=1.
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%I #15 Aug 06 2024 02:05:21

%S 1,2,3,2,5,3,7,3,5,11,3,13,7,5,17,3,19,5,7,11,23,5,13,7,29,5,31,11,17,

%T 7,3,37,19,13,41,7,43,11,5,23,47,7,5,17,13,53,11,19,29,59,5,61,31,7,

%U 13,11,67,17,23,7,71,73,37,5,19,11,13,79,41,83,7,17,43

%N a(n) is the greatest prime dividing the n-th cubefree number, for n >= 2; a(1)=1.

%H Amiram Eldar, <a href="/A370833/b370833.txt">Table of n, a(n) for n = 1..10000</a>

%H Jean-Marie De Koninck and Rafael Jakimczuk, <a href="https://doi.org/10.33044/revuma.3154">Summing the largest prime factor over integer sequences</a>, Revista de la Unión Matemática Argentina, Vol. 67, No. 1 (2024), pp. 27-35.

%F a(n) = A006530(A004709(n)).

%F Sum_{A004709(n) <= x} a(n) = Sum_{i=1..k} d_i * x^2/log(x)^i + O(x^2/log(x)^(k+1)), for any given positive integer k, where d_i are constants, d_1 = 315/(4*Pi^4) = 0.808446... (De Koninck and Jakimczuk, 2024).

%t s[n_] := Module[{f = FactorInteger[n]}, If[AllTrue[f[[;; , 2]], # < 3 &], f[[-1, 1]], Nothing]]; Array[s, 200]

%o (PARI) lista(kmax) = {my(f); print1(1, ", "); for(k = 2, kmax, f = factor(k); if(vecmax(f[, 2]) < 3, print1(f[#f~, 1], ", ")));}

%o (Python)

%o from sympy import mobius, integer_nthroot, primefactors

%o def A370833(n):

%o def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x,3)[0]+1))

%o m, k = n, f(n)

%o while m != k:

%o m, k = k, f(k)

%o return max(primefactors(m),default=1) # _Chai Wah Wu_, Aug 06 2024

%Y Cf. A004709, A006530, A073482, A370834, A370835.

%K nonn,easy

%O 1,2

%A _Amiram Eldar_, Mar 03 2024