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a(n) = Sum_{k=1..n} mu(k) * prime(floor(n/k)).
3

%I #12 Mar 31 2021 17:25:48

%S 2,1,1,2,4,5,7,7,9,12,12,15,17,16,16,20,24,22,26,23,21,26,28,28,32,33,

%T 31,32,32,29,41,39,43,40,44,40,44,45,45,47,51,52,60,55,53,52,62,64,64,

%U 56,54,55,55,65,67,69,69,70,74,73,73,70,80,80,76,69,81,84,90,81,83,87,93,94

%N a(n) = Sum_{k=1..n} mu(k) * prime(floor(n/k)).

%F Sum_{k=1..n} a(floor(n/k)) = prime(n).

%t Table[Sum[MoebiusMu[k] Prime[Floor[n/k]], {k, 1, n}], {n, 1, 74}]

%t g[1] = 2; g[n_] := Prime[n] - Prime[n - 1]; a[n_] := Sum[Sum[MoebiusMu[k/d] g[d], {d, Divisors[k]}], {k, 1, n}]; Table[a[n], {n, 1, 74}]

%o (PARI) a(n) = sum(k=1, n, moebius(k)*prime(n\k)); \\ _Michel Marcus_, Mar 22 2020

%o (Python)

%o from functools import lru_cache

%o from sympy import prime

%o @lru_cache(maxsize=None)

%o def A333450(n):

%o c, j = 2*(n+1)-prime(n), 2

%o k1 = n//j

%o while k1 > 1:

%o j2 = n//k1 + 1

%o c += (j2-j)*A333450(k1)

%o j, k1 = j2, n//j2

%o return 2*j-c # _Chai Wah Wu_, Mar 31 2021

%Y Cf. A000040, A001223, A007444, A007504, A008683, A333449.

%K nonn

%O 1,1

%A _Ilya Gutkovskiy_, Mar 21 2020