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A317625 a(n) = Sum_{k=1..n} phi(floor(n/k)) where phi is the Euler totient function.

%I A317625 #39 Dec 13 2024 16:43:30
%S A317625 1,2,4,5,8,8,13,12,16,17,24,18,27,26,32,31,40,32,45,36,46,51,64,42,57,
%T A317625 58,68,61,78,60,83,68,80,85,100,74,99,94,110,91,116,90,121,104,116,
%U A317625 127,152,100,131,122,144,137,166,130,161,136,162,171,202,126,171,164,182,163,190
%N A317625 a(n) = Sum_{k=1..n} phi(floor(n/k)) where phi is the Euler totient function.
%H A317625 Robert Israel, <a href="/A317625/b317625.txt">Table of n, a(n) for n = 1..10000</a>
%H A317625 Olivier Bordellès, Randell Heyman, and Igor E. Shparlinski, <a href="https://arxiv.org/abs/1808.00188">On a sum involving the Euler function</a>, arXiv:1808.00188 [math.NT], 2018.
%F A317625 a(n) <= (1/2)*(1 + 1/zeta(2))*n*log(n) + 4*n + sqrt(n)*log(n)/4 + sqrt(n), uniformly for n >= 3.
%F A317625 a(n) >= ((2629/4009)+o(1))*n*log(n)/zeta(2) as n approaches infinity.
%F A317625 Cautious conjecture: a(n) ~ n*log(n)/zeta(2).
%e A317625 a(4) = phi(floor(4/1))+phi(floor(4/2))+phi(floor(4/3))+phi(floor(4/4)) = phi(4)+phi(2)+phi(1)+phi(1) = 2+1+1+1 = 5.
%p A317625 with(numtheory): S:=0: for x to 30 do: for m to x do: S := S+phi(trunc(x/m)) end do; print(x, S); S := 0:end do:
%t A317625 Array[Sum[EulerPhi[Floor[#/k]], {k, #}] &, 65] (* _Michael De Vlieger_, Aug 02 2018 *)
%o A317625 (PARI) a(n) = sum(x=1, n, eulerphi(n\x)); \\ _Michel Marcus_, Aug 02 2018
%K A317625 nonn
%O A317625 1,2
%A A317625 _Randell G Heyman_, Aug 02 2018
%E A317625 More terms from _Michel Marcus_, Aug 02 2018