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Partial sums of the powerfree part function (A055231).
6

%I #9 Mar 05 2024 11:50:59

%S 1,3,6,7,12,18,25,26,27,37,48,51,64,78,93,94,111,113,132,137,158,180,

%T 203,206,207,233,234,241,270,300,331,332,365,399,434,435,472,510,549,

%U 554,595,637,680,691,696,742,789,792,793,795,846,859,912,914,969,976,1033

%N Partial sums of the powerfree part function (A055231).

%D Steven R. Finch, Mathematical Constants II, Cambridge University Press, 2018, p. 52.

%H Amiram Eldar, <a href="/A370900/b370900.txt">Table of n, a(n) for n = 1..10000</a>

%H Eckford Cohen, <a href="https://doi.org/10.1215/S0012-7094-61-02816-2">An elementary method in the asymptotic theory of numbers</a>, Duke Mathematical Journal, Vol. 28, No. 2 (1961), pp. 183-192.

%H Eckford Cohen, <a href="https://doi.org/10.1090/S0002-9947-1964-0166181-5">Some asymptotic formulas in the theory of numbers</a>, Transactions of the American Mathematical Society, Vol. 112, No. 2 (1964), pp. 214-227.

%H László Tóth, <a href="https://www.emis.de/journals/JIS/VOL20/Toth/toth25.html">Alternating Sums Concerning Multiplicative Arithmetic Functions</a>, Journal of Integer Sequences, Vol. 20 (2017), Article 17.2.1.

%F a(n) = Sum_{k=1..n} A055231(k).

%F a(n) = c * n^2 / 2 + O(R(n)), where c = Product_{p prime} (1 - (p^2+p-1)/(p^3*(p+1))) = 0.649606699337... (A191622), R(n) = x^(3/2) * exp(-c_1 * log(n)^(3/5) / log(log(n))^(1/5)) unconditionally, or x^(7/5) * exp(c_2 * log(n) / log(log(n))) assuming the Riemann hypothesis, and c_1 and c_2 are positive constants (Tóth, 2017).

%t f[p_, e_] := If[e == 1, p, 1]; pfp[n_] := Times @@ f @@@ FactorInteger[n]; pfp[1] = 1; Accumulate[Array[pfp[#] &, 100]]

%o (PARI) pfp(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] == 1, f[i, 1], 1));}

%o lista(kmax) = {my(s = 0); for(k = 1, kmax, s += pfp(k); print1(s, ", "))};

%Y Cf. A055231, A191622, A370901.

%K nonn,easy

%O 1,2

%A _Amiram Eldar_, Mar 05 2024