A370901 Partial alternating sums of the powerfree part function (A055231).

1, -1, 2, 1, 6, 0, 7, 6, 7, -3, 8, 5, 18, 4, 19, 18, 35, 33, 52, 47, 68, 46, 69, 66, 67, 41, 42, 35, 64, 34, 65, 64, 97, 63, 98, 97, 134, 96, 135, 130, 171, 129, 172, 161, 166, 120, 167, 164, 165, 163, 214, 201, 254, 252, 307, 300, 357, 299, 358, 343, 404, 342

Offset: 1

Author

Amiram Eldar, Mar 05 2024

Keywords

• sign
• easy

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• László Tóth, Alternating Sums Concerning Multiplicative Arithmetic Functions, Journal of Integer Sequences, Vol. 20 (2017), Article 17.2.1.

Crossrefs

Cf. A055231, A191622, A370900.

Similar sequences: A068762, A068773, A307704, A357817, A362028.

Programs

Mathematica

f[p_, e_] := If[e == 1, p, 1]; pfp[n_] := Times @@ f @@@ FactorInteger[n]; pfp[1] = 1; Accumulate[Array[(-1)^(# + 1) * pfp[#] &, 100]]

PARI

pfp(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] == 1, f[i, 1], 1));}
lista(kmax) = {my(s = 0); for(k = 1, kmax, s += (-1)^(k+1) * pfp(k); print1(s, ", "))};

Formula

a(n) = Sum_{k=1..n} (-1)^(k+1) * A055231(k).

a(n) = (5/38) * c * n^2 + O(R(n)), where c = Product_{p prime} (1 - (p^2+p-1)/(p^3*(p+1))) = 0.649606... (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).