A379583 Numerators of the partial sums of the reciprocals of the powerful part function (A057521).
1, 2, 3, 13, 17, 21, 25, 51, 467, 539, 611, 629, 701, 773, 845, 1699, 1843, 1859, 2003, 2039, 2183, 2327, 2471, 2489, 62369, 65969, 198307, 201007, 211807, 222607, 233407, 467489, 489089, 510689, 532289, 532889, 554489, 576089, 597689, 600389, 621989, 643589, 665189
Offset: 1
Examples
Fractions begin with 1, 2, 3, 13/4, 17/4, 21/4, 25/4, 51/8, 467/72, 539/72, 611/72, 629/72, ...
Links
- Amiram Eldar, Table of n, a(n) for n = 1..1000
- Maurice-Étienne Cloutier, Les parties k-puissante et k-libre d'un nombre, Thèse de doctorat, Université Laval, Québec (2018).
- Maurice-Étienne Cloutier, Jean-Marie De Koninck, and Nicolas Doyon, On the powerful and squarefree parts of an integer, Journal of Integer Sequences, Vol. 17 (2014), Article 14.6.6.
- László Tóth, Alternating Sums Concerning Multiplicative Arithmetic Functions, Journal of Integer Sequences, Vol. 20 (2017), Article 17.2.1. See section 4.12, p. 33.
Programs
-
Mathematica
f[p_, e_] := If[e > 1, p^e, 1]; powful[n_] := Times @@ f @@@ FactorInteger[n]; Numerator[Accumulate[Table[1/powful[n], {n, 1, 50}]]] -
PARI
powerful(n) = {my(f = factor(n)); prod(i=1, #f~, if(f[i, 2] > 1, f[i, 1]^f[i, 2], 1)); } list(nmax) = {my(s = 0); for(k = 1, nmax, s += 1 / powerful(k); print1(numerator(s), ", "))};
Formula
a(n) = numerator(Sum_{k=1..n} 1/A057521(k)).
a(n)/A379584(n) = c * n + O(n^(1/2)), where c = A191622 (Cloutier et al., 2014). The error term was improved by Tóth (2017) to O(n^(1/2) * exp(-c1 * log(n)^(3/5) / log(log(n))^(1/5))) unconditionally, and O(n^(2/5) * exp(c2 * log(n) / log(log(n)))) assuming the Riemann hypothesis, where c1 and c2 are positive constants.