login
A338559
Numerators of the fractions f(n) such that (6/Pi^2)*f(n) is the asymptotic density of the numbers k with A280292(k) = sopfr(k) - sopf(k) = n.
4
1, 0, 1, 1, 1, 17, 5, 17, 61, 559, 269, 5851, 5279, 954913, 1693849, 6394159, 1430687, 33257690117, 393069739, 330504317141, 146861034421, 3447587278559, 13150098373, 17185160160637123, 68404253084009, 219367146802450039, 527431007100952693, 2089195405327981487
OFFSET
0,6
COMMENTS
Alladi and Erdős (1977) proved that for all numbers n>=0, n!=1, the sequence of numbers k such that A280292(k) = n has a positive asymptotic density which is equal to a rational multiple of 1/zeta(2) = 6/Pi^2 (A059956).
LINKS
Krishnaswami Alladi and Paul Erdős, On an additive arithmetic function, Pacific Journal of Mathematics, Vol. 71, No. 2 (1977), pp. 275-294, alternative link. See Theorem 1.7, pp. 286-287.
Aleksandar Ivić, On certain large additive functions, arXiv:math/0311505 [math.NT], 2003.
FORMULA
Sum_{k>=0} a(k)/A338560(k) = zeta(2) = Pi^2/6 (A013661).
Let {P_j} be the set of partitions of n into prime parts, with j = 1..A000607(n). For a partition P_j = {p_i, b_i} of n = Sum_i b_i * p_i, where p_i are distinct primes, and b_i >= 1 are their multiplicities in the partition, let S(P_j) = Product_i p_i^(b_i + 1)) be a powerful number associated with the partition. f(n) = a(n)/A338560(n) = Sum_{j=1..r(n)} 1/psi(S(P_j)), where psi is the Dedekind psi function (A001615).
For any r>0, Sum_{n<=x, n nonsquarefree} 1/A280292(n)^r ~ c(r)*x + O(x^(1-r/2)*log(x)) + O(x^(1/2)*log(x)), where c(r) = (6/Pi^2) * Sum_{k>=2} (a(k)/A338560(k)) * (1/k^r) (Ivić, 2003).
EXAMPLE
1/1, 0/1, 1/6, 1/12, 1/12, 17/360, 5/72, 17/560, 61/2160, 559/30240, 269/12600, 5851/399168, ...
For n=0, the sequence of numbers k such that A280292(k) = 0 are the squarefree numbers (A005117), whose density is 6/Pi^2. Thus f(0) = a(0)/A338560(0) = 1 and a(0) = 1.
For n=1, there are no numbers k with A280292(k) = 1, thus a(1) = 0.
For n=2, the sequence of numbers k with A280292(k) = 2 is A081770 whose density is 1/Pi^2. Thus f(2) = a(2)/A338560(2) = 1/6 and a(2) = 1.
For n=7, there are A000607(7) = 3 partitions of 7 into prime parts: {{p_i, b_i}} = {{2, 2}, {3, 1}}, {{2, 1}, {5, 1}}, and {{7, 1}}. The powerful numbers associated with these partitions are 2^(2+1)*3^(1+1) = 72, 2^(1+1)*5^(1+1) = 100, and 7^(1+1) = 49. Thus, f(7) = a(7)/A338560(7) = 1/psi(72) + 1/psi(100) + 1/psi(49) = 1/144 + 1/180 + 1/56 = 17/560, and a(7) = 17.
MATHEMATICA
psi[1] = 1; psi[n_] := n*Times @@ (1 + 1/Transpose[FactorInteger[n]][[1]]); delta[ps_] := 1/psi[(Times @@ ps)*(Times @@ Union[ps])]; f[n_] := Total[delta /@ IntegerPartitions[n, Floor[n/2], Select[Range[n], PrimeQ]]]; Numerator @ Array[f, 30, 0]
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Amiram Eldar, Nov 02 2020
STATUS
approved