A379561 a(n) = A003418(n+1)*H(n), where H(n) = 1 + 1/2 + ... + 1/n is the n-th harmonic number.

2, 9, 22, 125, 137, 1029, 2178, 6849, 7129, 81191, 83711, 1118273, 1145993, 1171733, 2391514, 41421503, 42142223, 813635157, 825887397, 837527025, 848612385, 19761458895, 19994251455, 101086721625, 102157567401, 309561680403, 312536252003, 9146733078187

Offset: 1

Author

Miko Labalan, Dec 26 2024

Keywords

• nonn

Comments

abs(log(a(n)) - n - log(log(n))) < c*sqrt(n)*log(n)^(-1/2), where constant c = (2+A206431)*Pi/4. This also gives the upper bound of the squared error, (log(a(n)) - n - log(log(n)))^2 < (c^2)*n*log(n)^(-1).

A slightly better absolute error bound could be achieved by using the imaginary part of the nontrivial zeros of the Riemann zeta function, (zetazero(n)-1/2)/sqrt(-1) ~ (2*Pi)*n*LambertW(n/exp(1))^(-1). That bound would be, abs(log(a(n)) - n - log(log(n))) < sqrt(k)*sqrt(n)*LambertW(n/exp(1))^(-1/2), where constant k = 4*Pi/(1+2*A206431). This also gives the upper bound of the squared error, (log(a(n)) - n - log(log(n)))^2 < k*n*LambertW(n/exp(1))^(-1). The midline of the squared error would run along (4/(4+A206431))*n*LambertW(n/exp(1))^(-1).

Another slightly better absolute error bound but without relying on the properties of the zeta zeros would be, abs(log(a(n)) - n - log(log(n))) < n^(3/(9-10*A077761)).

log(a(n))-c*sqrt(n)*log(n)^(-1/2) is a lower bound of sigma_1(n) = A000203(n). Such that, n+log(log(n))-c*sqrt(n)*log(n)^(-1/2) < sigma_1(n) < H(n)+exp(H(n))*log(H(n)).

a(n) gives the total number of ordered pairs (k,m) where k in set {1,2,...,n}, m in set {1,2,...,A003418(n+1)}, and k divides m. Example: For n = 3, there are 22 ordered pairs (k,m) where k is {1,2,3} and m is a multiple of k up to 12. For k = 1, every m is a multiple of 1, m is {1,2,3,4,5,6,7,8,9,10,11,12} so there are 12 pairs. For k = 2, every m is a multiple of 2, m is {2,4,6,8,10,12} so there are 6 pairs. For k = 3, every m is a multiple of 3, m is {3,6,9,12} so there are 4 pairs. So the total ordered pairs is 12 + 6 + 4 = 22 = a(3). Each ordered pair (k,m) also represents an edge in a bipartite graph. Counting all such pairs gives the total number of edges in a graph.

Examples

			a(n)/A025558(n) = [ 2/1, 9/4, 22/9, 125/48, 137/50, 1029/360, 2178/735, ... ]
To evaluate the integral:
For n = 1: Integral_{x=0..1} Li_1(x^(1/2))/x^(1/2) dx = Integral_{x=0..1} -log(1-x^(1/2))/x^(1/2) dx = -2 * -(Sum_{x=1..oo} 1/(x*(x+1))) = -2 * -1 = 2.
For n = 2: Integral_{x=0..1} Li_1(x^(1/3))/x^(1/3) dx = Integral_{x=0..1} -log(1-x^(1/3))/x^(1/3) dx = -3 * -(Sum_{x=1..oo} 1/(x*(x+2))) = -3 * -((1/2)*(1+1/2)) = -3 * -3/4 = 9/4.
For n = 3: Integral_{x=0..1} Li_1(x^(1/4))/x^(1/4) dx = Integral_{x=0..1} -log(1-x^(1/4))/x^(1/4) dx = -4 * -(Sum_{x=1..oo} 1/(x*(x+3))) = -4 * -((1/3)*(1+1/2+1/3)) = -4 * -11/18 = 22/9.
		

Links

• Guilherme França and André LeClair, Statistical and other properties of Riemann zeros based on an explicit equation for the n-th zero on the critical line, arXiv:1307.8395 [math.NT], 2014.
• J. C. Lagarias, An elementary problem equivalent to the Riemann hypothesis, Am. Math. Monthly 109 (6) (2002) 534-543. arXiv preprint, arXiv:math/0008177 [math.NT], 2000-2001.

Crossrefs

Cf. A001008/A002805 (harmonic numbers).

Cf. A003418 (lcm).

Cf. A025558 (denominator).

Cf. A193758 (very similar sequence).

Cf. A000203, A025529, A027457, A077761, A152648, A206431.

Programs

PARI

a(n) = lcm(vector(n+1, i, i))*sum(i=1, n, 1/i); \\ Michel Marcus, Dec 28 2024

Formula

a(n) = A025558(n)*(Integral_{x=0..1} Li_1(x^(1/(n+1)))/x^(1/(n+1)) dx).

a(n) = A025558(n) + A027457(n+1).

Integral_{x=0..1} Li_1(x^(1/(n+1)))/x^(1/(n+1)) dx = ((n+1)/n)*H(n) = a(n)/A025558(n).

((n+1)/n)*H(n) ~ log(n) + gamma + (log(n)+gamma+1/2)/n + O(1/n^2).

log(a(n)) ~ n + log(log(n)) + O(c*sqrt(n)*log(n)^(-1/2)), (See comments for constant c).

G.f. for ((n+1)/n)*H(n): G(x) = Li_2(x)+(1/2)*log(1-x)^2-log(1-x)/(1-x), the lim_{x->oo} G(x) = -zeta(2).

Hyperbolic l.g.f. for ((n+1)/n)*H(n): LH(x) = Li_2(x)+(1/2)*log(1-x)^2+Li_3(x)-Li_3(1-x)+Li_2(1-x)*log(1-x)+(1/2)*log(x)*log(1-x)^2+zeta(3), the Integral_{x=0..1} LH(x) dx = 2*zeta(3) = A152648.

Dirichlet g.f. for ((n+1)/n)*H(n): zeta(s+1)*(zeta(s)+zeta(s+2)).