cp's OEIS Frontend

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

%I A379561 #131 Feb 05 2025 16:56:34
%S A379561 2,9,22,125,137,1029,2178,6849,7129,81191,83711,1118273,1145993,
%T A379561 1171733,2391514,41421503,42142223,813635157,825887397,837527025,
%U A379561 848612385,19761458895,19994251455,101086721625,102157567401,309561680403,312536252003,9146733078187
%N A379561 a(n) = A003418(n+1)*H(n), where H(n) = 1 + 1/2 + ... + 1/n is the n-th harmonic number.
%C A379561 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).
%C A379561 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).
%C A379561 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)).
%C A379561 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)).
%C A379561 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.
%H A379561 Guilherme França and André LeClair, <a href="https://doi.org/10.48550/arXiv.1307.8395">Statistical and other properties of Riemann zeros based on an explicit equation for the n-th zero on the critical line</a>, arXiv:1307.8395 [math.NT], 2014.
%H A379561 J. C. Lagarias, <a href="https://www.jstor.org/stable/2695443">An elementary problem equivalent to the Riemann hypothesis</a>, Am. Math. Monthly 109 (6) (2002) 534-543. <a href="https://arxiv.org/abs/math/0008177">arXiv preprint</a>, arXiv:math/0008177 [math.NT], 2000-2001.
%F A379561 a(n) = A025558(n)*(Integral_{x=0..1} Li_1(x^(1/(n+1)))/x^(1/(n+1)) dx).
%F A379561 a(n) = A025558(n) + A027457(n+1).
%F A379561 Integral_{x=0..1} Li_1(x^(1/(n+1)))/x^(1/(n+1)) dx = ((n+1)/n)*H(n) = a(n)/A025558(n).
%F A379561 ((n+1)/n)*H(n) ~ log(n) + gamma + (log(n)+gamma+1/2)/n + O(1/n^2).
%F A379561 log(a(n)) ~ n + log(log(n)) + O(c*sqrt(n)*log(n)^(-1/2)), (See comments for constant c).
%F A379561 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).
%F A379561 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.
%F A379561 Dirichlet g.f. for ((n+1)/n)*H(n): zeta(s+1)*(zeta(s)+zeta(s+2)).
%e A379561 a(n)/A025558(n) = [ 2/1, 9/4, 22/9, 125/48, 137/50, 1029/360, 2178/735, ... ]
%e A379561 To evaluate the integral:
%e A379561 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.
%e A379561 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.
%e A379561 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.
%o A379561 (PARI) a(n) = lcm(vector(n+1, i, i))*sum(i=1, n, 1/i); \\ _Michel Marcus_, Dec 28 2024
%Y A379561 Cf. A001008/A002805 (harmonic numbers).
%Y A379561 Cf. A003418 (lcm).
%Y A379561 Cf. A025558 (denominator).
%Y A379561 Cf. A193758 (very similar sequence).
%Y A379561 Cf. A000203, A025529, A027457, A077761, A152648, A206431.
%K A379561 nonn
%O A379561 1,1
%A A379561 _Miko Labalan_, Dec 26 2024