A386921 Numerators of the partial sums of 1/d(prime(k)+1), where d is the number of divisors function.

1, 5, 13, 4, 3, 7, 23, 25, 53, 7, 5, 11, 23, 73, 377, 49, 67, 18, 113, 77, 41, 21, 257, 131, 68, 559, 287, 73, 599, 307, 629, 213, 109, 83, 337, 689, 719, 739, 1493, 1523, 4609, 4699, 33253, 34513, 34933, 35353, 36193, 36613, 37033, 37663, 38083, 7667, 7835, 7891

Offset: 1

Author

Amiram Eldar, Aug 08 2025

Keywords

• nonn
• frac
• easy

Examples

			Fractions begin with 1/2, 5/6, 13/12, 4/3, 3/2, 7/4, 23/12, 25/12, 53/24, 7/3, 5/2, 11/4, ...
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Mikhail R. Gabdullin, Vitalii V. Iudelevich, and Sergei V. Konyagin, Karatsuba's divisor problem and related questions, arXiv:2304.04805 [math.NT], 2023.
• Vitalii V. Iudelevich, On the Karatsuba divisor problem, Izvestiya: Mathematics, Vol. 86, No. 5 (2022), pp. 992-1019; arXiv preprint, arXiv:2304.03049 [math.NT], 2023.

Crossrefs

Cf. A000005, A008329, A008864, A104528, A386922 (denominators).

Programs

Mathematica

Numerator[Accumulate[1/DivisorSigma[0, Prime[Range[100]] + 1]]]

PARI

list(lim) = {my(s = 0); forprime(p = 1, lim, s += (1/numdiv(p+1)); print1(numerator(s), ", "));}

Formula

a(n) = numerator(Sum_{k=1..n} 1/A008329(k)).

a(n)/A386922(n) <= 4 * K * n/log(n)^(3/2) + O(n*log(log(n))/log(n)^(5/2)), where K = (1/sqrt(Pi)) * Product_{p prime} sqrt(p/(p-1)) * (p * log(p/(p-1)) - 1/(p-1)) = 0.25320111501639846923... (Iudelevich, 2022). Gabdullin et al. (2023) conjectured that a(n)/A386922(n) ~ K * n/log(n)^(3/2).