A328684 Least k such that Sum_{m=1..k} 1/m > Product_{i=1..n} 1/(1 - 1/prime(i)).

4, 11, 24, 45, 69, 103, 143, 194, 254, 315, 390, 467, 553, 651, 759, 872, 990, 1121, 1258, 1405, 1566, 1734, 1912, 2097, 2285, 2483, 2696, 2921, 3161, 3415, 3659, 3915, 4178, 4457, 4736, 5030, 5332, 5642, 5964, 6295, 6633, 6988, 7343, 7715, 8099, 8501, 8900, 9296, 9704

Offset

1,1

Comments

This sequence compares partial sums of harmonic series with Euler's partial products.

Both Sum_{m=1..k} 1/m and Product_{i=1..n} 1/(1 - 1/prime(i)) are divergent.

Links

Table of n, a(n) for n=1..49.

Example

a(2)=11 because Sum_{m=1..10} 1/m = 7381/2520 = 2.92896... < Product_{i=1..2} 1/(1 - 1/prime(n)) = 3 < Sum_{m=1..11} 1/m = 83711/27720 = 3.01987...

Mathematica

s = 1; prec = 350; dd = {}; h = 1; hh = 1; k = 1; m = 1; Do[
k = k (1/(1 - Prime[n]^-s)); kk = N[k, prec];
While[kk > hh, h = h + 1/(m + 1)^s; hh = N[h, prec]; m++];
AppendTo[dd, m], {n, 1, 68}]; dd

Prog

(PARI) a(n) = my(k=1, pp = prod(i=1, n, 1/(1 - 1/prime(i))), s = 1); while (s <= pp, k++; s += 1/k); k; \\ Michel Marcus, Oct 29 2019
(PARI) apply( {A328684(n, p=1/prod(k=1, n, 1-1/prime(k)))=for(k=1, oo, (0 > p -= 1/k) && return(k))}, [1..49]) \\ M. F. Hasler, Oct 31 2019

Crossrefs

Cf. A328655.

Sequence in context: A306262 A099074 A014818 * A167875 A006527 A057304

Adjacent sequences: A328681 A328682 A328683 * A328685 A328686 A328687

Keyword

nonn

Author

Artur Jasinski, Oct 25 2019

Status

approved