

A328684


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


2



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
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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, 11/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 * A006527 A167875 A057304
Adjacent sequences: A328678 A328679 A328681 * A328685 A328687 A328688


KEYWORD

nonn


AUTHOR

Artur Jasinski, Oct 25 2019


STATUS

approved



