A374663 Lexicographically earliest sequence of positive integers a(1), a(2), a(3), ... such that for any n > 0, Sum_{k = 1..n} 1 / (k*a(k)) < 1.

2, 2, 2, 4, 10, 201, 34458, 1212060151, 1305857607493406801, 1534737681943564047120326770001682121, 2141290683979549415450148346297540185977813099483710032048213090481251382

Offset: 1

Author

Rémy Sigrist, Aug 04 2024

Keywords

• nonn

Comments

The harmonic series, Sum_{k > 0} 1/k, diverges. We divide each of its terms in such a way as to have a series bounded by 1.

Examples

			The initial terms, alongside the corresponding sums, are:
  n  a(n)        Sum_{k=1..n} 1/(k*a(k))
  -  ----------  -----------------------------------------
  1           2  1/2
  2           2  3/4
  3           2  11/12
  4           4  47/48
  5          10  1199/1200
  6         201  241199/241200
  7       34458  9696481199/9696481200
  8  1212060151  11752718467440661199/11752718467440661200
...
The denominators are in A375516.
		

References

• Rémy Sigrist and N. J. A. Sloane, Dampening Down a Divergent Series, Manuscript in preparation, September 2024.

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..14
• N. J. A. Sloane, A Nasty Surprise in a Sequence and Other OEIS Stories, Experimental Mathematics Seminar, Rutgers University, Oct 10 2024, Youtube video; Slides [Mentions this sequence]

Crossrefs

Cf. A000058, A001008/A002805, A002387, A374983, A375516, A375517.

Programs

Maple

s:= proc(n) s(n):= `if`(n=0, 0, s(n-1)+1/(n*a(n))) end:
a:= proc(n) a(n):= 1+floor(1/((1-s(n-1))*n)) end:
seq(a(n), n=1..11);  # Alois P. Heinz, Oct 18 2024

Mathematica

s[n_] := s[n] = If[n == 0, 0, s[n - 1] + 1/(n*a[n])];
a[n_] := 1 + Floor[1/((1 - s[n - 1])*n)];
Table[a[n], {n, 1, 11}] (* Jean-François Alcover, Jan 09 2025, after Alois P. Heinz *)

PARI

{ t = 0; for (n = 1, 11, for (v = ceil(1/(n*(1-t))), oo, if (t + 1/(n*v) < 1, t += 1/(n*v); print1 (v", "); break;););); }

Python

from itertools import count, islice
from math import gcd
def A374663_gen(): # generator of terms
    p, q = 0, 1
    for k in count(1):
        yield (m:=q//(k*(q-p))+1)
        p, q = p*k*m+q, k*m*q
        p //= (r:=gcd(p,q))
        q //= r
A374663_list = list(islice(A374663_gen(),11)) # Chai Wah Wu, Aug 28 2024

Formula

The ratios a(n)^2/a(n+1) are very close to the values 2, 2, 1, 8/5, 1/2, 7/6, 48/49, 9/8, 10/9, 11/10, 24/11^2, 13/12, 56/13^2, ... So it seems that often (but not always), a(n+1) is very close to (n/(n+1))*a(n)^2. - N. J. A. Sloane, Sep 08 2024