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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.
35
2, 2, 2, 4, 10, 201, 34458, 1212060151, 1305857607493406801, 1534737681943564047120326770001682121, 2141290683979549415450148346297540185977813099483710032048213090481251382
OFFSET
1,1
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.
REFERENCES
Rémy Sigrist and N. J. A. Sloane, Dampening Down a Divergent Series, Manuscript in preparation, September 2024.
LINKS
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]
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
EXAMPLE
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.
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
PROG
(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
KEYWORD
nonn,changed
AUTHOR
Rémy Sigrist, Aug 04 2024
STATUS
approved