%I #50 Jan 09 2025 15:03:36
%S 2,2,2,4,10,201,34458,1212060151,1305857607493406801,
%T 1534737681943564047120326770001682121,
%U 2141290683979549415450148346297540185977813099483710032048213090481251382
%N 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.
%C 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.
%D Rémy Sigrist and N. J. A. Sloane, Dampening Down a Divergent Series, Manuscript in preparation, September 2024.
%H N. J. A. Sloane, <a href="/A374663/b374663.txt">Table of n, a(n) for n = 1..14</a>
%H N. J. A. Sloane, <a href="https://www.youtube.com/watch?v=3RAYoaKMckM">A Nasty Surprise in a Sequence and Other OEIS Stories</a>, Experimental Mathematics Seminar, Rutgers University, Oct 10 2024, Youtube video; <a href="https://sites.math.rutgers.edu/~zeilberg/expmath/sloane85BD.pdf">Slides</a> [Mentions this sequence]
%F 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
%e The initial terms, alongside the corresponding sums, are:
%e n a(n) Sum_{k=1..n} 1/(k*a(k))
%e - ---------- -----------------------------------------
%e 1 2 1/2
%e 2 2 3/4
%e 3 2 11/12
%e 4 4 47/48
%e 5 10 1199/1200
%e 6 201 241199/241200
%e 7 34458 9696481199/9696481200
%e 8 1212060151 11752718467440661199/11752718467440661200
%e ...
%e The denominators are in A375516.
%p s:= proc(n) s(n):= `if`(n=0, 0, s(n-1)+1/(n*a(n))) end:
%p a:= proc(n) a(n):= 1+floor(1/((1-s(n-1))*n)) end:
%p seq(a(n), n=1..11); # _Alois P. Heinz_, Oct 18 2024
%t s[n_] := s[n] = If[n == 0, 0, s[n - 1] + 1/(n*a[n])];
%t a[n_] := 1 + Floor[1/((1 - s[n - 1])*n)];
%t Table[a[n], {n, 1, 11}] (* _Jean-François Alcover_, Jan 09 2025, after _Alois P. Heinz_ *)
%o (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;););); }
%o (Python)
%o from itertools import count, islice
%o from math import gcd
%o def A374663_gen(): # generator of terms
%o p, q = 0, 1
%o for k in count(1):
%o yield (m:=q//(k*(q-p))+1)
%o p, q = p*k*m+q, k*m*q
%o p //= (r:=gcd(p,q))
%o q //= r
%o A374663_list = list(islice(A374663_gen(),11)) # _Chai Wah Wu_, Aug 28 2024
%Y Cf. A000058, A001008/A002805, A002387, A374983, A375516, A375517.
%K nonn
%O 1,1
%A _Rémy Sigrist_, Aug 04 2024