cp's OEIS Frontend

A076538 Numerators a(n) of fractions slowly converging to e: let a(1) = 0, b(n) = n - a(n); if (a(n) + 1) / b(n) < e then a(n+1) = a(n) + 1, else a(n+1)= a(n).

%I A076538 #8 Jun 25 2022 00:38:05
%S A076538 0,1,2,2,3,4,5,5,6,7,8,8,9,10,10,11,12,13,13,14,15,16,16,17,18,19,19,
%T A076538 20,21,21,22,23,24,24,25,26,27,27,28,29,29,30,31,32,32,33,34,35,35,36,
%U A076538 37,38,38,39,40,40,41,42,43,43,44,45,46,46,47,48,48,49,50,51,51,52,53
%N A076538 Numerators a(n) of fractions slowly converging to e: let a(1) = 0, b(n) = n - a(n); if (a(n) + 1) / b(n) < e then a(n+1) = a(n) + 1, else a(n+1)= a(n).
%C A076538 a(n) + b(n) = n and as n -> +infinity, a(n) / b(n) converges to e. For all n, a(n) / b(n) < e.
%F A076538 a(1) = 0. b(n) = n - a(n). If (a(n) + 1) / b(n) < e, then a(n+1) = a(n) + 1, else a(n+1) = a(n).
%F A076538 a(n) = floor(n*exp(1)/(exp(1)+1)). - _Vladeta Jovovic_, Oct 04 2003
%e A076538 a(6)= 4 so b(6) = 6 - 4 = 2. a(7) = 5 because (a(6) + 1) / b(6) = 5/2 which is < e. So b(7) = 7 - 5 = 2. a(8) = 5 because (a(7) + 1) / b(7) = 6/2 which is not < e.
%o A076538 (PARI) a(n)=local(t); if(n<2,0,t=0; for(k=0,n-1,if(1+t<exp(1)*(k-t),t++)); t)
%Y A076538 Cf. A074840.
%Y A076538 Partial sums of A144610.
%K A076538 easy,frac,nonn
%O A076538 1,3
%A A076538 Robert A. Stump (bee_ess107(AT)msn.com), Oct 18 2002