cp's OEIS Frontend

A331030 Divide the terms of the harmonic series into groups sequentially so that the sum of each group is minimally greater than 1. a(n) is the number of terms in the n-th group.

%I A331030 #50 Jan 15 2020 00:24:54
%S A331030 2,5,13,36,98,266,723,1965,5342,14521,39472,107296,291661,792817,
%T A331030 2155100,5858169,15924154,43286339,117664468,319845186,869429357,
%U A331030 2363354022,6424262292,17462955450,47469234471,129034757473,350752836478,953445061679,2591732385596
%N A331030 Divide the terms of the harmonic series into groups sequentially so that the sum of each group is minimally greater than 1. a(n) is the number of terms in the n-th group.
%C A331030 a(n) = A046171(n+1) through a(5), and grows similarly for n > 5.
%C A331030 Let b(n) = Sum_{j=1..n} a(n); then for n >= 2 it appears that b(n) = round((b(n-1) + 1/2)*e). Verified through n = 10000 (using the approximation Sum_{j=1..k} 1/j = log(k) + gamma + 1/(2*k) - 1/(12*k^2) + 1/(120*k^4) - 1/(252*k^6) + 1/(240*k^8) - ..., where gamma is the Euler-Mascheroni constant, A001620). Cf. A081881. - _Jon E. Schoenfield_, Jan 10 2020
%F A331030 a(1)=2, a(n) = (min(p) : Sum_{s=r..p} 1/s > 1)-r+1, r=Sum_{k=1..n-1} a(k).
%e A331030 a(1)=2 because 1 + 1/2 = 1.5 > 1,
%e A331030 a(2)=5 because 1/3 + 1/4 + 1/5 + 1/6 + 1/7 = 1.0928... > 1,
%e A331030 etc.
%o A331030 (Python)
%o A331030 x = 0.0
%o A331030 y = 0.0
%o A331030 z = 0.0
%o A331030 for i in range(1,100000000000000000000000):
%o A331030   y += 1
%o A331030   x = x + 1/i
%o A331030   z = z + 1/i
%o A331030   if x > 1:
%o A331030     print(y)
%o A331030     y = 0
%o A331030     x = 0
%o A331030 (PARI) lista(lim=oo)={my(s=0, p=0); for(i=1, lim, s+=1/i; if(s>1, print1(i-p, ", "); s=0; p=i))} \\ _Andrew Howroyd_, Jan 08 2020
%Y A331030 Cf. A002387, A046171, A081881, A096005, A180224, A295572, A331028.
%K A331030 nonn
%O A331030 1,1
%A A331030 _Pablo Hueso Merino_ and _Alejandro Argüelles Trujillo_, Jan 07 2020
%E A331030 a(25)-a(29) from _Jon E. Schoenfield_, Jan 10 2020