A096005 -id:A096005 - 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.

2, 5, 13, 36, 98, 266, 723, 1965, 5342, 14521, 39472, 107296, 291661, 792817, 2155100, 5858169, 15924154, 43286339, 117664468, 319845186, 869429357, 2363354022, 6424262292, 17462955450, 47469234471, 129034757473, 350752836478, 953445061679, 2591732385596

Offset: 1

Pablo Hueso Merino and Alejandro Argüelles Trujillo, Jan 07 2020

nonn

a(n) = A046171(n+1) through a(5), and grows similarly for n > 5.

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

			a(1)=2 because 1 + 1/2 = 1.5 > 1,
a(2)=5 because 1/3 + 1/4 + 1/5 + 1/6 + 1/7 = 1.0928... > 1,
etc.

Cf. A002387, A046171, A081881, A096005, A180224, A295572, A331028.

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

x = 0.0
y = 0.0
z = 0.0
for i in range(1,100000000000000000000000):
  y += 1
  x = x + 1/i
  z = z + 1/i
  if x > 1:
    print(y)
    y = 0
    x = 0

a(1)=2, a(n) = (min(p) : Sum_{s=r..p} 1/s > 1)-r+1, r=Sum_{k=1..n-1} a(k).

a(25)-a(29) from Jon E. Schoenfield, Jan 10 2020

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.

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