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

1,1

Comments

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

Links

Table of n, a(n) for n=1..29.

Formula

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

Example

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.

Prog

(Python)
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
(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

Crossrefs

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

Sequence in context: A148295 A288540 A327014 * A046171 A022854 A356697

Adjacent sequences: A331027 A331028 A331029 * A331031 A331032 A331033

Keyword

nonn

Author

Pablo Hueso Merino and Alejandro Argüelles Trujillo, Jan 07 2020

Extensions

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

Status

approved