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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
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
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
KEYWORD
nonn
EXTENSIONS
a(25)-a(29) from Jon E. Schoenfield, Jan 10 2020
STATUS
approved