A331028 Partition the terms of the harmonic series into groups sequentially so that the sum of each group is equal to or minimally greater than 1; then a(n) is the number of terms in the n-th group.

1, 3, 8, 22, 60, 163, 443, 1204, 3273, 8897, 24184, 65739, 178698, 485751, 1320408, 3589241, 9756569, 26521104, 72091835, 195965925, 532690613, 1448003214, 3936080824, 10699376979, 29083922018, 79058296722, 214902731368, 584166189564, 1587928337892, 4316436745787

Offset: 1

Author

Alejandro Argüelles Trujillo and Pablo Hueso Merino, Jan 07 2020

Keywords

• nonn

Comments

a(n) is equal to A024581(n) through a(10), and grows very similarly for n > 10.

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). Cf. A331030. - Jon E. Schoenfield, Jan 14 2020

Examples

			a(1)=1 because 1 >= 1,
a(2)=3 because 1/2 + 1/3 + 1/4 = 1.0833... >= 1, etc.
		

Crossrefs

Cf. A004080, A024581, A136616, A331030.

Some sequences in the same spirit as this: A002387, A004080, A055980, A115515.

Programs

PARI

default(realprecision, 10^5); e=exp(1);
lista(nn) = {my(r=1); print1(r); for(n=2, nn, print1(", ", -r+(r=floor(e*r+(e+1)/2+(e-1/e)/(24*(r+1/2)))))); } \\ Jinyuan Wang, Mar 31 2020

Python

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

Formula

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

Extensions

a(20)-a(21) from Giovanni Resta, Jan 14 2020

More terms from Jinyuan Wang, Mar 31 2020