A245800 a(n) is the least number k such that Sum_{j=S(n)+1..S(n)+k} 1/j >= 1/2, where S(n) = Sum_{i=1..n-1} a(i) and S(1) = 0.

1, 1, 2, 3, 5, 9, 14, 24, 39, 64, 106, 175, 288, 475, 783, 1291, 2129, 3510, 5787, 9541, 15730, 25935, 42759, 70498, 116232, 191634, 315951, 520915, 858844, 1415994, 2334579, 3849070, 6346044, 10462858, 17250336, 28440996, 46891275, 77310643, 127463701, 210152115, 346482262

Offset

1,3

Comments

The limiting ratio of consecutive terms is sqrt(e) ~ 1.648721270700128.

Conjecture 1: If the summation threshold (which, for this sequence, is defined as 1/2) is set to any positive number R, then the limiting ratio of consecutive terms in the resulting sequence is e^R.

Conjecture 2: If the summation threshold is set to log(phi) = 0.4812118250..., then the Fibonacci sequence is generated.

Partition the harmonic series into the smallest-sized groups that sum to 1/2 or greater. You get 1, 1/2, 1/3 + 1/4, 1/5 + 1/6 + 1/7, 1/8 + 1/9 + 1/10 + 1/11 + 1/12, 1/13 + 1/14 + ... + 1/21, 1/22 + ... ; a(n) gives the number of terms in the n-th group. - Derek Orr, Nov 08 2014

A partition sums to exactly 1/2 for only n=2. - Jon Perry, Nov 09 2014

Links

Table of n, a(n) for n=1..41.

Example

Start with 1/1; 1/1 >= 1/2, so a(1) = 1.
1/1 has been used, so start a new sum with 1/2; 1/2 >= 1/2, so a(2) = 1.
1/1 and 1/2 have been used, so start a new sum with 1/3; 1/3 + 1/4 >= 1/2, so a(3) = 2.
1/1 through 1/4 have been used, so start a new sum with 1/5; 1/5 + 1/6 + 1/7 >= 1/2, so a(4) = 3, etc.

Prog

(Python)
import math
total = 0
prev = 1
count = 0
for n in range(1, 10**7):
....total = total + 1/n
....count = count + 1
....if (total >= 0.5):
........print(count, end=', ')
........prev = count
........total = 0
........count = 0
.
print("\ndone")
print(math.sqrt(math.e))
# Rob van den Tillaart, Aug 22 2014 [Corrected by Derek Orr, Oct 16 2014]
(Python)
def a(n):
..if n == 1:
....return 1
..tot, c, k = 0, 0, 0
..for x in range(1, 10**7):
....tot += 1/x
....if tot >= 0.5:
......k += 1
......tot = 0
....if k == n-1:
......c += 1
....if k == n:
......return c
n = 1
while n < 100:
..print(a(n), end=', ')
..n += 1
# Derek Orr, Oct 16 2014
(PARI) lista(nn) = {n = 1; while (n < nn, k = 1; while (sum(x=n, n+k-1, 1/x) < 1/2, k++); print1(k, ", "); n = n+k; ); } \\ Michel Marcus, Sep 14 2014
(PARI) a(n)=if(n==1, return(1)); p=sum(i=1, n-1, a(i)); k=1; while(sum(x=p+1, p+k, 1/x)<1/2, k++); k
n=1; while(n<100, print1(a(n), ", "); n++) \\ Derek Orr, Oct 16 2014

Crossrefs

Cf. A226161.

Cf. A019774, A002390.

Sequence in context: A173714 A026746 A004699 * A291896 A018155 A227375

Adjacent sequences: A245797 A245798 A245799 * A245801 A245802 A245803

Keyword

nonn

Author

Rob van den Tillaart, Aug 22 2014

Extensions

Edited by Derek Orr, Nov 08 2014

Status

approved