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

Author

Rob van den Tillaart, Aug 22 2014

Keywords

• nonn

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

Examples

			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.
		

Crossrefs

Cf. A226161.

Cf. A019774, A002390.

Programs

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

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

Extensions

Edited by Derek Orr, Nov 08 2014