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 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. 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 (list; graph; refs; listen; history; text; internal format)
 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 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

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Last modified May 30 08:43 EDT 2020. Contains 334712 sequences. (Running on oeis4.)