A226160
Least positive integer k such that 1 + 1/2 + ... + 1/k > n/tau, where tau = golden ratio = (1+sqrt(5))/2.
Original entry on oeis.org
1, 2, 4, 7, 12, 23, 42, 79, 146, 271, 503, 934, 1732, 3214, 5963, 11063, 20524, 38078, 70646, 131067, 243166, 451140, 836989, 1552846, 2880960, 5344978, 9916415, 18397696, 34132822, 63325839
Offset: 1
a(4) = 7 because 1 + 1/2 + ... + 1/6 < 4*tau < 1 + 1/2 + ... + 1/7.
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nn = 24; g = 1/GoldenRatio; f[n_] := 1/n; a[1] = 1; Do[s = 0; a[n] = NestWhile[# + 1 &, 1, ! (s += f[#]) > n*g &], {n, 1, nn}]; Map[a, Range[nn]]
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.
Original entry on oeis.org
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
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.
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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
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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
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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]
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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
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