A232111
Numerator of smallest nonnegative fraction of form +- 1 +- 1/2 +- 1/3 ... +- 1/n.
Original entry on oeis.org
0, 1, 1, 1, 1, 7, 1, 11, 13, 11, 11, 23, 23, 607, 251, 251, 25, 97, 97, 3767, 3767, 3767, 457, 24319, 24319, 7951, 4261, 13703, 13703, 872843, 872843, 17424097, 13828799, 902339, 7850449, 7850449, 7850449, 1526171, 68185267, 3429883, 3429883
Offset: 0
1-1/2-1/3-1/4+1/5 = 7/60. No other choice of term signs yields a smaller nonnegative fraction, so a(5) = 7.
0/1, 1/1, 1/2, 1/6, 1/12, 7/60, 1/20, 11/420, 13/840, 11/2520, 11/2520, 23/27720, 23/27720, 607/360360, 251/360360, 251/360360, 25/144144, 97/12252240, ...
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nMax = 19; d = {0}; Table[d = Flatten[{d + 1/n, d - 1/n}]; Numerator[Min[Abs[d]]], {n, nMax}] (* T. D. Noe, Nov 20 2013 *)
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a(n,t=0)=if(n==1,numerator(abs(n-t)),min(a(n-1,t-1/n),a(n-1,t+1/n))) \\ Charles R Greathouse IV, Apr 06 2014
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from itertools import product
from fractions import Fraction
def A232111(n): return min(x for x in (sum(d[i]*Fraction(1,i+1) for i in range(n)) for d in product((1,-1),repeat=n)) if x >= 0).numerator # Chai Wah Wu, Nov 24 2021
A349544
Smallest possible value of |Sum_{k=0..n} (+-) 2^k * 3^(n-k)|, where each (+-) can be either plus or minus sign, independently for each term in the sum.
Original entry on oeis.org
1, 1, 1, 5, 1, 19, 7, 5, 65, 61, 73, 227, 257, 5, 439, 1253, 2425, 2035, 833, 2677, 10591, 6509, 32071, 41173, 77263, 114323, 18145, 129685, 321151, 15757, 645449, 113957, 50735, 477653, 24295, 5089013, 3743881, 4809115, 12209455, 8216179, 32894927, 80299843, 45673913
Offset: 0
For n = 3, there are 2^3 = 8 possible choices of signs: 3^3 + 2*3^2 + 2^2*3 + 2^3 = 65, 3^3 + 2*3^2 + 2^2*3 - 2^3 = 49, 3^3 + 2*3^2 - 2^2*3 + 2^3 = 41, 3^3 + 2*3^2 - 2^2*3 - 2^3 = 25, 3^3 - 2*3^2 + 2^2*3 + 2^3 = 29, 3^3 - 2*3^2 + 2^2*3 - 2^3 = 13, 3^3 - 2*3^2 - 2^2*3 + 2^3 = 5, and 3^3 - 2*3^2 - 2^2*3 - 2^3 = -11. The smallest absolute value is 5, so a(3) = 5.
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b:= proc(k, n) option remember; `if`(k<0, {0}, map(x->
(t-> [x+t, abs(x-t)][])(2^(n-k)*3^k), b(k-1, n)))
end:
a:= n-> min(b(n$2)):
seq(a(n), n=0..18); # Alois P. Heinz, Nov 21 2021
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Min@*Abs/@FoldList[Join[3 #1 + 2^#2, 3 #1 - 2^#2] &, {1}, Range[25]]
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def f(k,n):
if k == 0 and n == 0: return (x for x in (1,))
if k < n: return (y*3 for y in f(k,n-1))
return (abs(x+y) for x in f(k-1,n) for y in (2**n,-2**n))
def A349544(n): return min(f(n,n)) # Chai Wah Wu, Nov 24 2021
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