A256220
Number of times that the numerator of a sum generated from the set 1, 1/2, 1/3,..., 1/n is a Fibonacci number.
Original entry on oeis.org
1, 3, 5, 9, 11, 22, 28, 37, 45, 62, 70, 125, 133, 172, 330, 421, 450, 840, 901, 1710, 2356, 2724, 2824, 5367, 6022, 7142, 8072, 18771, 19204, 35739, 36453, 42853, 82094, 88574, 155642, 264869
Offset: 1
a(3) = 5 because we obtain 5 following subsets {1}, {1/2}, {1/3}, {1,1/2} and {1/2, 1/3} having 5 sums with Fibonacci numerators: 1, 1, 1, 1+1/2 = 3/2 and 1/2+1/3 = 5/6.
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<<"DiscreteMath`Combinatorica`"; maxN=22; For[cnt=0; i=0; n=1, n<=maxN, n++, While[i<2^n-1, i++; s=NthSubset[i, Range[n]]; k=Numerator[Plus@@(1/s)]; If[IntegerQ[Sqrt[5*k^2+4]]||IntegerQ[Sqrt[5*k^2-4]], cnt++ ]]; Print[cnt]]
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from math import gcd, lcm
from itertools import combinations
def A256220(n):
m = lcm(*range(1,n+1))
fibset, mlist = set(), tuple(m//i for i in range(1,n+1))
a, b, c, k = 0, 1, 0, sum(mlist)
while b <= k:
fibset.add(b)
a, b = b, a+b
for l in range(1,n//2+1):
for p in combinations(mlist,l):
s = sum(p)
if s//gcd(s,m) in fibset:
c += 1
if 2*l != n and (k-s)//gcd(k-s,m) in fibset:
c += 1
return c+int(k//gcd(k,m) in fibset) # Chai Wah Wu, Feb 15 2022
A256221
Number of distinct nonzero Fibonacci numbers in the numerator of the 2^n sums generated from the set 1, 1/2, 1/3, ..., 1/n.
Original entry on oeis.org
1, 2, 3, 4, 5, 6, 8, 8, 8, 12, 12, 13, 13, 13, 13, 15, 15, 15, 17, 17, 17, 19, 21, 21, 23, 24, 25, 25, 25, 25, 25, 27
Offset: 1
a(4) = 4 because 4 sums yield distinct Fibonacci numerators: 1, 1 + 1/2 = 3/2, 1/2 + 1/3 = 5/6 and 1/2 + 1/3 + 1/4 = 13/12.
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S:= {0,1}: N:= {1}:
nfibs:= 10:
fibs:= {seq(combinat:-fibonacci(n),n=1..nfibs)}:
A[1]:= 1:
fibnums:= {1}:
for n from 2 to 24 do
Sp:= map(`+`,S,1/n);
N:= N union map(numer, Sp);
Nmax:= max(N);
S:= S union Sp;
while combinat:-fibonacci(nfibs) < Nmax do nfibs:= nfibs+1; fibs:= fibs union {combinat:-fibonacci(nfibs)} od;
newfibnums:= N intersect fibs;
fibnums:= newfibnums;
A[n]:= nops(fibnums);
od:
seq(A[n],n=1..24); # Robert Israel, Dec 09 2016
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<<"DiscreteMath`Combinatorica`";maxN=23; For[prms={}; i=0; n=1, n<=maxN, n++, While[i<2^n-1, i++; s=NthSubset[i, Range[n]]; k=Numerator[Plus@@(1/s)]; If[IntegerQ[Sqrt[5*k^2+4]]||IntegerQ[Sqrt[5*k^2-4]],prms=Union[prms, {k}]]]; Print[Length[prms]]]
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from math import gcd, lcm
from itertools import combinations
def A256221(n):
m = lcm(*range(1,n+1))
fset, fibset, mlist = set(), set(), tuple(m//i for i in range(1,n+1))
a, b, k = 0, 1, sum(mlist)
while b <= k:
fibset.add(b)
a, b = b, a+b
for l in range(1,n//2+1):
for p in combinations(mlist,l):
s = sum(p)
if (t := s//gcd(s,m)) in fibset:
fset.add(t)
if 2*l != n and (t := (k-s)//gcd(k-s,m)) in fibset:
fset.add(t)
if (t:= k//gcd(k,m)) in fibset: fset.add(t)
return len(fset) # Chai Wah Wu, Feb 15 2022
A256222
Largest Fibonacci number in the numerator of the 2^n sums generated from the set 1, 1/2, 1/3, ..., 1/n.
Original entry on oeis.org
0, 1, 3, 5, 13, 13, 13, 89, 89, 89, 1597, 1597, 1597, 1597, 1597, 1597, 17711, 17711, 17711, 28657, 28657, 28657, 28657, 1346269, 1346269, 1346269, 1346269, 24157817, 24157817, 24157817, 24157817, 24157817, 24157817, 39088169, 39088169, 39088169, 39088169
Offset: 0
a(3) = 5 because we obtain the 5 subsets {1}, {1/2}, {1/3}, {1,1/2} and {1/2, 1/3} having 5 sums with Fibonacci numerators: 1, 1, 1, 1+1/2 = 3/2 and 1/2+1/3 = 5/6 => the greatest Fibonacci number is 5.
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<<"DiscreteMath`Combinatorica`"; maxN=24; For[t={}; mx=0; i=0; n=0, n<=maxN, n++, While[i<2^n-1, i++; s=NthSubset[i, Range[n]]; k=Numerator[Plus@@(1/s)]; If[IntegerQ[Sqrt[5*k^2+4]]||IntegerQ[Sqrt[5*k^2-4]], If[k>mx, t=s]; mx=Max[mx, k]]]; Print[mx]]
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