When do unit fractions sum to an integer?












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Unit fractions are rational numbers in the form $frac{1}{n}$, where $n$ is an integer. In elementary number theory one can prove that the harmonic sum $$H_n= 1+ frac{1}{2}+...+frac{1}{n}$$ is never an integer. For elementary proof of this fact see the post: Elementary proof that the sum 1+...+1/n is not an integer



Take one step further, József Kürschák, in 1918, proved that the sum of the reciprocals of consecutive natural numbers (whether starting from 1 or not) is never an integer.



Unfortunately, the result does not hold with a border generalization when we consider the sum of arbitrary subsets of unit fractions. Examples include $$1=frac{1}{2}+frac{1}{3}+frac{1}{6}=frac{1}{2}+ frac{1}{3}+ frac{1}{7}+ frac{1}{43}+ frac{1}{1806}.$$ Erdős–Graham problem concerns with the existence of such unit fractions.



The question I ask now is: Without calculating the sum one by one, are there any nontrivial, characteristic properties that determines whether the given subset of unit fractions sums to an integer? One step back, is there any class of interesting subsets of unit fractions where the sum of each subset in the class has a non-integer sum?










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  • 1




    It's not hard to see that if $P$ is a perfect number, then $sum_{d|P, dneq 1} frac{1}{d} = 1$.
    – SAWblade
    Nov 27 '18 at 20:14












  • For any given finite subset of unit fractions, one obvious criterion is simply to sum them! You undoubtedly want something less obvious, but it might help to describe what you want in a "criterion."
    – Barry Cipra
    Nov 27 '18 at 20:32
















1














Unit fractions are rational numbers in the form $frac{1}{n}$, where $n$ is an integer. In elementary number theory one can prove that the harmonic sum $$H_n= 1+ frac{1}{2}+...+frac{1}{n}$$ is never an integer. For elementary proof of this fact see the post: Elementary proof that the sum 1+...+1/n is not an integer



Take one step further, József Kürschák, in 1918, proved that the sum of the reciprocals of consecutive natural numbers (whether starting from 1 or not) is never an integer.



Unfortunately, the result does not hold with a border generalization when we consider the sum of arbitrary subsets of unit fractions. Examples include $$1=frac{1}{2}+frac{1}{3}+frac{1}{6}=frac{1}{2}+ frac{1}{3}+ frac{1}{7}+ frac{1}{43}+ frac{1}{1806}.$$ Erdős–Graham problem concerns with the existence of such unit fractions.



The question I ask now is: Without calculating the sum one by one, are there any nontrivial, characteristic properties that determines whether the given subset of unit fractions sums to an integer? One step back, is there any class of interesting subsets of unit fractions where the sum of each subset in the class has a non-integer sum?










share|cite|improve this question




















  • 1




    It's not hard to see that if $P$ is a perfect number, then $sum_{d|P, dneq 1} frac{1}{d} = 1$.
    – SAWblade
    Nov 27 '18 at 20:14












  • For any given finite subset of unit fractions, one obvious criterion is simply to sum them! You undoubtedly want something less obvious, but it might help to describe what you want in a "criterion."
    – Barry Cipra
    Nov 27 '18 at 20:32














1












1








1







Unit fractions are rational numbers in the form $frac{1}{n}$, where $n$ is an integer. In elementary number theory one can prove that the harmonic sum $$H_n= 1+ frac{1}{2}+...+frac{1}{n}$$ is never an integer. For elementary proof of this fact see the post: Elementary proof that the sum 1+...+1/n is not an integer



Take one step further, József Kürschák, in 1918, proved that the sum of the reciprocals of consecutive natural numbers (whether starting from 1 or not) is never an integer.



Unfortunately, the result does not hold with a border generalization when we consider the sum of arbitrary subsets of unit fractions. Examples include $$1=frac{1}{2}+frac{1}{3}+frac{1}{6}=frac{1}{2}+ frac{1}{3}+ frac{1}{7}+ frac{1}{43}+ frac{1}{1806}.$$ Erdős–Graham problem concerns with the existence of such unit fractions.



The question I ask now is: Without calculating the sum one by one, are there any nontrivial, characteristic properties that determines whether the given subset of unit fractions sums to an integer? One step back, is there any class of interesting subsets of unit fractions where the sum of each subset in the class has a non-integer sum?










share|cite|improve this question















Unit fractions are rational numbers in the form $frac{1}{n}$, where $n$ is an integer. In elementary number theory one can prove that the harmonic sum $$H_n= 1+ frac{1}{2}+...+frac{1}{n}$$ is never an integer. For elementary proof of this fact see the post: Elementary proof that the sum 1+...+1/n is not an integer



Take one step further, József Kürschák, in 1918, proved that the sum of the reciprocals of consecutive natural numbers (whether starting from 1 or not) is never an integer.



Unfortunately, the result does not hold with a border generalization when we consider the sum of arbitrary subsets of unit fractions. Examples include $$1=frac{1}{2}+frac{1}{3}+frac{1}{6}=frac{1}{2}+ frac{1}{3}+ frac{1}{7}+ frac{1}{43}+ frac{1}{1806}.$$ Erdős–Graham problem concerns with the existence of such unit fractions.



The question I ask now is: Without calculating the sum one by one, are there any nontrivial, characteristic properties that determines whether the given subset of unit fractions sums to an integer? One step back, is there any class of interesting subsets of unit fractions where the sum of each subset in the class has a non-integer sum?







number-theory elementary-number-theory






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edited Nov 29 '18 at 1:36

























asked Nov 27 '18 at 20:02









William Sun

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  • 1




    It's not hard to see that if $P$ is a perfect number, then $sum_{d|P, dneq 1} frac{1}{d} = 1$.
    – SAWblade
    Nov 27 '18 at 20:14












  • For any given finite subset of unit fractions, one obvious criterion is simply to sum them! You undoubtedly want something less obvious, but it might help to describe what you want in a "criterion."
    – Barry Cipra
    Nov 27 '18 at 20:32














  • 1




    It's not hard to see that if $P$ is a perfect number, then $sum_{d|P, dneq 1} frac{1}{d} = 1$.
    – SAWblade
    Nov 27 '18 at 20:14












  • For any given finite subset of unit fractions, one obvious criterion is simply to sum them! You undoubtedly want something less obvious, but it might help to describe what you want in a "criterion."
    – Barry Cipra
    Nov 27 '18 at 20:32








1




1




It's not hard to see that if $P$ is a perfect number, then $sum_{d|P, dneq 1} frac{1}{d} = 1$.
– SAWblade
Nov 27 '18 at 20:14






It's not hard to see that if $P$ is a perfect number, then $sum_{d|P, dneq 1} frac{1}{d} = 1$.
– SAWblade
Nov 27 '18 at 20:14














For any given finite subset of unit fractions, one obvious criterion is simply to sum them! You undoubtedly want something less obvious, but it might help to describe what you want in a "criterion."
– Barry Cipra
Nov 27 '18 at 20:32




For any given finite subset of unit fractions, one obvious criterion is simply to sum them! You undoubtedly want something less obvious, but it might help to describe what you want in a "criterion."
– Barry Cipra
Nov 27 '18 at 20:32










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One sufficient condition for the sum to be a non-integer: there is some prime power that divides exactly one of the denominators.






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    1 Answer
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    One sufficient condition for the sum to be a non-integer: there is some prime power that divides exactly one of the denominators.






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      One sufficient condition for the sum to be a non-integer: there is some prime power that divides exactly one of the denominators.






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        One sufficient condition for the sum to be a non-integer: there is some prime power that divides exactly one of the denominators.






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        One sufficient condition for the sum to be a non-integer: there is some prime power that divides exactly one of the denominators.







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        answered Nov 27 '18 at 20:13









        Robert Israel

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