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A237051
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Smallest set of distinct numbers such that the sum of their reciprocals is 1 and each number is of the form p^i*q^j where p and q are distinct primes and i and j are positive.
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1
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6, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 33, 35, 36, 39, 40, 44, 52, 55
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OFFSET
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1,1
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COMMENTS
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In other words, smallest set of distinct numbers in A007774 (i.e., numbers that are divisible by exactly 2 different primes) whose reciprocals sum to 1.
This is the only set of 20 distinct numbers with the specified property. (The sum of the reciprocals of the first 19 numbers that are divisible by exactly 2 different primes is 229926/230945, which is less than 1, so no set of fewer than 20 such numbers can have reciprocals that sum to 1.)
There exist 8 such sets of 21 numbers, and 218 such sets of 22 numbers.
For a set of 25 distinct numbers having the specified property, see A201463.
The Magma program (as of Feb 12 2014) is very fast at n=20, but very slow if n is changed to a value of 22 or larger. The main opportunity for improving its efficiency may be in the way it handles the search for the last two terms of a set, given the first n-2 terms at the current point in the search tree. Specifically, if the remainder r = 1 - (1/i[1] + 1/i[2] + ... + 1/i[n-2]) is close to zero, e.g., 10^-8, then it tests about 10^8 values of i[n-1] (from about 10^8 to about 2*10^8), instead of applying some smarter approach that makes use of the factorization of the denominator of the remainder r.
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REFERENCES
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E. J. Barbeau, Expressing one as a sum of distinct reciprocals: comments and a bibliography, Eureka (Ottawa), 3 (1977), 178-181.
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LINKS
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PROG
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(Magma) n:=20; i:=[]; j:=1; i[j]:=0; r:=1; while true do i[j]+:=1; while #Factorization(i[j]) ne 2 do i[j]+:=1; end while; if (n-(j-1))/i[j] lt r then if j eq 1 then "done"; break; end if; j-:=1; r+:=1/i[j]; elif j eq n-1 then TestIntN:=Floor(1/(r-1/i[j])); if TestIntN le i[n-1] then j-:=1; r+:=1/i[j]; elif (r - 1/i[j] eq 1/TestIntN) and (#Factorization(TestIntN) eq 2) then i[n]:=TestIntN; i; end if; else r-:=1/i[j]; j+:=1; i[j]:=i[j-1]; if 1/(i[j]+1) ge r then i[j]:=Floor(1/r); end if; end if; end while; // Jon E. Schoenfield, Feb 12 2014
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CROSSREFS
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KEYWORD
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nonn,fini,full
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AUTHOR
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EXTENSIONS
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Improved Magma program and comments about opportunities for further improvement from Jon E. Schoenfield, Feb 12 2014
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STATUS
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approved
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