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Allan Johnson's set of 48 squarefree numbers whose reciprocals add to 1, with the property that each number has exactly two distinct prime factors.
+10
4
6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 77, 82, 85, 86, 87, 91, 93, 95, 115, 119, 123, 133, 155, 187, 203, 209, 215, 221, 247, 265, 287, 299, 319, 323, 391, 689, 731, 901
OFFSET
1,1
REFERENCES
R. K. Guy, Unsolved Problems in Number Theory (UPINT), Section D11.
CROSSREFS
Cf. A334342 (for a set of 47 terms).
KEYWORD
nonn,fini,full
AUTHOR
N. J. A. Sloane, Dec 03 2011
STATUS
approved
A list of 63 distinct numbers such that the sum of their reciprocals is 1 and each number is of the form p*q where p and q are distinct primes.
+10
3
6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 82, 91, 93, 95, 106, 118, 119, 122, 123, 133, 145, 155, 159, 161, 183, 187, 202, 213, 265, 287, 295, 299, 319, 355, 393, 413, 453, 497, 505, 583, 671, 835, 917, 1057, 1169, 1313, 1438, 1441, 1837, 1963, 3595, 5033, 7909
OFFSET
1,1
COMMENTS
At the time Burshtein's paper appeared, this was the smallest example known of this type. See A201514 for an even smaller example.
REFERENCES
E. J. Barbeau, Expressing one as a sum of distinct reciprocals: comments and a bibliography, Eureka (Ottawa), 3 (1977), 178-181.
CROSSREFS
KEYWORD
nonn,fini,full
AUTHOR
N. J. A. Sloane, Dec 01 2011
STATUS
approved
A list of 52 distinct numbers such that the sum of their reciprocals is 1 and each number is of the form p*q where p and q are distinct primes.
+10
3
6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 77, 82, 87, 91, 93, 95, 106, 119, 122, 123, 133, 155, 159, 161, 183, 187, 202, 203, 213, 265, 287, 299, 319, 355, 453, 497, 505, 583, 671, 1057, 1313, 1963
OFFSET
1,1
COMMENTS
This appears to be the smallest example known of this type.
Note that it has the property that no number divides another.
CROSSREFS
See A201464 for the previous record. Cf A201463, A201650.
KEYWORD
nonn,fini,full
AUTHOR
N. J. A. Sloane, Dec 02 2011
STATUS
approved
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.
+10
1
6, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 33, 35, 36, 39, 40, 44, 52, 55
OFFSET
1,1
COMMENTS
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.
REFERENCES
E. J. Barbeau, Expressing one as a sum of distinct reciprocals: comments and a bibliography, Eureka (Ottawa), 3 (1977), 178-181.
LINKS
N. Burshtein, Improving solutions of Sum_{i=1..k} 1/x_i = 1 ..., Discrete Math., 306 (2006), 1438-1439.
PROG
(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
CROSSREFS
KEYWORD
nonn,fini,full
AUTHOR
Jon E. Schoenfield, Feb 02 2014
EXTENSIONS
Improved Magma program and comments about opportunities for further improvement from Jon E. Schoenfield, Feb 12 2014
STATUS
approved

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