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A192787
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Number of distinct solutions of 4/n = 1/a + 1/b + 1/c in positive integers satisfying 1 <= a <= b <= c.
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16
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0, 1, 3, 3, 2, 8, 7, 10, 6, 12, 9, 21, 4, 17, 39, 28, 4, 26, 11, 36, 29, 25, 21, 57, 10, 20, 29, 42, 7, 81, 19, 70, 31, 25, 65, 79, 9, 32, 73, 96, 7, 86, 14, 62, 93, 42, 34, 160, 18, 53, 52, 59, 13, 89, 98, 136, 41, 33, 27, 196, 11, 37, 155, 128, 49, 103, 17, 73, 55, 185, 40, 211, 7, 32, 129, 80, 97, 160, 37, 292
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OFFSET
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1,3
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COMMENTS
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The Erdős-Straus conjecture is that a(n) > 0 for n > 1. Swett verified the conjecture for n < 10^14.
Vaughan shows that the number of n < x with a(n) = 0 is at most x exp(-c * (log x)^(2/3)) for some c > 0.
See A073101 for the 4/n conjecture due to Erdős and Straus.
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LINKS
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EXAMPLE
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a(1) = 0, since 4/1 = 4 cannot be expressed as the sum of three reciprocals.
a(2) = 1 because 4/2 = 1/1 + 1/2 + 1/2, and there are no other solutions.
a(3) = 3 since 4/3 = 1 + 1/4 + 1/12 = 1 + 1/6 + 1/6 = 1/2 + 1/2 + 1/3.
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MAPLE
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A192787 := proc(n) local t, a, b, t1, count; t:= 4/n; count:= 0; for a from floor(1/t)+1 to floor(3/t) do t1:= t - 1/a; for b from max(a, floor(1/t1)+1) to floor(2/t1) do if type( 1/(t1 - 1/b), integer) then count:= count+1; end if end do end do; count; end proc; # Robert Israel, Feb 19 2013
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MATHEMATICA
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PROG
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(PARI) a(n, show=0)=my(t=4/n, t1, s, c); for(a=1\t+1, 3\t, t1=t-1/a; for(b=max(1\t1+1, a), 2\t1, c=1/(t1-1/b); if(denominator(c)==1&&c>=b, s++; show&&print("4/", n, " = 1/", a, " + 1/", b, " + 1/", c)))); s \\ variant with print(...) added by Robert Munafo, Feb 19 2013, both combined through option "show" by M. F. Hasler, Jul 02 2022
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CROSSREFS
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A292581 counts the solutions with multiplicity. A073101 counts solutions with a, b, and c distinct.
Cf. A337432 (solutions with minimal c).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Examples and cross-references added by M. F. Hasler, Feb 19 2013
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STATUS
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approved
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