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A213062
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Minimal sum x(1) +...+ x(n) such that 1/x(1) +...+ 1/x(n) = 1, the x(i) being n distinct positive integers.
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3
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1, 0, 11, 24, 38, 50, 71, 87, 106, 127, 151, 185, 211, 249, 288, 325, 364, 406, 459, 508, 550, 613, 676, 728
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OFFSET
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1,3
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COMMENTS
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The term a(2)=0 corresponds to the fact that 1 cannot be written as Egyptian fraction with 2 (distinct) terms.
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LINKS
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EXAMPLE
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a(3) = 11 = 2 + 3 + 6, because 1/2+1/3+1/6 is the only Egyptian fraction with 3 terms having 1 as sum.
a(4) = 24 = 2 + 4 + 6 + 12 is the smallest sum of denominators among the six 4-term Egyptian fractions equal to 1.
a(5) = 38 = 3 + 4 + 5 + 6 + 20, least sum of denominators among 72 possible 5-term Egyptian fractions equal to 1.
a(6) = 50 = 3 + 4 + 6 + 10 + 12 + 15, least sum of denominators among 2320 possible 6-term Egyptian fractions equal to 1.
a(7) <= 71 = 3 + 5 + 20 + 6 + 10 + 12 + 15 (obtained from n=6 using 1/4 = 1/5 + 1/20).
a(8) <= 114 = 3 + 5 + 20 + 7 + 42 + 10 + 12 + 15 (obtained using 1/6 = 1/7 + 1/42).
a(9) <= 145 = 3 + 6 + 30 + 20 + 7 + 42 + 10 + 12 + 15 (obtained using 1/5 = 1/6 + 1/30).
a(10) <= 202 = 3 + 6 + 30 + 20 + 8 + 56 + 42 + 10 + 12 + 15 (obtained using 1/7 = 1/8 + 1/56).
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PROG
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(PARI) a(n, M=9e9, s=1, m=2)={ n==1 & return((numerator(s)==1 & 1 >= m*s || s==1)/s); sum( k=m, m+n-1, 1/k ) < s & return; for(x=max(m, 1\s+1), n\s, n*(x+(n-1)/2)>=M & break; (m=a(n-1, M-x, s-1/x, x+1)) & M=min(M, x+m)); M} /* For n>6, a good upper bound must be given as 2nd (optional) argument. Such a bound can be obtained using 1/x = 1/(x+1) + 1/x(x+1) in solutions for n-1, cf. Examples. */
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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