A213062 Minimal sum x(1) +...+ x(n) such that 1/x(1) +...+ 1/x(n) = 1, the x(i) being n distinct positive integers.

1, 0, 11, 24, 38, 50, 71, 87, 106, 127, 151, 185, 211, 249, 288, 325, 364, 406, 459, 508, 550, 613, 676, 728

Offset: 1

Author

M. F. Hasler, Jun 03 2012

Keywords

• nonn
• more

Comments

The term a(2)=0 corresponds to the fact that 1 cannot be written as Egyptian fraction with 2 (distinct) terms.

Examples

			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).
		

Links

• Eric Weisstein's World of Mathematics, Egyptian Fraction
• Wikipedia, Egyptian fraction
• Index entries for sequences related to Egyptian fractions

Crossrefs

Cf. A030659. - Alois P. Heinz, Sep 21 2012

Programs

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. */

Extensions

a(11)-a(24) from Robert Price, Aug 26 2012 - Sep 21 2012