cp's OEIS Frontend

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

%I A213062 #27 Feb 16 2025 08:33:17
%S A213062 1,0,11,24,38,50,71,87,106,127,151,185,211,249,288,325,364,406,459,
%T A213062 508,550,613,676,728
%N A213062 Minimal sum x(1) +...+ x(n) such that 1/x(1) +...+ 1/x(n) = 1, the x(i) being n distinct positive integers.
%C A213062 The term a(2)=0 corresponds to the fact that 1 cannot be written as Egyptian fraction with 2 (distinct) terms.
%H A213062 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/EgyptianFraction.html">Egyptian Fraction</a>
%H A213062 Wikipedia, <a href="http://en.wikipedia.org/wiki/Egyptian_fraction">Egyptian fraction</a>
%H A213062 <a href="/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>
%e A213062 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.
%e A213062 a(4) = 24 = 2 + 4 + 6 + 12 is the smallest sum of denominators among the six 4-term Egyptian fractions equal to 1.
%e A213062 a(5) = 38 = 3 + 4 + 5 + 6 + 20, least sum of denominators among 72 possible 5-term Egyptian fractions equal to 1.
%e A213062 a(6) = 50 = 3 + 4 + 6 + 10 + 12 + 15, least sum of denominators among 2320 possible 6-term Egyptian fractions equal to 1.
%e A213062 a(7) <= 71 = 3 + 5 + 20 + 6 + 10 + 12 + 15 (obtained from n=6 using 1/4 = 1/5 + 1/20).
%e A213062 a(8) <= 114 = 3 + 5 + 20 + 7 + 42 + 10 + 12 + 15 (obtained using 1/6 = 1/7 + 1/42).
%e A213062 a(9) <= 145 = 3 + 6 + 30 + 20 + 7 + 42 + 10 + 12 + 15 (obtained using 1/5 = 1/6 + 1/30).
%e A213062 a(10) <= 202 = 3 + 6 + 30 + 20 + 8 + 56 + 42 + 10 + 12 + 15 (obtained using 1/7 = 1/8 + 1/56).
%o A213062 (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. */
%Y A213062 Cf. A030659. - _Alois P. Heinz_, Sep 21 2012
%K A213062 nonn,more
%O A213062 1,3
%A A213062 _M. F. Hasler_, Jun 03 2012
%E A213062 a(11)-a(24) from _Robert Price_, Aug 26 2012 - Sep 21 2012