This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(6) = 1 since there is the only fraction 1 = 1/2+1/3+1/6.
n=20; try2[lev_, s_] := Module[{nmim, nmax, si, i}, AppendTo[soln, 0]; If[lev==1, nmin=2, nmin=1+soln[[ -2]]]; nmax=n-1; Do[If[iT. D. Noe, Mar 26 2004 *)
a(1) = 3 because 2 = prime(1) and 1/1 + 1/3 = 4/3, whose numerator is divisible by 2.
len=100; a=Table[0, {len}]; done=False; s={0}; n=0; While[ !done, n++; s=Join[s, s+1/n]; ns=Numerator[s]; done=True; Do[If[a[[i]]==0, p=Prime[i]; If[Count[ns, _?(#>0 && Mod[ #, p]==0&)]>0, a[[i]]=n, done=False]], {i, len}]]; a
a(3) = 6 counts numbers { 0, 1/3, 1/2, 2/3, 5/6, 1 }, each of which is can be represented as the sum of reciprocals 1/1, 1/2, and 1/3.
a(3) = 5 counts numbers { 0, 1/3, 1/2, 5/6, 1 }, each of which is can be represented as the sum of distinct reciprocals 1/1, 1/2, and 1/3.
s:= proc(n) option remember;
`if`(n=0, {0}, map(x-> `if`(n-1 nops(s(n)):
seq(a(n), n=0..20); # Alois P. Heinz, May 23 2012
s[_] := s[n] = If[n == 0, {0}, If[n-1 < n*#, #, {#, # + 1/n}]& /@ s[n-1] // Flatten];
a[n_] := Length[s[n]];
Table[Print[n, " ", a[n]]; a[n], {n, 0, 32}] (* Jean-François Alcover, May 13 2019, after Alois P. Heinz *)
{ A212657(n) = my(L=lcm(vector(n,i,i))); polcoeff( prod(i=1,n, 1+x^(L/i)+O(x^(L+1)) )/(1-x), L); }
a(1) = 1 via [1]
a(2) = 6 via [2, 3, 6]
a(3) = 6 via [2, 3, 6]
a(4) = 12 via [2, 4, 6, 12]
a(5) = 20 via [2, 4, 5, 20]
a(6) = 6 via [2, 3, 6]
a(7) = 28 via [2, 4, 7, 14, 28]
a(8) = 24 via [2, 3, 8, 24]
a(9) = 18 via [2, 3, 9, 18]
a(10) = 15 via [2, 3, 10, 15]
a(11) > 30
a(12) = 12 via [2, 4, 6, 12]
a(13) > 30
a(14) = 28 via [2, 4, 7, 14, 28]
a(15) = 15 via [2, 3, 10, 15]
a(16) > 30
a(17) > 30
a(18) = 18 via [2, 3, 9, 18]
From _Jon E. Schoenfield_, Feb 15 2020: (Start)
For n=31, the largest prime or prime power divisor of n is P=31, and the set of unit fractions {1/1, 1/2, 1/3, 1/4, 1/5, 1/6} has no subset sum that includes 1/(n/P) = 1/1 and has a numerator divisible by 31, but the set of unit fractions {1/1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7} does have such a subset sum, namely, 1/1 + 1/3 + 1/7 = 31/21, so a(31) >= 7*31 = 217. In fact, the numbers 1*31=31, 3*31=93, and 7*31=217 are elements of many sets of integers that include n=31, include no element > 217, and have a reciprocal sum of 1 (one such set is {2,3,12,28,31,93,217}), so a(31)=217.
For n=62, the largest prime or prime power divisor of n is again P=31, and the set of unit fractions {1/1, 1/2, 1/3, 1/4} has no subset sum that includes 1/(n/P) = 1/2 and has a numerator divisible by 31, but the set of unit fractions {1/1, 1/2, 1/3, 1/4, 1/5} does have such a subset sum, namely, 1/2 + 1/3 + 1/5 = 31/30, so a(62) >= 5*31 = 155. In fact, the numbers 2*31=62, 3*31=93, and 5*31=155 are elements of many sets of integers that include n=62, include no element > 155, and have a reciprocal sum of 1 (one such set is {2,3,12,20,62,93,155}), so a(62)=155.
(End)
Table[SelectFirst[Range@ 20, MemberQ[Map[Total, 1/DeleteCases[Rest@ Subsets[Range@ #, #], w_ /; FreeQ[w, n]]], 1] &] /. k_ /; MissingQ@ k -> 0, {n, 12}] (* Michael De Vlieger, Aug 18 2016, Version 10.2, values of a(n) > 20 appear as 0 *)
10 is in this sequence because any Egyptian fraction with 1/10 as its term with largest denominator either contains 1/5 as well or not; either way, the resulting sum will have a factor 5 in its denominator (any other term will contribute a multiple of 5 to the numerator of the sum), hence cannot equal 1.
Comments