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.
From _Jon E. Schoenfield_, Apr 09 2017: (Start) 6 is in the sequence because 1/2 + 1/3 + 1/6 = 1. (Note that the prime factorization of 6 is 2*3, and if we start with 1/6, adding 1/3 yields 1/2, which removes the factor 3 from the denominator; then adding 1/2 removes the 2.) 23 cannot be in the sequence because it is a prime: for any positive integer j1 < 23, 1/j1 + 1/23 = (23 + j1)/(23*j1), which cannot be reduced; adding another 1/j2 to the sum (with j2 < 23) will give (23*(j1 + j2) + j1*j2)/(23*j1*j2), from which the factor of 23 in the denominator still cannot be removed by reduction (since 23 does not divide j1*j2, so 23 cannot divide the numerator); and adding any further reciprocals of integers < 23 similarly cannot remove the factor of 23 from the denominator. 25 cannot be in the sequence because it is a prime power: for any positive integer j1 < 25, 1/j1 + 1/25 = (25 + j1)/(25*j1), which cannot be reduced unless 5 divides j1, but even then the denominator would remain divisible by 25 (and, as above, this would continue to be the case after the addition any number of reciprocals of other integers < 25). For additional examples, including some ideas for heuristics for obtaining solutions for numbers n that are in the sequence, see the Links. (End) For an inelegantly written Magma program that computes the first 1000 terms in about 0.3 seconds on the Magma Calculator, see the Links. - _Jon E. Schoenfield_, Apr 19 2017
n=55; try3[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 30 2004 *)
a(4) = 2 since there are two fractions 1=1/2+1/4+1/4 and 1=1/4+1/4+1/4+1/4.
a(6)=2 since there are two fractions 1=1/1 and 1=1/2+1/3+1/6.
a(4) = 4 since there are four fractions 1=1/2+1/4+1/4, 1=1/4+1/2+1/4, 1=1/4+1/4+1/2 and 1=1/4+1/4+1/4+1/4.
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); }
For n = 4 the a(4) = 7 subsets are:
{} because 0 < 1,
{2} because 1/2 < 1,
{2, 3} because 1/2 + 1/3 = 5/6 < 1,
{2, 4} because 1/2 + 1/4 = 3/4 < 1,
{3} because 1/3 < 1,
{3, 4} because 1/3 + 1/4 = 7/12 < 1, and
{4} because 1/4 < 1.
a[n_] := 1 + Length@ Select[Subsets[Range[2,n], {1, n-1}], Total[1/#] < 1 &]; Array[a, 15] (* Giovanni Resta, Jun 01 2018 *)
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