A278638 Numbers n such that 1/n is a difference of Egyptian fractions with all denominators < n.
6, 12, 15, 18, 20, 21, 24, 28, 30, 33, 35, 36, 40, 42, 44, 45, 48, 52, 54, 55, 56, 60, 63, 65, 66, 68, 70, 72, 75, 76, 77, 78, 80, 84, 85, 88, 90, 91, 95, 96, 99, 100, 102, 104, 105, 108, 110, 112, 114, 115, 117, 119, 120, 126, 130, 132, 133, 135, 136, 138, 140, 143, 144, 145, 147, 150, 152, 153
Offset: 1
Comments
Examples
44 is in the sequence because 1/44 = (1/12 + 1/33) - 1/11. 4 is not in the sequence because 1/4 can't be written as the difference of sums of two subsets of {1, 1/2, 1/3}.Links
Crossrefs
Programs
Maple
N:= 200: # to get all terms <= N V:= Vector(N): f:= proc(n) option remember; local F,E,p,e,k,m,L,L1,i,s,t,sg,Maybe; global Rep; F:= numtheory:-factorset(n); if nops(F) = 1 then return false fi; if ormap(m -> n < m^2 and m^2 < 2*n, numtheory:-divisors(n)) then for m in numtheory:-divisors(n) do if n < m^2 and m^2 < 2*n then k:= n/m; Rep[n]:= [m*(k-m),-k*(k-m)]; return true fi od fi; F:= convert(F,list); E:= map(p -> padic:-ordp(n,p), F); i:= max[index](zip(`^`,F,E)); p:= F[i]; e:= E[i]; k:= n/p^e; Maybe:= false; for i from 3^(k-1) to 2*3^(k-1)-1 do L:= (-1) +~ convert(i,base,3); s:= 1/k - add(L[i]/i,i=1..k-1); if numer(s) mod p = 0 then Maybe:= true; t:= abs(s/p^e); sg:= signum(s); if (numer(t) <= 1 and (denom(t) < n or (denom(t) < N and V[denom(t)] = 1))) or (numer(t) = 2 and denom(t) < N and V[denom(t)] = 1) then L1:= subs(0=NULL, [seq(L[i]*i*p^e,i=1..k-1)]); if t = 0 then ; elif numer(t) = 1 and denom(t) < n then L1:= [op(L1),sg/t] elif procname(2/t) then L1:= ([op(L1), 2*sg/t, op(expand(sg*Rep[2/t]))]) else next fi; if max(abs~(L1)) < n then Rep[n]:= L1; return true fi; fi; fi od: if Maybe then printf("Warning: %d is uncertain\n",n) else false fi; end proc: for n from 6 to N do if V[n] = 0 and f(n) then V[n] := 1; for j from 2*n to N by n do if not assigned(Rep[j]) then V[j]:= 1; Rep[j] := map(`*`,Rep[n],j/n); f(j):= true; fi od; fi; od: select(t -> V[t]=1,[$6..N]);Mathematica
sol[n_] := Module[{c, cc}, cc = Array[c, n-1]; FindInstance[AllTrue[cc, -1 <= # <= 1&] && 1/n == Total[cc/Range[n-1]], cc, Integers, 1]]; Reap[For[n = 6, n <= 200, n++, If[sol[n] != {}, Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Jul 29 2020 *)