OFFSET
1,1
COMMENTS
Numbers n such that we can write 1/n = Sum_{1<=k<n} c(k)/k with all c(k) in {-1,0,1}.
If n is in the sequence, so is k*n for all k>1 (cf. A328226).
Contains A001284, because 1/(m*k) = 1/(m*(k-m))-1/(k*(k-m)).
Disjoint from A000961.
2*p^k with p prime is in the sequence if and only if p=3.
3*p^k with p prime is in the sequence if and only if p=2,5,7 or 11.
4*p^k with p prime is in the sequence if and only if p=3,5,7,11,13,17 or 19.
For each m that is not a term, there are only finitely many primes p such that some m*p^k is a term. [Corrected by Max Alekseyev, Oct 08 2019]
LINKS
Robert Israel, Table of n, a(n) for n = 1..431
Robert Israel, Examples for n = 1..431
Robert Israel, 1/n as a difference of Egyptian fractions with all denominators < n, Math StackExchange, 2017.
EXAMPLE
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}.
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert Israel, Nov 24 2016
STATUS
approved