OFFSET
1,1
COMMENTS
Suppose that B1 and B2 are increasing sequences of positive integers, and let B be the increasing sequence of numbers in the union of B1 and B2. Every positive integer n has a unique representation given by the greedy algorithm with B1 as base, and likewise for B2 and B. For many n, the number of terms in the B-representation of n is less than the number of terms in the B1-representation, as well as the B2-representation, but not for all n, as in the example 45 = 27 + 18 (ternary) and 45 = 32 + 9 + 4 (mixed).
Note that 1 and 2 = 10_2 = 2_3 are each representable as terms in both binary and ternary. - Michael S. Branicky, Jan 06 2022
EXAMPLE
7 = 6 + 1 = 21_3 is not a term;
11 = 9 + 2 = 102_3 is not a term;
13 = 9 + 4 = 3^2 + 2^2 is a term;
22 = 18 + 4 = 2*3^2 + 2^2 is a term.
MATHEMATICA
u = Table[2^n, {n, 0, 50}]; v = Table[3^n, {n, 0, 40}];
uQ[n_] := MemberQ[u, n]; vQ[n_] := MemberQ[v, n];
Attributes[uQ] = {Listable}; Attributes[vQ] = {Listable};
s = Reverse[Union[Flatten[Table[{2^(n - 1), 3^n}, {n, 1, 30}]]]];
w = Map[#[[1]] &, Select[Map[{#[[1]], {Apply[And, uQ[#[[2]]]],
Apply[And, vQ[#[[2]]]]}} &, Map[{#, DeleteCases[
s Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #,
s]][[2, 1]], 0]} &,
Range[700]]], #[[2]] == {False, False} &]]
(* Peter J. C. Moses, Jun 14 2020 *)
PROG
(Python)
from itertools import count, takewhile
N = 10**6
B1 = list(takewhile(lambda x: x[0] <= N, ((2**i, 2) for i in count(0))))
B21 = list(takewhile(lambda x: x[0] <= N, ((3**i, 3) for i in count(0))))
B22 = list(takewhile(lambda x: x[0] <= N, ((2*3**i, 3) for i in count(0))))
B = sorted(set(B1 + B21 + B22), reverse=True)
def ok(n):
r, bases = [], set()
for t, b in B:
if t <= n:
r.append(t)
if t != 1 and t != 2:
bases.add(b)
n -= t
if n == 0:
return len(bases) == 2
print([k for k in range(1, 132) if ok(k)]) # Michael S. Branicky, Jan 06 2022
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Clark Kimberling, Jul 06 2020
EXTENSIONS
Terms and examples corrected by Michael S. Branicky, Jan 06 2022
STATUS
approved