A336004 Numbers whose mixed binary-ternary representation is not a binary or ternary representation. See Comments.

13, 14, 22, 23, 26, 31, 35, 38, 39, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 53, 58, 59, 62, 67, 70, 71, 73, 74, 75, 76, 77, 78, 79, 85, 86, 89, 94, 95, 97, 98, 103, 104, 107, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 131

Offset: 1

Author

Clark Kimberling, Jul 06 2020

Keywords

• nonn
• base

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

Examples

			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.
		

Crossrefs

Cf. A000079, A000244, A336005, A336006, A336007.

Programs

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 *)

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

Extensions

Terms and examples corrected by Michael S. Branicky, Jan 06 2022