A336005 a(n) is the number of terms in the mixed binary-ternary representation of n. See Comments.

1, 1, 1, 1, 2, 1, 2, 1, 1, 2, 2, 2, 2, 3, 2, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 3, 2, 3, 2, 2, 3, 3, 3, 3, 4, 3, 2, 3, 2, 3, 3, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 2, 1, 2, 2, 2, 2, 3, 2, 3, 2, 2, 3, 3, 3, 3, 4, 3, 2, 1, 2, 2, 2, 2, 3

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).

Examples

			7 = 6 + 1, so a(7) = 2.
45 = 32 + 9 + 4, so a(45) = 3.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A336004, A336006.

Programs

Mathematica

z = 20; zz = 100;
b1 = Sort[Table[2^k, {k, 0, z}], Greater];
b2 = Sort[Union[Table[3^k, {k, 0, z}], Table[2*3^k, {k, 0, z}]],
   Greater]; b = Sort[Union[b1, b2], Greater];
g1 = Map[{#, DeleteCases[b1  Reap[
         FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, b1]][[2,
         1]], 0]} &, Range[zz]];
m1 = Map[Length[#[[2]]] &, g1];
g2 = Map[{#, DeleteCases[b2 Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, b2]][[2, 1]], 0]} &, Range[zz]];
m2 = Map[Length[#[[2]]] &, g2];
g = Map[{#, DeleteCases[
     b Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #,
         b]][[2, 1]], 0]} &, Range[zz]]
m = Map[Length[#[[2]]] &, g];
m1 (* # terms in binary representation *)
m2 (* # terms in ternary representation *)
m  (* # terms in mixed base representation *)  (* A336005 *)

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 gbt(n, B): # greedy binary-ternary representation
    r = []
    for t, b in B:
        if t <= n:
            r.append(t)
            n -= t
            if n == 0:
                return r
def a(n): return len(gbt(n, B))
print([a(n) for n in range(1, 87)]) # Michael S. Branicky, Jan 06 2022