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

1,5

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).

Links

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

Example

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

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 trinary representation *)
m  (* # terms in mixed base representation *)  (* A336005 *)

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 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

Crossrefs

Cf. A336004, A336006.

Sequence in context: A110627 A205107 A228667 * A298674 A211354 A211352

Adjacent sequences: A336002 A336003 A336004 * A336006 A336007 A336008

Keyword

nonn,base

Author

Clark Kimberling, Jul 06 2020

Status

approved