A336006 a(n) = the least k such that the mixed binary-ternary representation of k has n terms. See Comments.

1, 5, 14, 46, 127, 383, 1407, 3594, 11786, 31469, 97005, 451299, 982740, 3079892, 7862861, 24640077, 110733519, 244951247, 1019792225, 3344315159, 13804668362, 48164406730, 185603360202, 468032896683, 1567544524459, 4109410352788, 12905503374996, 58659088284918

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

			1 = 1 (1 term);
5 = 4 + 1 (2 terms);
14 = 9 + 4 + 1 (3 terms);
46 = 32 + 9 + 4 + 1 (4 terms);
127 = 81 + 32 + 9 + 4 + 1 (5 terms).
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..2030
• Michael S. Branicky, Proof of formula

Crossrefs

Cf. A336004, A336005.

Programs

Mathematica

z = 20; zz = 100000;
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];
(* Peter J. C. Moses, Jul 05 2020 *)
Table[First[Flatten[Position[m, k]]], {k, 1, 11}]

Python

from itertools import count, takewhile, islice
def big_greedy(k, B, start=0):
    idx = start
    while idx < len(B) and B[idx] <= k: idx += 1
    return B[idx - 1]
def agen(limit=10**1001):
    an, idx, t = 1, 0, 2
    B1 = list(takewhile(lambda x: x <= limit, (2**i for i in count(0))))
    B21 = list(takewhile(lambda x: x <= limit, (3**i for i in count(0))))
    B22 = list(takewhile(lambda x: x <= limit, (2*3**i for i in count(0))))
    B = sorted(set(B1 + B21 + B22))
    while an <= limit:
        yield an
        while t != big_greedy(an+t, B, start=idx):
            idx, t = idx+1, B[idx+1]
        an += t
print(list(islice(agen(), 28))) # Michael S. Branicky, Jan 06 2022

Formula

a(n+1) = a(n) + t, where t is the least element in B such that the largest element of B in the interval (a(n), a(n) + t) is t; see link for proof. - Michael S. Branicky, Jan 06 2022

Extensions

a(12) and beyond from Michael S. Branicky, Jan 06 2022