Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.
%I #26 Jan 28 2025 00:49:32
%S 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,
%T 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,
%U 3,2,3,2,2,3,3,3,3,4,3,2,1,2,2,2,2,3
%N a(n) is the number of terms in the mixed binary-ternary representation of n. See Comments.
%C 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).
%H Michael S. Branicky, <a href="/A336005/b336005.txt">Table of n, a(n) for n = 1..10000</a>
%e 7 = 6 + 1, so a(7) = 2.
%e 45 = 32 + 9 + 4, so a(45) = 3.
%t z = 20; zz = 100;
%t b1 = Sort[Table[2^k, {k, 0, z}], Greater];
%t b2 = Sort[Union[Table[3^k, {k, 0, z}], Table[2*3^k, {k, 0, z}]],
%t Greater]; b = Sort[Union[b1, b2], Greater];
%t g1 = Map[{#, DeleteCases[b1 Reap[
%t FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, b1]][[2,
%t 1]], 0]} &, Range[zz]];
%t m1 = Map[Length[#[[2]]] &, g1];
%t g2 = Map[{#, DeleteCases[b2 Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, b2]][[2, 1]], 0]} &, Range[zz]];
%t m2 = Map[Length[#[[2]]] &, g2];
%t g = Map[{#, DeleteCases[
%t b Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #,
%t b]][[2, 1]], 0]} &, Range[zz]]
%t m = Map[Length[#[[2]]] &, g];
%t m1 (* # terms in binary representation *)
%t m2 (* # terms in ternary representation *)
%t m (* # terms in mixed base representation *) (* A336005 *)
%o (Python)
%o from itertools import count, takewhile
%o N = 10**6
%o B1 = list(takewhile(lambda x: x[0] <= N, ((2**i, 2) for i in count(0))))
%o B21 = list(takewhile(lambda x: x[0] <= N, ((3**i, 3) for i in count(0))))
%o B22 = list(takewhile(lambda x: x[0] <= N, ((2*3**i, 3) for i in count(0))))
%o B = sorted(set(B1 + B21 + B22), reverse=True)
%o def gbt(n, B): # greedy binary-ternary representation
%o r = []
%o for t, b in B:
%o if t <= n:
%o r.append(t)
%o n -= t
%o if n == 0:
%o return r
%o def a(n): return len(gbt(n, B))
%o print([a(n) for n in range(1, 87)]) # _Michael S. Branicky_, Jan 06 2022
%Y Cf. A336004, A336006.
%K nonn,base,changed
%O 1,5
%A _Clark Kimberling_, Jul 06 2020