A336006 - OEIS

%I #18 Jan 06 2022 12:34:40

%S 1,5,14,46,127,383,1407,3594,11786,31469,97005,451299,982740,3079892,

%T 7862861,24640077,110733519,244951247,1019792225,3344315159,

%U 13804668362,48164406730,185603360202,468032896683,1567544524459,4109410352788,12905503374996,58659088284918

%N a(n) = the least k such that the mixed binary-ternary representation of k has n terms. 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="/A336006/b336006.txt">Table of n, a(n) for n = 1..2030</a>

%H Michael S. Branicky, <a href="/A336006/a336006.txt">Proof of formula</a>

%F 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

%e 1 = 1 (1 term);

%e 5 = 4 + 1 (2 terms);

%e 14 = 9 + 4 + 1 (3 terms);

%e 46 = 32 + 9 + 4 + 1 (4 terms);

%e 127 = 81 + 32 + 9 + 4 + 1 (5 terms).

%t z = 20; zz = 100000;

%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[

%t       b2 Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #,

%t           b2]][[2, 1]], 0]} &, Range[zz]];

%t m2 = Map[Length[#[[2]]] &, g2];

%t g = Map[{#, DeleteCases[b Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #,

%t           b]][[2, 1]], 0]} &, Range[zz]];

%t m = Map[Length[#[[2]]] &, g];

%t (* _Peter J. C. Moses_, Jul 05 2020 *)

%t Table[First[Flatten[Position[m, k]]], {k, 1, 11}]

%o (Python)

%o from itertools import count, takewhile, islice

%o def big_greedy(k, B, start=0):

%o     idx = start

%o     while idx < len(B) and B[idx] <= k: idx += 1

%o     return B[idx - 1]

%o def agen(limit=10**1001):

%o     an, idx, t = 1, 0, 2

%o     B1 = list(takewhile(lambda x: x <= limit, (2**i for i in count(0))))

%o     B21 = list(takewhile(lambda x: x <= limit, (3**i for i in count(0))))

%o     B22 = list(takewhile(lambda x: x <= limit, (2*3**i for i in count(0))))

%o     B = sorted(set(B1 + B21 + B22))

%o     while an <= limit:

%o         yield an

%o         while t != big_greedy(an+t, B, start=idx):

%o             idx, t = idx+1, B[idx+1]

%o         an += t

%o print(list(islice(agen(), 28))) # _Michael S. Branicky_, Jan 06 2022

%Y Cf. A336004, A336005.

%K nonn,base

%O 1,2

%A _Clark Kimberling_, Jul 06 2020

%E a(12) and beyond from _Michael S. Branicky_, Jan 06 2022