0,2

Since A049354(1) = 11 it follows that a(0) = 1.

The numbers 2 through 10 have no divisors in A049354 and A049354(1) = 11 = 102_3, so a(1) = 11.

105 has only two divisors in A049354, 15 = A049354(2) and 21 = A049354(4) and is the smallest with exactly two divisors in A049354, so a(2) = 105.

N:= 15: # for terms before the first term >= 3^(N+1)

db:= proc(n) option remember; local L, d, m;

L:= convert(n, base, 3);

d:= nops(L);

d mod 3 = 0 and 3*numboccur(0, L) = d and 3*numboccur(1, L) = d

end proc:

W:= Vector(3^(N+1), datatype=integer[4]):

for d from 3 to N by 3 do

for t from 3^(d-1) to 3^d-1 do

if db(t) then

J:= [seq(i, i=t..3^(N+1), t)];

W[J]:= W[J] +~ 1;

fi

od od:

M:= max(W):

V:= Array(0..M): count:= 0:

for i from 1 to 3^(N+1) while count < M+1 do

if V[W[i]] = 0 then V[W[i]]:= i; count:= count+1 fi;

od:

L:= convert(V, list):

if not member(0, L, 'm') then m:= M+2 fi:

L[1..m-1]; # Robert Israel, Jun 03 2024

balQ[n_, b_] := balQ[n, b] = MinMax@ Differences@ DigitCount[n, b] == {0, 0}; f[n_] := DivisorSum[n, 1 &, balQ[#, 3] &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n] + 1; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[12, 10^5] (* Amiram Eldar, Jun 03 2024 *)

(Magma) bal:=func<n|Multiplicity(Intseq(n, 3), 1) eq Multiplicity(Intseq(n, 3), 0) and Multiplicity(Intseq(n, 3), 1) eq Multiplicity(Intseq(n, 3), 2)>; a:=[]; for n in [0..34] do k:=1; while #[d:d in Divisors(k)|bal(d)] ne n do k:=k+1; end while; Append(~a, k); end for; a;

Cf. A049354, A357035.

Sequence in context: A358340 A295840 A158470 * A163933 A359987 A377641

Adjacent sequences: A372143 A372144 A372145 * A372147 A372148 A372149

nonn,base

Marius A. Burtea, May 23 2024

approved