A325112 Integers k such that no nonzero subsequence of the decimal representation of k is divisible by 3.

1, 2, 4, 5, 7, 8, 10, 11, 14, 17, 20, 22, 25, 28, 40, 41, 44, 47, 50, 52, 55, 58, 70, 71, 74, 77, 80, 82, 85, 88, 100, 101, 104, 107, 110, 140, 170, 200, 202, 205, 208, 220, 250, 280, 400, 401, 404, 407, 410, 440, 470, 500, 502, 505, 508, 520, 550, 580, 700

Offset: 1

Author

Jonathan Kal-El Peréz, Mar 27 2019

Keywords

• nonn
• base

Comments

Integers whose decimal representation contains either just one nonzero digit, which is congruent to 1 or 2 (mod 3), or just two nonzero digits, which are either both == 1 (mod 3) or both == 2 (mod 3). - Robert Israel, Dec 25 2019

Examples

			From _David A. Corneth_, Sep 09 2024: (Start) 404 is in the sequence as its nonzero digits are (4,4). The nonzero subsequences of digits are (), (4), (4,4) with respective sums 0, 4, 8. None of these subsequences have a sum that is divisible by 3.
4160 is not in the sequence as one of its nonzero subsequences is (6) which sums to 6. As 6 is divisible by 3, 4160 is not in the sequence. (End)
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Chai Wah Wu, Algorithms for complementary sequences, arXiv:2409.05844 [math.NT], 2024.

Crossrefs

Cf. A014261 (for 2), A325113 (for 4), A261189 (for 5), A325114 (for 7).

A261188 is a subsequence.

A376045 is the complement.

Programs

Maple

F:= proc(d) local i,j,k, g;
   g:= [1,2,4,5,7,8];
   op(sort([seq(i*10^(d-1),i=g), seq(seq(seq(i*10^(d-1) + j*10^k, j = select(t -> (t-i) mod 3 = 0, g)),k=0..d-2),i=g)]));
end proc:
seq(F(d),d=1..4); # Robert Israel, Dec 25 2019

Mathematica

With[{k = 3}, Select[Range@ 700, NoneTrue[DeleteCases[FromDigits /@ Rest@ Subsequences[IntegerDigits@ #], 0], Mod[#, k] == 0 &] &]] (* Michael De Vlieger, Mar 31 2019 *)

Python

from itertools import combinations
def A325112(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x):
        c, l = 0, len(str(x))
        for i in range(l):
            k = 10**i
            for j in (1,2,4,5,7,8):
                if j*k<=x:
                    c += 1
        for a in combinations((10**i for i in range(l)),2):
            for b in ((1, 1), (1, 4), (1, 7), (2, 2), (2, 5), (2, 8), (4, 1), (4, 4), (4, 7), (5, 2), (5, 5), (5, 8), (7, 1), (7, 4), (7, 7), (8, 2), (8, 5), (8, 8)):
                if a[0]*b[0]+a[1]*b[1] <= x:
                    c += 1
        return n+x-c
    return bisection(f,n,n) # Chai Wah Wu, Sep 10 2024