A350692 Smallest number with exactly n zeros in its digits both in decimal and binary representation.

0, 1003, 130007, 1003007, 50200030, 100007900, 1000300030, 102000001007, 1080007000030, 30090004000500, 100004000300030, 1020070000000500, 9000050000003000, 1000000800080003007, 4003700000000300000, 100000070002000003007, 1000006000000010027000

Offset

1,2

Links

Scott R. Shannon, Table of n, a(n) for n = 1..40 (terms 1..23 from Michael S. Branicky)

Example

a(2) = 1003 which in binary is 1111101011. Both representations contain exactly 2 zeros. And there is no smaller number satisfying this constraint.

Mathematica

Join[{0}, Table[
t=0; While[!IntegerQ[k=Min@Flatten[Select[FromDigits/@ Select[Permutations[#], First@#!=0&], Count[IntegerDigits[#, 2], 0]==n&]&/@(Join[Table[0, n], #]&/@Tuples[Range@9, ++t])]]]; k, {n, 2, 7}]] (* Giorgos Kalogeropoulos, Jan 13 2022 *)

Prog

(Python)
from itertools import count
def A350692_helper(n, m): # generator in order of numbers with n decimal digits and m 0's. Leading zeros are allowed.
    if n >= m:
        if n == 1:
            if m == 1:
                yield 0
            else:
                yield from range(1, 10)
        elif n == m:
            yield 0
        else:
            for b in A350692_helper(n-1, m-1):
                yield b
            r = 10**(n-1)
            for a in range(1, 10):
                k = a*r
                for b in A350692_helper(n-1, m):
                    yield k+b
def A350692(n):
    if n == 1:
        return 0
    for l in count(n):
        r = 10**(l-1)
        for a in range(1, 10):
            k = a*r
            for s in A350692_helper(l-1, n):
                m = k+s
                if bin(m)[2:].count('0') == n:
                    return m # Chai Wah Wu, Jan 13 2022

Crossrefs

Cf. A023416, A055641, A350691.

Sequence in context: A151956 A111349 A182935 * A013686 A153226 A202147

Adjacent sequences: A350689 A350690 A350691 * A350693 A350694 A350695

Keyword

nonn,base

Author

Ruediger Jehn, Jan 12 2022

Extensions

a(8)-a(17) from Michael S. Branicky, Jan 12 2022

Status

approved