A350691 Smallest number with exactly n 1's in its digits both in decimal and binary representation.

1, 1152, 131081, 131110, 131111, 11011112, 111151110, 111151111, 1111511111, 111711111181, 1111416111111, 11151111111171, 1131141411111111, 10111111111115011, 117111183111111111, 1011411111118111111, 11111111181111141121, 111112811111101111111

Offset

1,2

Links

Scott R. Shannon, Table of n, a(n) for n = 1..45 (terms 1..25 from Michael S. Branicky, terms 26..35 from Chai Wah Wu)

Example

a(2) = 1152 which in binary is 10010000000. Both representations contain exactly 2 1's. And there is no smaller number satisfying this constraint.

Mathematica

Join[{1}, Table[t=0; While[!IntegerQ[k=Min@Flatten[Select[FromDigits/@Select[Permutations[#], First@#!=0&], Count[IntegerDigits[#, 2], 1]==n&]&/@(Join[Table[1, n], #]&/@Tuples[Join[{0}, Range[2, 9]], ++t])]]]; k, {n, 2, 12}]] (* Giorgos Kalogeropoulos, Jan 13 2022 *)

Prog

(Python)
numbers = [0, 2, 3, 4, 5, 6, 7, 8, 9]
terms = []
for ones in range(1, 13):
  if ones == 3: # needs more than 2 digits differing from "1"
    terms.append(131081)
  else:
    x = 1
    min_solution = 999999999999999999
    for i in range(ones+1):
      x += 10 ** (i + 1)
    for i in range(ones+2):
      for j in numbers:
        candidatej = x + (j - 1) * 10**i
        for k in range(i+1, ones+2):
          for l in numbers:
            candidate = candidatej + (l - 1) * 10 ** k
            if bin(candidate)[2:].count('1') == ones:
              if candidate < min_solution:
                min_solution = candidate
    terms.append(min_solution)
(Python)
from itertools import count
def A350691_helper(n, m): # generator in order of numbers with n decimal digits and m 1's. Leading zeros are allowed.
    if n >= m:
        if n == 1:
            if m == 1:
                yield 1
            else:
                yield 0
                yield from range(2, 10)
        elif n == m:
            yield (10**m-1)//9
        else:
            for b in A350691_helper(n-1, m):
                yield b
            r = 10**(n-1)
            for b in A350691_helper(n-1, m-1):
                yield r+b
            for a in range(2, 10):
                k = a*r
                for b in A350691_helper(n-1, m):
                    yield k+b
def A350691(n):
    for l in count(n):
        r = 10**(l-1)
        for a in range(1, 10):
            n2 = n-1 if a == 1 else n
            k = a*r
            for s in A350691_helper(l-1, n2):
                m = k+s
                if bin(m)[2:].count('1') == n:
                    return m # Chai Wah Wu, Jan 13 2022

Crossrefs

Cf. A000120, A268643, A350692.

Sequence in context: A269216 A339926 A170775 * A166490 A035761 A107557

Adjacent sequences: A350688 A350689 A350690 * A350692 A350693 A350694

Keyword

nonn,base

Author

Ruediger Jehn, Jan 12 2022

Extensions

a(13)-a(18) from Michael S. Branicky, Jan 12 2022

Status

approved