A126933 -id:A126933 - cp's OEIS frontend

A053312 a(n) contains n digits (either '1' or '2') and is divisible by 2^n.

2, 12, 112, 2112, 22112, 122112, 2122112, 12122112, 212122112, 1212122112, 11212122112, 111212122112, 1111212122112, 11111212122112, 211111212122112, 1211111212122112, 11211111212122112, 111211111212122112, 2111211111212122112, 12111211111212122112

Offset: 1

Henry Bottomley, Mar 06 2000

Corresponding quotients a(n) / 2^n are in A126933. - Bernard Schott, Mar 15 2023

			a(5) = 22112 since 22112 = 2^5 * 691 and 22112 contains 5 digits.

Alois P. Heinz, Table of n, a(n) for n = 1..1000

Cf. A000079, A023396, A050621, A050622, A035014, A126933, A207778.

a:= proc(n) option remember; `if`(n=1, 2, (t-> parse(
      cat(`if`(irem(t, 2^n)=0, 2, 1), t)))(a(n-1)))
    end:
seq(a(n), n=1..20);  # Alois P. Heinz, Feb 02 2025

Select[Flatten[Table[FromDigits/@Tuples[{1,2},n],{n,20}]],Divisible[ #,2^IntegerLength[#]]&] (* Harvey P. Dale, Jul 01 2019 *)
RecurrenceTable[{a[1]==2, a[n]==a[n-1]+10^(n-1)(2-Mod[a[n-1]/2^(n-1),2])}, a[n], {n, 1, 20}] (* Paul C Abbott, Feb 03 2025 *)

from itertools import count, islice
def A053312_gen(): # generator of terms
    a = 0
    for n in count(0):
        yield (a:=a+(10**n if (a>>n)&1 else 10**n<<1))
A053312_list = list(islice(A053312_gen(),20)) # Chai Wah Wu, Mar 15 2023

a(n) = a(n-1) + 10^(n-1)*(2-[a(n-1)/2^(n-1) mod 2]), i.e., a(n) ends with a(n-1); if the (n-1)-th term is divisible by 2^n then the n-th term begins with a 2; if not, then the n-th term begins with a 1.

A023396 If any odd power of 2 ends with k 1's and 2's, they must be the first k terms of this sequence in reverse order.

Original entry on oeis.org

2, 1, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1

Offset: 1

David W. Wilson

			2^1 ends in 2;
2^5 ends in 32;
2^9 ends in 512;
2^13 ends in 8192;
2^89 ends in ...562112.
There exists a power of two ending in 12, so for n = 3 the choice for a(3) = 1 or a(3) = 2 comes from the existence of a power of two ending in either 112 or 212. As 112 is divisible by 2^n = 8 (and 212 is not) a(3) = 1. - _David A. Corneth_, Jun 11 2020

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A053312, A126933.

first(n) = my(f = 2, pow10 = 1, pow2 = 2); { for(i = 2, n, pow10*=10; pow2<<=1; c1 = pow10 + f; if(c1 % pow2 == 0, f = c1, c2 = 2*pow10 + f; if(c2 % pow2 == 0, f = c2 ) ) ); Vecrev(digits(f)) } \\ David A. Corneth, Jun 11 2020

Definition corrected by Gerry Leversha, Mar 17 2007

A207778 Smallest multiple of 2^n using only 1's and 2's.

Original entry on oeis.org

1, 2, 12, 112, 112, 2112, 2112, 122112, 122112, 12122112, 12122112, 12122112, 111212122112, 1111212122112, 11111212122112, 11111212122112, 11111212122112, 11111212122112, 111211111212122112, 111211111212122112, 111211111212122112, 111211111212122112

Offset: 0

Lekraj Beedassy, Feb 20 2012

An induction-based argument can be used to show that this sequence is actually infinite.

Problem 1, proposed during the 5th All-Soviet-Union Mathematical Competition in 1971 at Riga (Pertsel link), asks for a proof that this sequence is infinite. - Bernard Schott, Mar 20 2023

J. B. Tabov and P. J. Taylor, Methods of Problem Solving, Book 1, Australian Mathematics Trust, 1996.

Alois P. Heinz, Table of n, a(n) for n = 0..300
Vladimir A. Pertsel, Problems of the All-Soviet-Union Mathematical Competitions 1961-1986, the 5th competition, Riga, 1971, problem 144.
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A000079, A007931, A023396, A053312, A126933.

Table[m = 1; p = 2^k; While[Total@ DigitCount[m p][[3 ;; -1]] > 0, m++]; m p, {k, 0, 11}] (* Michael De Vlieger, Mar 17 2023 *)

a(n) = my(k=1, d=digits(k*2^n)); while (!((vecmin(d)>=1) && (vecmax(d)<=2)), k++; d=digits(k*2^n)); k*2^n; \\ Michel Marcus, Mar 15 2023

a(n) <= A053312(n).

More terms from Alois P. Heinz, Feb 20 2012

A342932 The unique sequence {a(1), a(2), a(3), a(4), ...} of digits 1, 2, or 3 such that the number a(n)a(n-1)...a(2)a(1), read in base 6, is divisible by 3^n.

Original entry on oeis.org

3, 1, 2, 3, 3, 1, 1, 3, 1, 2, 2, 1, 1, 3, 1, 3, 1, 3, 3, 3, 2, 3, 3, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 3, 3, 2, 2, 3, 1, 3, 2, 2, 2, 3, 3, 1, 2, 2, 2, 2, 2, 1, 3, 2, 2, 3, 2, 2, 1, 2, 1, 3, 2, 2, 3, 1, 1, 1, 1, 1, 3, 2, 2, 2, 3, 3, 2, 1, 3, 1, 1, 2, 2, 3, 1, 3, 2, 3, 2, 3, 1, 1, 3, 1, 2, 3, 3, 2, 3, 2, 3, 1, 1, 2, 3

Offset: 1

Eugen Ionascu, Mar 29 2021

The distribution seems to be uniform but random (empirical observation).

To prove that such a digit sequence exists and is unique is a good (but uncommon) example of a proof by induction.

			3 is divisible by 3^1;
13_6 = 1*6 + 3 = 9, which is divisible by 3^2,
213_6 = 2*6^2 + 1*6 + 3 = 81, which is divisible by 3^3.

Rémy Sigrist, Colored scatterplot of (n, 3*o(n)-n) for = 1..1000000 (where o corresponds to the ordinal transform of the sequence)
Eugen Ionascu, Mathematica File

Cf. A023396, A053312, A126933.

nd[n_] := Module[{k, i, s, ss, L, a}, L = Array[f, n]; f[1] = 3;
  Do[s = Sum[6^(k - 1)*f[k], {k, 1, i - 1}];
   ss = Mod[2^(i - 1)*s/3^(i - 1), 3];
   If[ss == 0, f[i] = 3, If[ss == 1, f[i] = 2, f[i] = 1]], {i, 2, n}];
  s = Sum[6^(k - 1)*f[k], {k, 1, n}];
  {L, s/3^n}]

{ q=0; t=1; for (n=1, 105, print1 (d=[3,1,2][1+lift(-q/Mod(t,3))]", "); q=(t*d+q)/3; t*=2) } \\ Rémy Sigrist, Apr 15 2021

n, div, divnum = 0, 1, 0
while n < 87:
    div, a = 3*div, 1
    while (a*6**n+divnum)%div != 0:
        a = a+1
    divnum, n = divnum+a*6**n, n+1
    print(a, end=', ') # A.H.M. Smeets, Apr 13 2021

cp's OEIS Frontend

A053312 a(n) contains n digits (either '1' or '2') and is divisible by 2^n.

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A023396 If any odd power of 2 ends with k 1's and 2's, they must be the first k terms of this sequence in reverse order.

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A207778 Smallest multiple of 2^n using only 1's and 2's.

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Mathematica

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Extensions

A342932 The unique sequence {a(1), a(2), a(3), a(4), ...} of digits 1, 2, or 3 such that the number a(n)a(n-1)...a(2)a(1), read in base 6, is divisible by 3^n.

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Examples

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Crossrefs

Programs

Mathematica

PARI

Python