A023396 -id:A023396 - 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.

A126933 Quotients arising from sequence A053312.

Original entry on oeis.org

1, 3, 14, 132, 691, 1908, 16579, 47352, 414301, 1183713, 5474669, 27151397, 135646011, 678174568, 6442602909, 18480090517, 85533990571, 424236721848, 4026815626549, 11550150977337, 53458791308981, 265147974756053, 1324666882885839, 6622797918981982, 62916043734881616, 329481245744393933

Offset: 1

Gerry Leversha, Mar 18 2007

Take the decimal number formed by the first n digits of A023396 in reverse order and divide by 2^n.
The sequence A053312 gives n-digit numbers consisting entirely of 1s and 2s which are divisible by 2^n. The quotients upon division form the present sequence. The parity of the n-th term here determines the next term in A023396; if odd, it is a 1 and if even, a 2.
This was set as a problem in the All Union Mathematical Olympiad of 1971 and can be found in the reference cited here.

			a(4) = A053312(4) / 2^4 = 2112 / 16 = 132. - _David A. Corneth_, Jun 11 2020

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

David A. Corneth, Table of n, a(n) for n = 1..1431 (terms <= 10^1000)
Index to sequences related to Olympiads.

Cf. A023396, A053312.

from itertools import count, islice
def A126933_gen(): # generator of terms
    a, b = 2, 10
    for n in count(1):
        a+=b if (c:=a>>n)&1 else b<<1
        b *= 10
        yield c
A126933_list = list(islice(A126933_gen(),20)) # Chai Wah Wu, Mar 16 2023

a(n) < 0.3 * 5^n. - David A. Corneth, Jun 11 2020

Name changed and other minor edits by Ray Chandler, Jun 17 2020

A023407 If any power of 2 ends with k 3's and 6's, they must be the first k terms of this sequence in reverse order.

Original entry on oeis.org

6, 3, 3, 6, 6, 3, 6, 3, 6, 3, 3, 3, 3, 3, 6, 3, 3, 3, 6, 3, 3, 3, 3, 3, 6, 6, 3, 3, 6, 3, 3, 6, 3, 6, 3, 3, 3, 6, 3, 6, 3, 6, 6, 3, 3, 6, 3, 6, 6, 6, 3, 3, 3, 3, 6, 6, 3, 6, 3, 3, 3, 6, 6, 6, 3, 6, 3, 3, 6, 6, 6, 3, 6, 6, 3, 3, 6, 3, 3, 3, 6, 3, 3, 6, 3, 6, 3, 3, 6, 6, 3, 3, 6, 3, 3, 6, 3, 3, 6, 3, 6, 3, 6, 3, 6, 3, 3, 3

Offset: 1

David W. Wilson

Ray Chandler, Table of n, a(n) for n = 1..10000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets

Cf. A023396.

a(n) = 3*A023396(n).

A023409 If any power of 2 ends with k 6's and 7's, they must be the first k terms of this sequence in reverse order.

Original entry on oeis.org

6, 7, 7, 7, 6, 6, 6, 6, 7, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6, 7, 7, 6, 6, 7, 7, 7, 6, 6, 6, 6, 7, 6, 7, 7, 7, 7, 7, 6, 6, 6, 6, 7, 6, 7, 7, 6, 6, 6, 6, 7, 6, 7, 6, 6, 7, 6, 7, 7, 7, 6, 6, 6, 7, 6, 7, 7, 7, 6, 6, 6, 6, 6, 7, 6, 6, 6, 7, 7, 6, 7, 7, 6, 7, 7, 6, 7, 6, 6, 7, 7, 6, 7, 6, 7, 7, 6, 6, 7, 7, 6, 6, 7

Offset: 0

David W. Wilson

From Robert Israel, Mar 30 2018: (Start)
a(0)=6. If the concatenation 6a(n)...a(0) (as a decimal number) is divisible by 2^(n+2) then a(n+1)=6, otherwise a(n+1)=7.
Pomerance (see link) shows the sequence is not eventually periodic. (End)

Robert Israel, Table of n, a(n) for n = 0..10000
C. Pomerance, Sixes and sevens, Missouri J. Math. Sci. 6 (1994), 62-63.

Cf. A023396, A023397, A023398, A023399, A023400, A023401, A023402, A023403, A023404, A023405, A023406, A023407, A023408, A023410, A023411, A023412, A023413, A023414, A023415.

a[0]:= 6: v:= 6:
for n from 1 to 100 do
  if 6*10^n+v mod 2^(n+1)=0 then a[n]:= 6 else a[n]:= 7 fi;
  v:= v + a[n]*10^n
od:
seq(a[i],i=0..100); # Robert Israel, Mar 30 2018

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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A126933 Quotients arising from sequence A053312.

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A023407 If any power of 2 ends with k 3's and 6's, they must be the first k terms of this sequence in reverse order.

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A023409 If any power of 2 ends with k 6's and 7's, they must be the first k terms of this sequence in reverse order.

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Maple

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

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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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Python