A053312 -id:A053312 - cp's OEIS frontend

A126933 Quotients arising from sequence A053312.

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

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

A140288 The least n-digit multiple of 5^n using the decimal digits {1, 2, 3, 4, 5} exclusively.

Original entry on oeis.org

5, 25, 125, 3125, 53125, 453125, 4453125, 14453125, 314453125, 2314453125, 22314453125, 122314453125, 4122314453125, 44122314453125, 444122314453125, 4444122314453125, 54444122314453125, 254444122314453125, 1254444122314453125, 21254444122314453125

Offset: 1

Michel Criton (mcriton(AT)wanadoo.fr), May 24 2008

			a(5) = 53125 = 17 * 5^5.

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

Cf. A053312, A035014.

lista(nn) = {v = vector(nn); v[1] = 5; for(k = 2, nn, j = 1; while((j*10^(k-1) + v[k-1]) % 5^k > 0, j++); v[k] = j*10^(k-1) + v[k-1]); for(k = 1, nn, print1(v[k], ", "));} \\ Jinyuan Wang, Aug 27 2019

To obtain the (n+1)-th term, write the n-th term as k * 5^n. Multiply k by a multiple of 2^n to get a multiple of 5. Add the multiplicator of 2^n to the left of the n-th term.

More terms from Alois P. Heinz, Apr 05 2017

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

A147884 a(n) is the smallest positive integer k such that the last n digits of 2^k are 1 or 2.

Original entry on oeis.org

1, 9, 89, 89, 589, 3089, 3089, 3089, 315589, 315589, 8128089, 164378089, 945628089, 1922190589, 11687815589, 109344065589, 231414378089, 1452117503089, 4503875315589, 65539031565589, 141832976878089, 1667711883128089, 3575060515940589

Offset: 1

Max Alekseyev, Nov 17 2008

Minfeng Wang, Table of n, a(n) for n = 1..1429

Cf. A053312

{ m=2; for(n=1,50, print1(znlog(m,Mod(2,5^n)),", "); m+=10^n; if(m%(2^(n+1)), m+=10^n); ) }

from itertools import count, islice
from sympy import discrete_log
def A147884_gen(): # generator of terms
    a, b, c = 0, 1, 1
    for n in count(0):
        a+=b*c if (a>>n)&1 else b*c<<1
        c *= 5
        yield int(discrete_log(c,a,2))
        b <<= 1
A147884_list = list(islice(A147884_gen(),20)) # Chai Wah Wu, Mar 16 2023

a(n) = the smallest degree k such that 2^k == A053312(n) (mod 5^n).

Extended by Minfeng Wang, Dec 15 2024

A241882 Numbers with d digits that are divisible by 2^d and have at most 2 distinct digits: exactly one even digit and at most one odd digit.

Original entry on oeis.org

2, 4, 6, 8, 12, 16, 32, 36, 44, 52, 56, 72, 76, 88, 92, 96, 112, 144, 232, 272, 336, 344, 544, 552, 616, 656, 696, 744, 776, 888, 944, 992, 1616, 1888, 2112, 2272, 2992, 3232, 3344, 3888, 4144, 4544, 4944, 5552, 5888, 6336, 6656, 7744, 7776, 7888, 9696, 9888

Offset: 1

J. Lowell, Apr 30 2014

Union of 20 different sequences, all of which are defined as "a(n) contains n digits (either [any odd digit] or [any nonzero even digit] and is divisible by 2^n)."

Subsequence of A050622. - Michel Marcus, May 07 2014

			24 is not in the sequence because it has distinct even digits.

Ray Chandler, Table of n, a(n) for n = 1..10000

Cf. A035014, A053312-A053318, A053332-A053338, A053376-A053380 (sequences whose union is this sequence).

isok(n) = {digs = digits(n); d = #digs; if (n % 2^d, return (0)); sd = Set(digs); if (#sd > 2, return (0)); if (#sd < 2, return (1)); ((sd[1] % 2) + (sd[2] % 2)) == 1;} \\ Michel Marcus, May 02 2014

More terms from Michel Marcus, May 02 2014

A284924 a(n) is the least n-digit multiple of 5^n using the decimal digits {5, 6, 7, 8, 9} exclusively.

Original entry on oeis.org

5, 75, 875, 6875, 96875, 796875, 6796875, 66796875, 966796875, 5966796875, 65966796875, 865966796875, 8865966796875, 68865966796875, 868865966796875, 8868865966796875, 78868865966796875, 578868865966796875, 7578868865966796875, 87578868865966796875

Offset: 1

J. Lowell, Apr 05 2017

			a(2) = 75 because 75 is divisible by 5^2 (25*3=75) and uses only digits in the 5-9 interval.

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

Cf. A140288, A035014, A053312.

More terms from Alois P. Heinz, Apr 06 2017

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

A126933 Quotients arising from sequence A053312.

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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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A140288 The least n-digit multiple of 5^n using the decimal digits {1, 2, 3, 4, 5} exclusively.

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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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A147884 a(n) is the smallest positive integer k such that the last n digits of 2^k are 1 or 2.

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A241882 Numbers with d digits that are divisible by 2^d and have at most 2 distinct digits: exactly one even digit and at most one odd digit.

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PARI

Extensions

A284924 a(n) is the least n-digit multiple of 5^n using the decimal digits {5, 6, 7, 8, 9} exclusively.

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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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