A035014 -id:A035014 - cp's OEIS frontend

A050620 Quotients arising from sequence A035014.

2, 11, 43, 209, 1042, 6771, 26823, 130599, 651237, 3255306, 21158903, 83821639, 408121757, 2035115566, 13224589033, 52388661704, 331370112102, 1691563962301, 6567827879588, 41430886596044, 211450306579272, 1059399469695886, 5298071316879193, 20530429091056784

Offset: 1

Patrick De Geest, Jun 15 1999

Ray Chandler, Table of n, a(n) for n = 1..1431 (terms < 10^1000, first 1000 terms from Alois P. Heinz)

Cf. A023402, A035014.

a(n) = if (n==1, 2, a(n-1)/2 + 5^(n-1)*(4-(a(n-1) % 2))/2);

a(n) = a(n-1)/2 + 5^(n-1)*(4-[a(n-1) mod 2])/2. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jun 17 2001

a(n) = A035014(n)/2^n. - Michel Marcus, Apr 07 2017

More terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jun 17 2001

More terms from Michel Marcus, Apr 07 2017

A241687 Numbers k such that A035014(k) begins with a 3.

Original entry on oeis.org

3, 4, 5, 7, 8, 9, 10, 12, 13, 14, 16, 19, 24, 27, 28, 30, 31, 32, 34, 35, 36, 37, 38, 39, 41, 42, 44, 46, 47, 51, 53, 55, 56, 59, 61, 63, 64, 65, 67, 68, 69, 71, 72, 76, 82, 83, 84, 85, 87, 89, 92, 93, 96, 97, 99, 100, 102, 104, 105, 106, 109, 111, 113, 114

Offset: 1

J. Lowell, Apr 27 2014

Numbers k such that A023402(k) = 3.

			3 is a term of this sequence because A035014(3) = 344 begins with a 3.

Jon E. Schoenfield, Table of n, a(n) for n = 1..1000

Cf. A035014, A023402.

lista(nn) = {my(v=vector(nn), list=List()); v[1] = 4; for (n=2, nn, v[n] = v[n-1] + 10^(n-1)*(4-(v[n-1]/2^(n-1) % 2)); if (v[n]\10^(n-1) == 3, listput(list, n));); Vec(list);} \\ Michel Marcus, Feb 27 2022

More terms from Jon E. Schoenfield, May 05 2014

A241672 Numbers k such that A035014(k) begins with a 4.

Original entry on oeis.org

1, 2, 6, 11, 15, 17, 18, 20, 21, 22, 23, 25, 26, 29, 33, 40, 43, 45, 48, 49, 50, 52, 54, 57, 58, 60, 62, 66, 70, 73, 74, 75, 77, 78, 79, 80, 81, 86, 88, 90, 91, 94, 95, 98, 101, 103, 107, 108, 110, 112, 115, 117, 118, 120, 123, 124, 126, 127, 128, 129, 131, 136, 138, 139, 140, 143, 144, 145, 146

Offset: 1

J. Lowell, Apr 26 2014

A023402(n) = 4 for numbers in the sequence, 3 for numbers not in the sequence.

			3 is not in the sequence because A035014(3) begins with a 3.

Cf. A035014, A023402, A241687 (complement).

More terms from R. J. Mathar, May 02 2014

A050621 Smallest n-digit number divisible by 2^n.

Original entry on oeis.org

2, 12, 104, 1008, 10016, 100032, 1000064, 10000128, 100000256, 1000000512, 10000001024, 100000002048, 1000000004096, 10000000008192, 100000000016384, 1000000000032768, 10000000000065536, 100000000000131072

Offset: 1

Patrick De Geest, Jun 15 1999

Quotients arising from this sequence give A034478 ((5^(n-1)+1)/2).

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (12,-20).

Cf. A035014, A034478, A050622.

[2^(n-1)+10^(n-1): n in [1..21]]; // Vincenzo Librandi, Sep 12 2014

CoefficientList[Series[2 (1 - 6 x)/((1 - 2 x) (1 - 10 x)), {x, 0, 30}], x] (* Vincenzo Librandi, Sep 12 2014 *)

a(n) = 10^(n-1) + 2^(n-1) \\ Charles R Greathouse IV, Jun 11 2015

a(n) = 10^(n-1) + 2^(n-1).
G.f.: Q(0) where Q(k)= 1 + 5^k/(1 - 2*x/(2*x + 5^k/Q(k+1) )); (continued fraction ). - Sergei N. Gladkovskii, Apr 10 2013
G.f.: 2*x*(1-6*x)/((1-2*x)*(1-10*x)). - Vincenzo Librandi, Sep 12 2014
a(n) = 12*a(n-1) - 20*a(n-2) for n>1. - Vincenzo Librandi, Sep 12 2014

A050622 Numbers m that are divisible by 2^k, where k is the digit length of m.

Original entry on oeis.org

2, 4, 6, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, 248, 256, 264, 272, 280, 288, 296, 304, 312, 320, 328, 336, 344, 352, 360

Offset: 1

Patrick De Geest, Jun 15 1999

The number of terms of length k is equal to (9*5^(k-1) - 1)/2. - Bernard Schott, Apr 06 2020

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A035014, A050621, A055642.

seq(seq(j*2^k, j=(5^(k-1)+1)/2 .. 5^k-1),k=1..3); # Robert Israel, Apr 05 2020

Select[Range[360], IntegerQ[#/2^IntegerLength[#]] &] (* Jayanta Basu, May 25 2013 *)

isok(n) = n % (2^#Str(n)) == 0; \\ Michel Marcus, Sep 17 2015

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

Original entry on oeis.org

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.

A023402 If any power of 2 ends with k 3's and 4's, they must be the first k terms of this sequence in reverse order.

Original entry on oeis.org

4, 4, 3, 3, 3, 4, 3, 3, 3, 3, 4, 3, 3, 3, 4, 3, 4, 4, 3, 4, 4, 4, 4, 3, 4, 4, 3, 3, 4, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 4, 3, 3, 4, 3, 4, 3, 3, 4, 4, 4, 3, 4, 3, 4, 3, 3, 4, 4, 3, 4, 3, 4, 3, 3, 3, 4, 3, 3, 3, 4, 3, 3, 4, 4, 4, 3, 4, 4, 4, 4, 4, 3, 3, 3, 3, 4, 3, 4, 3, 4, 4, 3, 3, 4, 4, 3, 3, 4, 3, 3, 4, 3, 4, 3, 3, 3, 4, 4

Offset: 1

David W. Wilson

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

Cf. A035014.

A053318 a(n) contains n digits (either '2' or '7') and is divisible by 2^n.

Original entry on oeis.org

2, 72, 272, 2272, 22272, 222272, 7222272, 27222272, 727222272, 2727222272, 72727222272, 772727222272, 7772727222272, 77772727222272, 277772727222272, 2277772727222272, 72277772727222272, 272277772727222272

Offset: 1

Henry Bottomley, Mar 06 2000

Robert Israel, Table of n, a(n) for n = 1..999

Cf. A023399, A050621, A050622, A035014.

A[1]:= 2:
for n from 2 to 100 do
   if A[n-1] mod 2^n = 0 then A[n]:= A[n-1]+2*10^(n-1)
   else A[n]:= A[n-1]+7*10^(n-1)
     fi
od:
seq(A[i],i=1..100); # Robert Israel, Oct 27 2019

Table[Select[FromDigits/@Tuples[{2,7},n],Mod[#,2^n]==0&],{n,18}]//Flatten (* Harvey P. Dale, Jul 14 2025 *)

a(n) = a(n-1) + 10^(n-1)*(2 + 5*(a(n-1)/2^(n-1) mod 2)), i.e., a(n) ends with a(n-1); if a(n-1) is divisible by 2^n then a(n) begins with a 2, if not then a(n) begins with a 7.

Formula corrected by Robert Israel, Oct 27 2019

A053316 a(n) contains n digits (either '2' or '3') and is divisible by 2^n.

Original entry on oeis.org

2, 32, 232, 3232, 23232, 223232, 2223232, 32223232, 232223232, 3232223232, 23232223232, 323232223232, 3323232223232, 23323232223232, 323323232223232, 3323323232223232, 33323323232223232, 333323323232223232

Offset: 1

Henry Bottomley, Mar 06 2000

Robert Israel, Table of n, a(n) for n = 1..999

Cf. A023397, A050621, A050622, A035014.

A[1]:= 2:
for n from 2 to 100 do
   if A[n-1] mod 2^n = 0 then A[n]:= A[n-1]+2*10^(n-1)
   else A[n]:= A[n-1]+3*10^(n-1)
     fi
od:
seq(A[i],i=1..100); # Robert Israel, Oct 27 2019

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 a(n-1) is divisible by 2^n then a(n) begins with a 2, if not then a(n) begins with a 3.

Formula corrected by Robert Israel, Oct 27 2019

A053317 a(n) contains n digits (either '2' or '5') and is divisible by 2^n.

Original entry on oeis.org

2, 52, 552, 5552, 55552, 255552, 5255552, 55255552, 255255552, 2255255552, 22255255552, 222255255552, 5222255255552, 55222255255552, 255222255255552, 2255222255255552, 22255222255255552, 222255222255255552

Offset: 1

Henry Bottomley, Mar 06 2000

Robert Israel, Table of n, a(n) for n = 1..999

Cf. A023398, A050621, A050622, A035014, A055880.

A[1]:= 2:
for n from 2 to 100 do
   if A[n-1] mod 2^n = 0 then A[n]:= A[n-1]+2*10^(n-1)
   else A[n]:= A[n-1]+5*10^(n-1)
fi
od:
seq(A[i],i=1..100); # Robert Israel, Oct 27 2019

Table[Select[FromDigits/@Tuples[{2,5},n],Divisible[#,2^n]&],{n,18}]//Flatten (* Harvey P. Dale, Oct 12 2022 *)

a(n) = a(n-1) + 10^(n-1)*(2 + 3*(a(n-1)/2^(n-1) mod 2)), i.e., a(n) ends with a(n-1); if a(n-1) is divisible by 2^n then a(n) begins with a 2, if not then a(n) begins with a 5.

Formula corrected by Robert Israel, Oct 27 2019

cp's OEIS Frontend

A050620 Quotients arising from sequence A035014.

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A241687 Numbers k such that A035014(k) begins with a 3.

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A241672 Numbers k such that A035014(k) begins with a 4.

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A050621 Smallest n-digit number divisible by 2^n.

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A050622 Numbers m that are divisible by 2^k, where k is the digit length of m.

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

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A023402 If any power of 2 ends with k 3's and 4's, they must be the first k terms of this sequence in reverse order.

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A053318 a(n) contains n digits (either '2' or '7') and is divisible by 2^n.

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Maple