A050621 -id:A050621 - cp's OEIS frontend

A035014 a(n) contains n digits (either '3' or '4') and is divisible by 2^n.

4, 44, 344, 3344, 33344, 433344, 3433344, 33433344, 333433344, 3333433344, 43333433344, 343333433344, 3343333433344, 33343333433344, 433343333433344, 3433343333433344, 43433343333433344, 443433343333433344, 3443433343333433344, 43443433343333433344

Offset: 1

J. Lowell

If (n-1)st term is divisible by 2^n, then n-th term begins with a 4. If not, then n-th term begins with a 3.

Proof of conjecture that a(n) ends with a(n-1): If a(n) is divisible by 2^n, then a(n) is divisible by 2^(n-1), so a(n)-k*10^(n-1) is divisible by 2^(n-1) for integer k, but if k is first digit of a(n) then a(n)-k*10^(n-1) is an (n-1)-digit number made up of 3s and 4s and divisible by 2^(n-1) and so must be a(n-1). - Henry Bottomley, Feb 14 2000

Ray Chandler, Table of n, a(n) for n = 1..1000 (first 100 terms from Jon E. Schoenfield)

Cf. A050620, A050621, A050622, A023402.

A035014 := proc(n)
    option remember ;
      local pre;
      if n = 1 then
        4;
    else
        pre := procname(n-1) ;
        pre+10^(n-1)*(4-modp(pre/2^(n-1),2)) ;
    end if;
end proc: # R. J. Mathar, May 02 2014

a(n) = if (n==1, 4, a(n-1) + 10^(n-1)*(4-(a(n-1)/2^(n-1) % 2))); \\ Michel Marcus, Apr 07 2017

a(n) = a(n-1) + 10^(n-1)*(4-[a(n-1)/2^(n-1) mod 2]), i.e., a(n) ends with a(n-1). - Henry Bottomley, Feb 14 2000

Corrected and extended by Patrick De Geest, Jun 15 1999

More terms from Henry Bottomley, Feb 14 2000

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

A155594 10^n+2^n-1.

Original entry on oeis.org

1, 11, 103, 1007, 10015, 100031, 1000063, 10000127, 100000255, 1000000511, 10000001023, 100000002047, 1000000004095, 10000000008191, 100000000016383, 1000000000032767, 10000000000065535, 100000000000131071

Offset: 0

Mohammad K. Azarian, Jan 24 2009

Index entries for linear recurrences with constant coefficients, signature (13,-32,20).

Cf. A074501, A074600, A005057, A005056, A099393, A155588, A155590, A155592, A155593.

LinearRecurrence[{13,-32,20},{1,11,103},20] (* Harvey P. Dale, Mar 07 2015 *)

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

G.f.: 1/(1-10*x)+1/(1-2*x)-1/(1-x). E.g.f.: e^(10*x)+e^(2*x)-e^x.
a(n)=12*a(n-1)-20*a(n-2)-9 with a(0)=1, a(1)=11 - Vincenzo Librandi, Jul 21 2010
a(0)=1, a(1)=11, a(2)=103, a(n)=13*a(n-1)-32*a(n-2)+20*a(n-3). - Harvey P. Dale, Mar 07 2015
a(n) = A050621(n+1)-1. - R. J. Mathar, Mar 10 2022

A034478 a(n) = (5^n + 1)/2.

Original entry on oeis.org

1, 3, 13, 63, 313, 1563, 7813, 39063, 195313, 976563, 4882813, 24414063, 122070313, 610351563, 3051757813, 15258789063, 76293945313, 381469726563, 1907348632813, 9536743164063, 47683715820313, 238418579101563

Offset: 0

N. J. A. Sloane

Terms (with the offset changed to 1) are also the quotients arising from sequence A050621.
Partial sums of A020699. - Paul Barry, Sep 03 2003
Binomial transform of A081294. - Paul Barry, Jan 13 2005

			G.f. = 1 + 3*x + 13*x^2 + 63*x^3 + 313*x^4 + 1563*x^5 + 7813*x^6 + ...

Nathaniel Johnston, Table of n, a(n) for n = 0..250
Index entries for linear recurrences with constant coefficients, signature (6,-5).

Cf. A020699, A034474, A050621, A081294.

seq((5^n + 1)/2, n=0..20); # Zerinvary Lajos, Jun 16 2007

LinearRecurrence[{6, -5},{1, 3},22] (* Ray Chandler, May 25 2021 *)

[lucas_number2(n,6,5)/2 for n in range(0,22)] # Zerinvary Lajos, Jul 08 2008

E.g.f.: exp(3*x)*cosh(2*x). - Paul Barry, Mar 17 2003
G.f.: (1-3*x)/((1-x)*(1-5*x)). - Paul Barry, Sep 03 2003
a(n) = Sum_{k=0..n} Sum_{j=0..k} binomial(n, k)*binomial(2*k, 2*j). - Paul Barry, Jan 13 2005
a(n) = 6*a(n-1) - 5*a(n-2) for n>1, a(0)=1, a(1)=3. - Philippe Deléham, Jul 11 2005
a(n)^2 + (a(n) - 1)^2 = a(2*n). E.g., 63^2 + 62^2 = 7813 = a(6). - Gary W. Adamson, Jun 17 2006
a(n) = 5*a(n-1) - 2 for n>0, a(0)=1. - Vincenzo Librandi, Aug 01 2010
a(n) = A034474(n)/2. - Elmo R. Oliveira, Dec 10 2023

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.

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

A050623 Smallest n-digit number divisible by 3^n.

Original entry on oeis.org

3, 18, 108, 1053, 10206, 100602, 1001646, 10005525, 100009323, 1000053864, 10000125297, 100000190088, 1000000837998, 10000002002688, 100000000895004, 1000000037647854, 10000000032104772, 100000000321047720

Offset: 1

Patrick De Geest, Jun 15 1999

Cf. A050621, A050624.

a(n) = 3^n*ceiling(10^(n-1)/3^n). - Vladeta Jovovic, Feb 14 2003

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

A053332 a(n) contains n digits (either '4' or '7') and is divisible by 2^n.

Original entry on oeis.org

4, 44, 744, 7744, 47744, 447744, 4447744, 44447744, 444447744, 4444447744, 74444447744, 474444447744, 4474444447744, 44474444447744, 444474444447744, 7444474444447744, 77444474444447744, 477444474444447744

Offset: 1

Henry Bottomley, Mar 06 2000

Ray Chandler, Table of n, a(n) for n = 1..1000

Cf. A023404, A050621, A050622, A035014.

Flatten[Table[Select[FromDigits/@Tuples[{4,7},n],Divisible[#,2^n]&], {n,20}]] (* Harvey P. Dale, Jul 25 2011 *)

a(n) = a(n-1)+10^(n-1)*(4+3*[a(n-1)/2^(n-1) mod 2]) i.e. a(n) ends with a(n-1); if (n-1)-th term is divisible by 2^n then n-th term begins with a 4, if not then n-th term begins with a 7.

cp's OEIS Frontend

A035014 a(n) contains n digits (either '3' or '4') and is 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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A155594 10^n+2^n-1.

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A034478 a(n) = (5^n + 1)/2.

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

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

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A050623 Smallest n-digit number divisible by 3^n.

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

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