A023402 -id:A023402 - 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

A023408 If any power of 2 ends with k 5's and 6's, they must be the first k terms of this sequence in reverse order.

Original entry on oeis.org

6, 5, 6, 6, 6, 5, 6, 6, 6, 6, 5, 6, 6, 6, 5, 6, 5, 5, 6, 5, 5, 5, 5, 6, 5, 5, 6, 6, 5, 6, 6, 6, 5, 6, 6, 6, 6, 6, 6, 5, 6, 6, 5, 6, 5, 6, 6, 5, 5, 5, 6, 5, 6, 5, 6, 6, 5, 5, 6, 5, 6, 5, 6, 6, 6, 5, 6, 6, 6, 5, 6, 6, 5, 5, 5, 6, 5, 5, 5, 5, 5, 6, 6, 6, 6, 5, 6, 5, 6, 5, 5, 6, 6, 5, 5, 6, 6, 5, 6, 6, 5, 6, 5, 6, 6, 6, 5, 5

Offset: 1

David W. Wilson

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

Cf. A023402.

Conjecture: a(n) + A023402(n) = 9, for n > 1. - J. Lowell, Nov 16 2020

A050620 Quotients arising from sequence A035014.

Original entry on oeis.org

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

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

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

cp's OEIS Frontend

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

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A023408 If any power of 2 ends with k 5's and 6's, they must be the first k terms of this sequence in reverse order.

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A050620 Quotients arising from sequence A035014.

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

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PARI

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

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