cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A300428 a(n) is the least positive k such that the binary representation of n appears as a substring in the binary representation of 1/k (ignoring the radix point and adding trailing zeros if necessary in case of a terminating expansion).

Original entry on oeis.org

1, 1, 5, 1, 3, 5, 9, 1, 5, 3, 11, 5, 11, 9, 17, 1, 9, 7, 5, 11, 3, 13, 11, 9, 5, 11, 13, 9, 11, 17, 33, 1, 17, 11, 9, 7, 19, 5, 13, 11, 29, 3, 19, 13, 27, 11, 19, 17, 9, 19, 5, 11, 19, 13, 29, 9, 19, 11, 13, 17, 25, 33, 65, 1, 33, 23, 17, 13, 11, 29, 9, 23, 7
Offset: 1

Views

Author

Rémy Sigrist, Mar 05 2018

Keywords

Comments

In other words, a(n) is the least k > 0 such that floor((2^i) / k) mod A062383(n) = n for some integer i >= 0.
This sequence is similar to A035335 for the base 2.
All terms are odd.
All terms appears infinitely many times (as a(n) equals at least a(2*n) or a(2*n + 1)).
See also A300475 for a similar sequence.

Examples

			The first terms, alongside the binary representation of 1/a(n) with the earliest occurrence of the binary representation of n in parentheses, are:
  n  a(n)    bin(1/a(n))
  -- ----    -----------
   1    1    (1).000...
   2    1    (1.0)000...
   3    5    0.00(11)001...
   4    1    (1.00)000...
   5    3    0.0(101)010...
   6    5    0.00(110)011...
   7    9    0.000(111)000...
   8    1    (1.000)000...
   9    5    0.001(1001)100...
  10    3    0.0(1010)101...
  11   11    0.000(1011)101...
  12    5    0.00(1100)110...
  13   11    0.000101(1101)000...
  14    9    0.000(1110)001...
  15   17    0.0000(1111)000...
  16    1    (1.0000)000...
  17    9    0.00011(10001)110...
  18    7    0.00(10010)010...
  19    5    0.001(10011)001...
  20   11    0.0001011(10100)010...
		

Crossrefs

Programs

  • PARI
    See Links section.

Formula

a(2^k) = 1 for any k >= 0.
a(2^k - 1) = 2^k + 1 for any k > 1.
a(A000975(k)) = 3 for any k > 2.
a(A033138(k)) = 7 for any k > 4.
a(n) <= A300475(n) for any n > 0.

A309668 a(n) is the least positive number of the form floor(2^k/n) for some k >= 0 not yet in the sequence.

Original entry on oeis.org

1, 2, 5, 4, 3, 10, 9, 8, 7, 6, 11, 21, 19, 18, 17, 16, 15, 14, 13, 12, 24, 23, 22, 42, 20, 39, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 52, 25, 49, 48, 47, 46, 45, 44, 43, 85, 41, 40, 80, 78, 38, 75, 74, 73, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61
Offset: 1

Views

Author

Rémy Sigrist, Aug 11 2019

Keywords

Comments

The sequence is well defined as for any n > 0, there are infinitely many positive numbers of the form floor(2^k/n) with k >= 0.
The sequence is a permutation of the natural numbers, with inverse A309734:
- for any m > 0, floor(2^k/A300475(m)) = m for some k,
- also, for any u > 0, floor(2^(k-u)/(A300475(m)*2^u)) = m,
- so the set S_m = { v such that floor(2^k/v) = m for some k >= 0 } is infinite
- and eventually a(n) = m for some n in S_m, QED.

Examples

			The first terms, alongside the binary representations of a(n) and of 1/n (with that of a(n) in parentheses), are:
  --  ----  ---------  ---------------------
   1     1          1  (1).00000000000000...
   2     2         10  0.(10)000000000000...
   3     5        101  0.0(101)0101010101...
   4     4        100  0.0(100)0000000000...
   5     3         11  0.00(11)0011001100...
   6    10       1010  0.00(1010)10101010...
   7     9       1001  0.00(1001)00100100...
   8     8       1000  0.00(1000)00000000...
   9     7        111  0.000(111)00011100...
  10     6        110  0.000(110)01100110...
  11    11       1011  0.000(1011)1010001...
  12    21      10101  0.000(10101)010101...
  13    19      10011  0.000(10011)101100...
  14    18      10010  0.000(10010)010010...
  15    17      10001  0.000(10001)000100...
		

Crossrefs

Cf. A070939, A300475, A309734 (inverse).

Programs

  • PARI
    s=1; for (n=1, 67, q=1/n; while (bittest(s, f=floor(q)), q*=2); print1 (f ", "); s+=2^f)
Showing 1-2 of 2 results.