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.

Showing 1-3 of 3 results.

A036991 Numbers k with the property that in the binary expansion of k, reading from right to left, the number of 0's never exceeds the number of 1's.

Original entry on oeis.org

0, 1, 3, 5, 7, 11, 13, 15, 19, 21, 23, 27, 29, 31, 39, 43, 45, 47, 51, 53, 55, 59, 61, 63, 71, 75, 77, 79, 83, 85, 87, 91, 93, 95, 103, 107, 109, 111, 115, 117, 119, 123, 125, 127, 143, 151, 155, 157, 159, 167, 171, 173, 175, 179, 181, 183, 187, 189, 191, 199, 203
Offset: 1

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Author

Keywords

Comments

List of binary words that correspond to a valid pairing of parentheses. - Joerg Arndt, Nov 27 2004
This sequence includes as subsequences A000225, A002450, A007583, A036994, A052940, A112627, A113836, A113841, A290114; and also A015521 (without 0), A083713 (without 0), A086224 (without 6), A182512 (without 0). - Gennady Eremin, Nov 27 2021 and Aug 26 2023
Partial differences are powers of 2 (cf. A367626, A367627). - Gennady Eremin, Dec 23 2021
This is the sequence A030101(A014486(n)), n >= 0, sorted into ascending order. See A014486 for more references, illustrations, etc., concerning Dyck paths and other associated structures enumerated by the Catalan numbers. - Antti Karttunen, Sep 25 2023
The terms in this sequence with a given length in base 2 are counted by A001405. For example, the number of terms of bit length k=5 (these are 19, 21, 23, 27, 29, and 31) is equal to A001405(k-1) = A001405(4) = 6. - Gennady Eremin, Nov 07 2023

Examples

			From _Joerg Arndt_, Dec 05 2021: (Start)
List of binary words with parentheses for those in the sequence (indicated by P). The binary words are scanned starting from the least significant bit, while the parentheses words are written left to right:
     Binary   Parentheses (if the value is in the sequence)
00:  ..... P  [empty string]
01:  ....1 P   ()
02:  ...1.
03:  ...11 P   (())
04:  ..1..
05:  ..1.1 P   ()()
06:  ..11.
07:  ..111 P   ((()))
08:  .1...
09:  .1..1
10:  .1.1.
11:  .1.11 P   (()())
12:  .11..
13:  .11.1 P   ()(())
14:  .111.
15:  .1111 P   (((())))
16:  1....
17:  1...1
18:  1..1.
19:  1..11 P   (())()
(End)
		

Crossrefs

Cf. A350577 (primes subsequence).
See also A014486, A030101, A036988, A036990, A036992. A036994 is a subset (requires the count of zeros to be strictly less than the count of 1's).
See also A030308, A000225, A002450, A007583, A350346, A367625, A367626 & A367627 (first differences).

Programs

  • Haskell
    a036991 n = a036991_list !! (n-1)
    a036991_list = filter ((p 1) . a030308_row) [0..] where
       p     []    = True
       p ones (0:bs) = ones > 1 && p (ones - 1) bs
       p ones (1:bs) = p (ones + 1) bs
    -- Reinhard Zumkeller, Jul 31 2013
    
  • Maple
    q:= proc(n) local l, t, i; l:= Bits[Split](n); t:=0;
          for i to nops(l) do t:= t-1+2*l[i];
            if t<0 then return false fi
          od: true
        end:
    select(q, [$0..300])[];  # Alois P. Heinz, Oct 09 2019
  • Mathematica
    moreOnesRLQ[n_Integer] := Module[{digits, len, flag = True, iter = 1, ones = 0, zeros = 0}, digits = Reverse[IntegerDigits[n, 2]]; len = Length[digits]; While[flag && iter < len, If[digits[[iter]] == 1, ones++, zeros++]; flag = ones >= zeros; iter++]; flag]; Select[Range[0, 203], moreOnesRLQ] (* Alonso del Arte, Sep 21 2011 *)
    Join[{0},Select[Range[210],Min[Accumulate[Reverse[IntegerDigits[#,2]]/.{0->-1}]]>-1&]] (* Harvey P. Dale, Apr 18 2014 *)
  • PARI
    select( {is_A036991(n,c=1)=!n||!until(!n>>=1,(c-=(-1)^bittest(n,0))||return)}, [0..99]) \\ M. F. Hasler, Nov 26 2021
  • Python
    def ok(n):
        if n == 0: return True # by definition
        count = {"0": 0, "1": 0}
        for bit in bin(n)[:1:-1]:
            count[bit] += 1
            if count["0"] > count["1"]: return False
        return True
    print([k for k in range(204) if ok(k)]) # Michael S. Branicky, Nov 25 2021
    
  • Python
    from itertools import count, islice
    def A036991_gen(): # generator of terms
        yield 0
        for n in count(1):
            s = bin(n)[2:]
            c, l = 0, len(s)
            for i in range(l):
                c += int(s[l-i-1])
                if 2*c <= i:
                    break
            else:
                yield n
    A036991_list = list(islice(A036991_gen(),20)) # Chai Wah Wu, Dec 30 2021
    

Formula

If a(n) = A000225(k) for some k, then a(n+1) = a(n) + A060546(k). - Gennady Eremin, Nov 07 2023

Extensions

More terms from Erich Friedman
Edited by N. J. A. Sloane, Sep 14 2008 at the suggestion of R. J. Mathar
Offset corrected and example adjusted accordingly by Reinhard Zumkeller, Jul 31 2013

A094014 Expansion of (1-2*x)/(1-8*x^2).

Original entry on oeis.org

1, -2, 8, -16, 64, -128, 512, -1024, 4096, -8192, 32768, -65536, 262144, -524288, 2097152, -4194304, 16777216, -33554432, 134217728, -268435456, 1073741824, -2147483648, 8589934592, -17179869184, 68719476736, -137438953472
Offset: 0

Views

Author

Paul Barry, Apr 21 2004

Keywords

Comments

Second inverse binomial transform of A094013. Third inverse binomial transform of A000129(2n-1).
The unsigned sequence has g.f. (1+2*x)/(1-8*x^2) and abs(a(n)) = 2^(3*n/2)*(1/2 + sqrt(2)/4 + (1/2 - sqrt(2)/4)*(-1)^n).

Crossrefs

Programs

  • Magma
    [n le 2 select (-2)^(n-1) else 8*Self(n-2): n in [1..41]]; // G. C. Greubel, Dec 04 2021
    
  • Mathematica
    LinearRecurrence[{0,8}, {1,-2}, 40] (* G. C. Greubel, Dec 04 2021 *)
  • Sage
    [(-2)^n*2^(n//2) for n in (0..40)] # G. C. Greubel, Dec 04 2021

Formula

a(n) = (2*sqrt(2))^n*(1/2 - sqrt(2)/4) + (-2*sqrt(2))^n*(1/2 + sqrt(2)/4).
a(n) = (-2)^n * A016116(n). - R. J. Mathar, Apr 28 2008
Abs(a(n)) = A113836(n+1) - A113836(n) for n > 0. - Reinhard Zumkeller, Feb 22 2010
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = Sum_{k=0..n} A158020(n,k)*3^k. - Philippe Deléham, Dec 01 2011
E.g.f.: cosh(2*sqrt(2)*x) - (1/sqrt(2))*sinh(2*sqrt(2)*x). - G. C. Greubel, Dec 04 2021

A173593 Numbers having in binary representation exactly two ones in three consecutive digits.

Original entry on oeis.org

3, 5, 6, 11, 13, 22, 27, 45, 54, 91, 109, 182, 219, 365, 438, 731, 877, 1462, 1755, 2925, 3510, 5851, 7021, 11702, 14043, 23405, 28086, 46811, 56173, 93622, 112347, 187245, 224694, 374491, 449389, 748982, 898779, 1497965, 1797558, 2995931, 3595117
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 22 2010

Keywords

Comments

a(2*n-1) = A033129(n+1);
a(3*n-2) = A113836(n+1);
a(6*n-5) = A083713(n);
a(2*n) - a(2*n-1) = A077947(n+1);
a(2*n+1) - a(2*n) = A077947(n).

Examples

			a(10) =   91 =      1011011_2
a(11) =  109 =      1101101_2
a(12) =  182 =     10110110_2
a(13) =  219 =     11011011_2
a(14) =  365 =    101101101_2
a(15) =  438 =    110110110_2
a(16) =  731 =   1011011011_2
a(17) =  877 =   1101101101_2
a(18) = 1462 =  10110110110_2
a(19) = 1755 =  11011011011_2
a(20) = 2925 = 101101101101_2
		

Crossrefs

Cf. A007088.
Bisections A033129, A033120.

Programs

  • Mathematica
    LinearRecurrence[{0, 2, 1, 0, -2}, {3, 5, 6, 11, 13}, 50] (* Jean-François Alcover, Feb 17 2018 *)

Formula

From R. J. Mathar, Feb 24 2010: (Start)
a(n) = 2*a(n-2) + a(n-3) - 2*a(n-5).
G.f.: x*(-3-5*x+2*x^3+4*x^4)/ ((1-x) * (1+x+x^2) * (2*x^2-1)). (End)
Showing 1-3 of 3 results.