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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A020987 Zero-one version of Golay-Rudin-Shapiro sequence (or word).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0
Offset: 0

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Author

Keywords

Comments

This is (1-A020985(n))/2. See A020985, which is the main entry for this sequence, for more information. N. J. A. Sloane, Jun 06 2012
A word that is uniform primitive morphic, but not pure morphic. - N. J. A. Sloane, Jul 14 2018

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 78.
  • Dekking, Michel, Michel Mendes France, and Alf van der Poorten. "Folds." The Mathematical Intelligencer, 4.3 (1982): 130-138 & front cover, and 4:4 (1982): 173-181 (printed in two parts).
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • Lipshitz, Leonard, and A. van der Poorten. "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990): 339-358.

Crossrefs

Cf. A020985.
A014081(n) mod 2. Characteristic function of A022155.
Sequences mentioned in the Allouche et al. "Taxonomy" paper, listed by example number: 1: A003849, 2: A010060, 3: A010056, 4: A020985 and A020987, 5: A191818, 6: A316340 and A273129, 18: A316341, 19: A030302, 20: A063438, 21: A316342, 22: A316343, 23: A003849 minus its first term, 24: A316344, 25: A316345 and A316824, 26: A020985 and A020987, 27: A316825, 28: A159689, 29: A049320, 30: A003849, 31: A316826, 32: A316827, 33: A316828, 34: A316344, 35: A043529, 36: A316829, 37: A010060.

Programs

A245563 Table read by rows: row n gives list of lengths of runs of 1's in binary expansion of n, starting with low-order bits.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 3, 4, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 2, 2, 2, 3, 1, 3, 4, 5, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 1, 4, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1, 2, 2, 2, 3, 2, 3, 1, 3, 1, 3, 2, 3, 4, 1, 4, 5, 6, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1
Offset: 0

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Author

N. J. A. Sloane, Aug 10 2014

Keywords

Comments

A formula for A071053(n) depends on this table.

Examples

			Here are the run lengths for the numbers 0 through 21:
0, []
1, [1]
2, [1]
3, [2]
4, [1]
5, [1, 1]
6, [2]
7, [3]
8, [1]
9, [1, 1]
10, [1, 1]
11, [2, 1]
12, [2]
13, [1, 2]
14, [3]
15, [4]
16, [1]
17, [1, 1]
18, [1, 1]
19, [2, 1]
20, [1, 1]
21, [1, 1, 1]
		

Crossrefs

Row sums = A000120 (the binary weight).
Row lengths are A069010.
The version for prime indices (instead of binary indices) is A124010.
Numbers with distinct run-lengths are A328592.
Numbers with equal run-lengths are A164707.

Programs

  • Haskell
    import Data.List (group)
    a245563 n k = a245563_tabf !! n !! k
    a245563_row n = a245563_tabf !! n
    a245563_tabf = [0] : map
       (map length . (filter ((== 1) . head)) . group) (tail a030308_tabf)
    -- Reinhard Zumkeller, Aug 10 2014
    
  • Maple
    for n from 0 to 128 do
    lis:=[]; t1:=convert(n,base,2); L1:=nops(t1); out1:=1; c:=0;
    for i from 1 to L1 do
    if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
    elif out1 = 0 and t1[i] = 1 then c:=c+1;
    elif out1 = 1 and t1[i] = 0 then c:=c;
    elif out1 = 0 and t1[i] = 0 then lis:=[op(lis),c]; out1:=1; c:=0;
    fi;
    if i = L1 and c>0 then lis:=[op(lis),c]; fi;
    od:
    lprint(n,lis);
    od:
  • Mathematica
    Join@@Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2==#1+1&],{n,0,100}] (* Gus Wiseman, Nov 03 2019 *)
  • Python
    from re import split
    A245563_list = [0]
    for n in range(1,100):
        A245563_list.extend(len(d) for d in split('0+',bin(n)[:1:-1]) if d != '')
    # Chai Wah Wu, Sep 07 2014

A121016 Numbers whose binary expansion is properly periodic.

Original entry on oeis.org

3, 7, 10, 15, 31, 36, 42, 45, 54, 63, 127, 136, 153, 170, 187, 204, 221, 238, 255, 292, 365, 438, 511, 528, 561, 594, 627, 660, 682, 693, 726, 759, 792, 825, 858, 891, 924, 957, 990, 1023, 2047, 2080, 2145, 2184, 2210, 2275, 2340, 2405, 2457, 2470, 2535
Offset: 1

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Author

Jacob A. Siehler, Sep 08 2006

Keywords

Comments

A finite sequence is aperiodic if its cyclic rotations are all different. - Gus Wiseman, Oct 31 2019

Examples

			For example, 204=(1100 1100)_2 and 292=(100 100 100)_2 belong to the sequence, but 30=(11110)_2 cannot be split into repeating periods.
From _Gus Wiseman_, Oct 31 2019: (Start)
The sequence of terms together with their binary expansions and binary indices begins:
   3:         11 ~ {1,2}
   7:        111 ~ {1,2,3}
   10:      1010 ~ {2,4}
   15:      1111 ~ {1,2,3,4}
   31:     11111 ~ {1,2,3,4,5}
   36:    100100 ~ {3,6}
   42:    101010 ~ {2,4,6}
   45:    101101 ~ {1,3,4,6}
   54:    110110 ~ {2,3,5,6}
   63:    111111 ~ {1,2,3,4,5,6}
  127:   1111111 ~ {1,2,3,4,5,6,7}
  136:  10001000 ~ {4,8}
  153:  10011001 ~ {1,4,5,8}
  170:  10101010 ~ {2,4,6,8}
  187:  10111011 ~ {1,2,4,5,6,8}
  204:  11001100 ~ {3,4,7,8}
  221:  11011101 ~ {1,3,4,5,7,8}
  238:  11101110 ~ {2,3,4,6,7,8}
  255:  11111111 ~ {1,2,3,4,5,6,7,8}
  292: 100100100 ~ {3,6,9}
(End)
		

Crossrefs

A020330 is a subsequence.
Numbers whose binary expansion is aperiodic are A328594.
Numbers whose reversed binary expansion is Lyndon are A328596.
Numbers whose binary indices have equal run-lengths are A164707.

Programs

  • Mathematica
    PeriodicQ[n_, base_] := Block[{l = IntegerDigits[n, base]}, MemberQ[ RotateLeft[l, # ] & /@ Most@ Divisors@ Length@l, l]]; Select[ Range@2599, PeriodicQ[ #, 2] &]
  • PARI
    is(n)=n=binary(n);fordiv(#n,d,for(i=1,#n/d-1, for(j=1,d, if(n[j]!=n[j+i*d], next(3)))); return(d<#n)) \\ Charles R Greathouse IV, Dec 10 2013

A034931 Triangle read by rows: Pascal's triangle (A007318) mod 4.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 0, 2, 0, 1, 1, 1, 2, 2, 1, 1, 1, 2, 3, 0, 3, 2, 1, 1, 3, 1, 3, 3, 1, 3, 1, 1, 0, 0, 0, 2, 0, 0, 0, 1, 1, 1, 0, 0, 2, 2, 0, 0, 1, 1, 1, 2, 1, 0, 2, 0, 2, 0, 1, 2, 1, 1, 3, 3, 1, 2, 2, 2, 2, 1, 3, 3, 1, 1, 0, 2, 0, 3, 0, 0, 0, 3, 0, 2, 0, 1, 1, 1, 2, 2, 3, 3, 0, 0, 3, 3, 2, 2, 1, 1
Offset: 0

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Author

Keywords

Comments

The number of 3's in row n is given by 2^(A000120(n)-1) if A014081(n) is nonzero, else by 0 [Davis & Webb]. - R. J. Mathar, Jul 28 2017

Examples

			Triangle begins:
                 1
                1 1
               1 2 1
              1 3 3 1
             1 0 2 0 1
            1 1 2 2 1 1
           1 2 3 0 3 2 1
          1 3 1 3 3 1 3 1
         1 0 0 0 2 0 0 0 1
        1 1 0 0 2 2 0 0 1 1
       1 2 1 0 2 0 2 0 1 2 1
      1 3 3 1 2 2 2 2 1 3 3 1
  ...
		

Crossrefs

Cf. A007318, A047999, A083093, A034930, A008975, A034932, A163000 (# 2's), A270438 (# 1's), A249732 (# 0's).
Sequences based on the triangles formed by reading Pascal's triangle mod m: A047999 (m = 2), A083093 (m = 3), (this sequence) (m = 4), A095140 (m = 5), A095141 (m = 6), A095142 (m = 7), A034930(m = 8), A095143 (m = 9), A008975 (m = 10), A095144 (m = 11), A095145 (m = 12), A275198 (m = 14), A034932 (m = 16).

Programs

  • Haskell
    a034931 n k = a034931_tabl !! n !! k
    a034931_row n = a034931_tabl !! n
    a034931_tabl = iterate
       (\ws -> zipWith ((flip mod 4 .) . (+)) ([0] ++ ws) (ws ++ [0])) [1]
    -- Reinhard Zumkeller, Mar 14 2015
    
  • Maple
    A034931 := proc(n,k)
        modp(binomial(n,k),4) ;
    end proc:
    seq(seq(A034931(n,k),k=0..n),n=0..10); # R. J. Mathar, Jul 28 2017
  • Mathematica
    Mod[ Flatten[ Table[ Binomial[n, k], {n, 0, 13}, {k, 0, n}]], 4] (* Robert G. Wilson v, May 26 2004 *)
  • PARI
    C(n, k)=binomial(n, k)%4 \\ Charles R Greathouse IV, Aug 09 2016
    
  • PARI
    f(n,k)=2*(bitand(n-k, k)==0);
    T(n,j)=if(j==0,return(1)); my(k=logint(n,2),K=2^k,K1=K/2,L=n-K); if(LCharles R Greathouse IV, Aug 11 2016
    
  • Python
    from math import isqrt, comb
    def A034931(n):
        g = (m:=isqrt(f:=n+1<<1))-(f<=m*(m+1))
        k = n-comb(g+1,2)
        if k.bit_count()+(g-k).bit_count()-g.bit_count()>1: return 0
        s, c, d = bin(g)[2:], 1, 0
        w = (bin(k)[2:]).zfill(l:=len(s))
        for i in range(0,l-1):
            r, t = s[i:i+2], w[i:i+2]
            if (x:=int(r,2)) < (y:=int(t,2)):
                d += (t[0]>r[0])+(t[1]>r[1])
            else:
                c = c*comb(x,y)&3
        d -= sum(1 for i in range(1,l-1) if w[i]>s[i])
        return (c<Chai Wah Wu, Jul 19 2025

Formula

T(n+1,k) = (T(n,k) + T(n,k-1)) mod 4. - Reinhard Zumkeller, Mar 14 2015

A033264 Number of blocks of {1,0} in the binary expansion of n.

Original entry on oeis.org

0, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 1, 2, 2, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 2, 2, 2, 1, 1, 1, 2, 1, 2, 2, 2
Offset: 1

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Author

Keywords

Comments

Number of i such that d(i) < d(i-1), where Sum_{d(i)*2^i: i=0,1,....,m} is base 2 representation of n.
This is the base-2 down-variation sequence; see A297330. - Clark Kimberling, Jan 18 2017

Crossrefs

a(n) = A005811(n) - ceiling(A005811(n)/2) = A005811(n) - A069010(n).
Equals (A072219(n+1)-1)/2.
Cf. also A175047, A030308.
Essentially the same as A087116.

Programs

  • Haskell
    a033264 = f 0 . a030308_row where
       f c [] = c
       f c (0 : 1 : bs) = f (c + 1) bs
       f c (_ : bs) = f c bs
    -- Reinhard Zumkeller, Feb 20 2014, Jun 17 2012
    
  • Maple
    f:= proc(n) option remember; local k;
    k:= n mod 4;
    if k = 2 then procname((n-2)/4) + 1
    elif k = 3 then procname((n-3)/4)
    else procname((n-k)/2)
    fi
    end proc:
    f(1):= 0: f(0):= q:
    seq(f(i),i=1..100); # Robert Israel, Aug 31 2015
  • Mathematica
    Table[Count[Partition[IntegerDigits[n, 2], 2, 1], {1, 0}], {n, 102}] (* Michael De Vlieger, Aug 31 2015, after Robert G. Wilson v at A014081 *)
    Table[SequenceCount[IntegerDigits[n,2],{1,0}],{n,110}] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jan 26 2017 *)
  • PARI
    a(n) = { hammingweight(bitand(n>>1, bitneg(n))) }; \\ Gheorghe Coserea, Aug 30 2015
    
  • Python
    def A033264(n): return ((n>>1)&~n).bit_count() # Chai Wah Wu, Jun 25 2025

Formula

G.f.: 1/(1-x) * Sum_(k>=0, t^2/(1+t)/(1+t^2), t=x^2^k). - Ralf Stephan, Sep 10 2003
a(n) = A069010(n) - (n mod 2). - Ralf Stephan, Sep 10 2003
a(4n) = a(4n+1) = a(2n), a(4n+2) = a(n)+1, a(4n+3) = a(n). - Ralf Stephan, Aug 20 2003
a(n) = A087116(n) for n > 0, since strings of 0's alternate with strings of 1's, which end in (1,0). - Jonathan Sondow, Jan 17 2016
Sum_{n>=1} a(n)/(n*(n+1)) = Pi/4 - log(2)/2 (A196521) (Allouche and Shallit, 1990). - Amiram Eldar, Jun 01 2021

A037800 Number of occurrences of 01 in the binary expansion of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 1, 2, 2, 2, 1, 2, 1, 1, 0, 1, 1, 1, 1, 2, 1
Offset: 0

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Author

Keywords

Comments

Number of i such that d(i)>d(i-1), where Sum{d(i)*2^i: i=0,1,...,m} is base 2 representation of n.
This is the base-2 up-variation sequence; see A297330. - Clark Kimberling, Jan 18 2017

Crossrefs

Programs

  • Haskell
    a037800 = f 0 . a030308_row where
       f c [_]          = c
       f c (1 : 0 : bs) = f (c + 1) bs
       f c (_ : bs)     = f c bs
    -- Reinhard Zumkeller, Feb 20 2014
    
  • Mathematica
    Table[SequenceCount[IntegerDigits[n,2],{0,1}],{n,0,120}] (* Harvey P. Dale, Aug 10 2023 *)
  • PARI
    a(n) = { if(n == 0, 0, -1 + hammingweight(bitnegimply(n, n>>1))) };  \\ Gheorghe Coserea, Aug 31 2015

Formula

a(2n) = a(n), a(2n+1) = a(n) + [n is even]. - Ralf Stephan, Aug 21 2003
G.f.: 1/(1-x) * Sum_{k>=0} t^5/(1+t)/(1+t^2) where t=x^2^k. - Ralf Stephan, Sep 10 2003
a(n) = A069010(n) - 1, n>0. - Ralf Stephan, Sep 10 2003
Sum_{n>=1} a(n)/(n*(n+1)) = log(2)/2 + Pi/4 - 1 = A231902 - 1 (Allouche and Shallit, 1990). - Amiram Eldar, Jun 01 2021

A014082 Number of occurrences of '111' in binary expansion of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 3, 4, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 1, 0
Offset: 0

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Author

Keywords

Comments

a(n) = A213629(n,7) for n > 6. - Reinhard Zumkeller, Jun 17 2012

Crossrefs

Programs

  • Haskell
    import Data.List (tails, isPrefixOf)
    a014082 = sum . map (fromEnum . ([1,1,1] `isPrefixOf`)) .
                        tails . a030308_row
    -- Reinhard Zumkeller, Jun 17 2012
    
  • Maple
    See A014081.
    f:= proc(n) option remember;
      if n::even then procname(n/2)
      elif n mod 8 = 7 then 1 + procname((n-1)/2)
      else procname((n-1)/2)
    fi
    end proc:
    f(0):= 0:
    map(f, [$0..1000]); # Robert Israel, Sep 11 2015
  • Mathematica
    f[n_] := Count[ Partition[ IntegerDigits[n, 2], 3, 1], {1, 1, 1}]; Table[f@n, {n, 0, 104}] (* Robert G. Wilson v, Apr 02 2009 *)
    a[0] = a[1] = 0; a[n_] := a[n] = If[EvenQ[n], a[n/2], a[(n - 1)/2] + Boole[Mod[(n - 1)/2, 4] == 3]]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Oct 22 2012, after Ralf Stephan *)
    Table[SequenceCount[IntegerDigits[n,2],{1,1,1},Overlaps->True],{n,0,110}] (* Harvey P. Dale, Mar 05 2023 *)
  • PARI
    a(n) = hammingweight(bitand(n, bitand(n>>1, n>>2))); \\ Gheorghe Coserea, Aug 30 2015

Formula

a(2n) = a(n), a(2n+1) = a(n) + [n congruent to 3 mod 4]. - Ralf Stephan, Aug 21 2003
G.f.: 1/(1-x) * Sum_{k>=0} t^7(1-t)/(1-t^8), where t=x^2^k. - Ralf Stephan, Sep 08 2003

A213629 In binary representation: T(n,k) = number of (possibly overlapping) occurrences of k in n, triangle read by rows, 1<=k<=n.

Original entry on oeis.org

1, 1, 1, 2, 0, 1, 1, 1, 0, 1, 2, 1, 0, 0, 1, 2, 1, 1, 0, 0, 1, 3, 0, 2, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 2, 1, 0, 1, 0, 0, 0, 0, 1, 2, 2, 0, 0, 1, 0, 0, 0, 0, 1, 3, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 2, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 3, 1, 1, 0, 1, 1, 0, 0
Offset: 1

Views

Author

Reinhard Zumkeller, Jun 17 2012

Keywords

Comments

The definition is based on the definition of pattern functions in the paper of Allouche and Shallit;
sum of n-th row = A029931(n);
T(n,1) = A000120(n);
T(n,2) = A033264(n) for n > 1;
T(n,3) = A014081(n) for n > 2;
T(n,4) = A056978(n) for n > 3;
T(n,5) = A056979(n) for n > 4;
T(n,6) = A056980(n) for n > 5;
T(n,7) = A014082(n) for n > 6;
T(n,k) = 0 for k with floor(n/2) < k < n;
T(n,n) = 1;
A122953(n) = Sum_{k=1..n} A057427(T(n,k));
A005811(n) = T(n,1) + T(n,2) - T(n,3);
A007302(n) = A000120(n) - sum (A213629(n,A136412(k))).

Examples

			The triangle begins:
.   1:                        1
.   2:                      1   1
.   3:                    2   0   1
.   4:                  1   1   0   1
.   5:                2   1   0   0   1
.   6:              2   1   1   0   0   1
.   7:            3   0   2   0   0   0   1
.   8:          1   1   0   1   0   0   0   1
.   9:        2   1   0   1   0   0   0   0   1
.  10:      2   2   0   0   1   0   0   0   0   1
.  11:    3   1   1   0   1   0   0   0   0   0   1
.  12:  2   1   1   1   0   1   0   0   0   0   0   1.
		

Crossrefs

Programs

  • Haskell
    import Data.List (inits, tails, isPrefixOf)
    a213629 n k = a213629_tabl !! (n-1) !! (k-1)
    a213629_row n = a213629_tabl !! (n-1)
    a213629_tabl = map f $ tail $ inits $ tail $ map reverse a030308_tabf where
       f xss = map (\xs ->
               sum $ map (fromEnum . (xs `isPrefixOf`)) $ tails $ last xss) xss
  • Mathematica
    t[n_, k_] := (idn = IntegerDigits[n, 2]; idk = IntegerDigits[k, 2]; ln = Length[idn]; lk = Length[idk]; For[cnt = 0; i = 1, i <= ln - lk + 1, i++, If[idn[[i ;; i + lk - 1]] == idk, cnt++]]; cnt); Table[t[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Oct 22 2012 *)

A257250 Numbers n for which A256999(n) = n; numbers that cannot be made any larger by rotating (by one or more steps) the non-msb bits of their binary representation (with A080541 or A080542).

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 24, 26, 28, 30, 31, 32, 48, 52, 56, 58, 60, 62, 63, 64, 96, 100, 104, 106, 112, 114, 116, 118, 120, 122, 124, 126, 127, 128, 192, 200, 208, 212, 224, 226, 228, 232, 234, 236, 240, 242, 244, 246, 248, 250, 252, 254, 255, 256, 384, 392, 400, 416, 420, 424, 426, 448, 450
Offset: 0

Views

Author

Antti Karttunen, May 16 2015

Keywords

Comments

These correspond to the maximal (lexicographically largest) representatives selected from each equivalence class of binary necklaces. See the last example.
Indexing starts from zero, because a(0) = 0 is a special case.
If k is a member then so also is 2*k, i.e., k with 0 appended to the end of its binary representation.
If k is a member then so also is A004755(k), i.e., k with 1 prepended to the front of its binary representation.
One obtains A065609 if one erases the most significant bit of each term [as A053645(a(n))] and then discards any zero-terms produced from the terms that originally were powers of two (A000079).
First differs from A328607 in lacking 108, with binary expansion 1101100. If we define a dual-necklace to be a finite sequence that is lexicographically maximal (not minimal) among all of its cyclic rotations, these are numbers whose binary expansion, without the most significant digit, is a dual-necklace. - Gus Wiseman, Nov 04 2019

Examples

			For n = 5, with binary representation "101", if we rotate other bits than the most significant bit (that is, only the two rightmost digits "01") one step to either direction, we get "110" = 6 > 5, so 5 can be made larger by such rotations, and thus is NOT included in this sequence.
For n = 6, with binary representation "110", no such rotation will yield a larger number, and thus 6 is included in this sequence.
For n = 28, with binary representation "11100", if we rotate non-msb bits towards right, we get additional numbers 22, 19 and 25 (with binary representations "10110", "10011", "11001") before coming to 28 again, and 28 is the largest of these numbers, thus 28 is included in this sequence.
  Also, if we discard the most significant bit of each and consider them just as binary strings, then A053645(28) = 12 is the lexicographically largest representative of {"1100", "0110", "0011", "1001"}, which is the complete set of representatives for a particular equivalence class of binary necklaces, obtained by rotating all bits of binary string "1100" successively towards right or left.
		

Crossrefs

Complement: A257739.
Odd terms: A000225.
Subsequence of A065609.
Subsequence: A258003.
The non-dual version is A328668.
The version involving all digits is A065609.
The non-dual reversed version is A328607.
Numbers whose reversed binary expansion is a necklace are A328595.
Binary necklaces are A000031.
Necklace compositions are A008965.

Programs

  • Mathematica
    reckQ[q_]:=Array[OrderedQ[{RotateRight[q,#],q}]&,Length[q]-1,1,And];
    Select[Range[0,110],#<=1||reckQ[Rest[IntegerDigits[#,2]]]&] (* Gus Wiseman, Nov 04 2019 *)

A328593 Numbers whose binary indices have no consecutive divisible parts.

Original entry on oeis.org

0, 1, 2, 4, 6, 8, 12, 14, 16, 18, 20, 22, 24, 28, 30, 32, 40, 44, 46, 48, 50, 52, 54, 56, 60, 62, 64, 66, 68, 70, 72, 76, 78, 80, 82, 84, 86, 88, 92, 94, 96, 104, 108, 110, 112, 114, 116, 118, 120, 124, 126, 128, 132, 134, 144, 146, 148, 150, 152, 156, 158, 160
Offset: 1

Views

Author

Gus Wiseman, Oct 21 2019

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

Examples

			The sequence of terms together with their binary expansions and binary indices begins:
   0:      0 ~ {}
   1:      1 ~ {1}
   2:     10 ~ {2}
   4:    100 ~ {3}
   6:    110 ~ {2,3}
   8:   1000 ~ {4}
  12:   1100 ~ {3,4}
  14:   1110 ~ {2,3,4}
  16:  10000 ~ {5}
  18:  10010 ~ {2,5}
  20:  10100 ~ {3,5}
  22:  10110 ~ {2,3,5}
  24:  11000 ~ {4,5}
  28:  11100 ~ {3,4,5}
  30:  11110 ~ {2,3,4,5}
  32: 100000 ~ {6}
  40: 101000 ~ {4,6}
  44: 101100 ~ {3,4,6}
  46: 101110 ~ {2,3,4,6}
  48: 110000 ~ {5,6}
  50: 110010 ~ {2,5,6}
		

Crossrefs

The version for prime indices is A328603.
Numbers with no successive binary indices are A003714.
Partitions with no consecutive divisible parts are A328171.
Compositions without consecutive divisible parts are A328460.

Programs

  • Mathematica
    Select[Range[0,100],!MatchQ[Join@@Position[Reverse[IntegerDigits[#,2]],1],{_,x_,y_,_}/;Divisible[y,x]]&]
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