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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A038189 Bit to left of least significant 1-bit in binary expansion of n.

Original entry on oeis.org

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

Views

Author

Fred Lunnon, Dec 11 1999

Keywords

Comments

Characteristic function of A091067.
Image, under the coding i -> floor(i/2), of the fixed point, starting with 0, of the morphism 0 -> 01, 1 -> 02, 2 -> 32, 3 -> 31. - Jeffrey Shallit, May 15 2016
Restricted to the positive integers, completely additive modulo 2. - Peter Munn, Jun 20 2022
If a(n) is defined as 1-a(-n) for all n<0, then a(n) = a(2*n), a(4*n+1) = 0, a(4*n+3) = 1 for all n in Z. - Michael Somos, Oct 05 2024

Examples

			a(6) = 1 since 6 = 110 and bit before rightmost 1 is a 1.
		

References

  • Jean-Paul Allouche and Jeffrey O. Shallit, Automatic sequences, Cambridge, 2003, sect. 5.1.6

Crossrefs

A014707(n)=a(n+1). A014577(n)=1-a(n+1).
The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410. - N. J. A. Sloane, Jul 27 2012
Related sequences A301848, A301849, A301850. - Fred Lunnon, Mar 27 2018

Programs

  • C
    int a(int n) { return (n & ((n&-n)<<1)) ? 1 : 0; } /* from Russ Cox */
    
  • Magma
    function a (n)
      if n eq 0 then return 0; // alternatively,  return 1;
      else while IsEven(n) do n := n div 2; end while; end if;
      return n div 2 mod 2; end function;
      nlo := 0; nhi := 32;
      [a(n) : n in [nlo..nhi] ]; //  Fred Lunnon, Mar 27 2018
    
  • Maple
    A038189 := proc(n)
        option remember;
        if n = 0 then
            0 ;
        elif type(n,'even') then
            procname(n/2) ;
        elif modp(n,4) = 1 then
            0 ;
        else
            1 ;
        end if;
    end proc:
    seq(A038189(n),n=0..100) ; # R. J. Mathar, Mar 30 2018
  • Mathematica
    f[n_] := Block[{id2 = Join[{0}, IntegerDigits[n, 2]]}, While[ id2[[-1]] == 0, id2 = Most@ id2]; id2[[-2]]]; f[0] = 0; Array[f, 105, 0] (* Robert G. Wilson v, Apr 14 2009 and fixed Feb 27 2014 *)
    f[n_] := f[n] = Switch[Mod[n, 4], 0, f[n/2], 1, 0, 2, f[n/2], 3, 1]; f[0] = 0; Array[f, 105, 0] (* Robert G. Wilson v, Apr 14 2009, fixed Feb 27 2014 *)
    a[ n_] := If[n==0, 0, Mod[(n/2^IntegerExponent[n, 2]-1)/2, 2]]; (* Michael Somos, Oct 05 2024 *)
  • PARI
    {a(n) = if(n==0, 0, ((n/2^valuation(n,2)-1)/2)%2)}; /* Michael Somos, Sep 22 2005 */
    
  • PARI
    a(n) = if(n<3, 0, prod(m=1,n, kronecker(-n,m)==kronecker(m,n))) /* Michael Somos, Sep 22 2005 */
    
  • PARI
    A038189(n)=bittest(n,valuation(n,2)+1) \\ M. F. Hasler, Aug 06 2015
    
  • PARI
    a(n)=my(h=bitand(n,-n));n=bitand(n,h<<1);n!=0; \\ Joerg Arndt, Apr 09 2021
    
  • Python
    def A038189(n):
        s = bin(n)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return int(s[m-i-2]) if m-i-2 >= 0 else 0 # Chai Wah Wu, Apr 08 2021
    
  • Python
    def a(n): return 1 if (n & ((n&-n)<<1)) else 0
    print([a(n) for n in range(108)]) # Michael S. Branicky, Feb 06 2025 after Russ Cox

Formula

a(0) = 0, a(2*n) = a(n) for n>0, a(4*n+1) = 0, a(4*n+3) = 1.
G.f.: Sum_{k>=0} t^3/(1-t^4), where t=x^2^k. Parity of A025480. For n >= 1, a(n) = 1/2 * (1 - (-1)^A025480(n-1)). - Ralf Stephan, Jan 04 2004 [index adjusted by Peter Munn, Jun 22 2022]
a(n) = 1 if Kronecker(-n,m)=Kronecker(m,n) for all m, otherwise a(n)=0. - Michael Somos, Sep 22 2005
a(n) = 1 iff A164677(n) < 0. - M. F. Hasler, Aug 06 2015
For n >= 1, a(n) = A065339(n) mod 2. - Peter Munn, Jun 20 2022
From A.H.M. Smeets, Mar 08 2023: (Start)
a(n+1) = 1 - A014577(n) for n >= 0.
a(n+1) = 2 - A014710(n) for n >= 0.
a(n) = (1 - A034947(n))/2 for n > 0. (End)
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/2. - Amiram Eldar, Aug 30 2024

Extensions

More terms from David W. Wilson
Definition corrected by Russ Cox and Ralf Stephan, Nov 08 2004

A103391 "Even" fractal sequence for the natural numbers: Deleting every even-indexed term results in the same sequence.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 3, 5, 2, 6, 4, 7, 3, 8, 5, 9, 2, 10, 6, 11, 4, 12, 7, 13, 3, 14, 8, 15, 5, 16, 9, 17, 2, 18, 10, 19, 6, 20, 11, 21, 4, 22, 12, 23, 7, 24, 13, 25, 3, 26, 14, 27, 8, 28, 15, 29, 5, 30, 16, 31, 9, 32, 17, 33, 2, 34, 18, 35, 10, 36, 19, 37, 6, 38, 20, 39, 11, 40, 21, 41, 4, 42, 22, 43, 12, 44, 23, 45, 7, 46, 24, 47, 13, 48, 25, 49, 3, 50, 26, 51, 14, 52, 27, 53, 8
Offset: 1

Views

Author

Eric Rowland, Mar 20 2005

Keywords

Comments

A003602 is the "odd" fractal sequence for the natural numbers.
Lexicographically earliest infinite sequence such that a(i) = a(j) => A348717(A005940(i)) = A348717(A005940(j)) for all i, j >= 1. A365718 is an analogous sequence related to A356867 (Doudna variant D(3)). - Antti Karttunen, Sep 17 2023

Crossrefs

Cf. A003602, A005940, A025480, A220466, A286387, A353368 (Dirichlet inverse).
Cf. also A110962, A110963, A365718.
Differs from A331743(n-1) for the first time at n=192, where a(192) = 97, while A331743(191) = 23.
Differs from A351460.

Programs

  • Haskell
    -- import Data.List (transpose)
    a103391 n = a103391_list !! (n-1)
    a103391_list = 1 : ks where
       ks = concat $ transpose [[2..], ks]
    -- Reinhard Zumkeller, May 23 2013
    
  • Maple
    nmax := 82: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 2 to ceil(nmax/(p+2))+1 do a((2*n-3)*2^p+1) := n od: od: a(1) := 1: seq(a(n), n=1..nmax); # Johannes W. Meijer, Jan 28 2013
  • Mathematica
    a[n_] := ((n-1)/2^IntegerExponent[n-1, 2] + 3)/2; a[1] = 1; Array[a, 100] (* Amiram Eldar, Sep 24 2023 *)
  • PARI
    A003602(n) = (n/2^valuation(n, 2)+1)/2; \\ From A003602
    A103391(n) = if(1==n,1,(1+A003602(n-1))); \\ Antti Karttunen, Feb 05 2020
    
  • Python
    def v(n): b = bin(n); return len(b) - len(b.rstrip("0"))
    def b(n): return (n//2**v(n)+1)//2
    def a(n): return 1 if n == 1 else 1 + b(n-1)
    print([a(n) for n in range(1, 106)]) # Michael S. Branicky, May 29 2022
    
  • Python
    def A103391(n): return (n-1>>(n-1&-n+1).bit_length())+2 if n>1 else 1 # Chai Wah Wu, Jan 04 2024

Formula

For n > 1, a(n) = A003602(n-1) + 1. - Benoit Cloitre, May 26 2007, indexing corrected by Antti Karttunen, Feb 05 2020
a((2*n-3)*2^p+1) = n, p >= 0 and n >= 2, with a(1) = 1. - Johannes W. Meijer, Jan 28 2013
Sum_{k=1..n} a(k) ~ n^2/6. - Amiram Eldar, Sep 24 2023

Extensions

Data section extended up to a(105) (to better differentiate from several nearby sequences) by Antti Karttunen, Feb 05 2020

A087117 Number of zeros in the longest string of consecutive zeros in the binary representation of n.

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, Aug 14 2003

Keywords

Comments

The following four statements are equivalent: a(n) = 0; n = 2^k - 1 for some k > 0; A087116(n) = 0; A023416(n) = 0.
The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. Then a(k) is the maximum part of this composition, minus one. The maximum part is A333766(k). - Gus Wiseman, Apr 09 2020

Crossrefs

Positions of zeros are A000225.
Positions of terms <= 1 are A003754.
Positions of terms > 0 are A062289.
Positions of first appearances are A131577.
The version for prime indices is A252735.
The proper maximum is A333766.
The version for minimum is A333767.
Maximum prime index is A061395.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Runs are counted by A124767.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Runs-resistance is A333628.
- Weakly decreasing compositions are A114994.
- Weakly increasing compositions are A225620.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.

Programs

  • Haskell
    import Data.List (unfoldr, group)
    a087117 0       = 1
    a087117 n
      | null $ zs n = 0
      | otherwise   = maximum $ map length $ zs n where
      zs = filter ((== 0) . head) . group .
           unfoldr (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2)
    -- Reinhard Zumkeller, May 01 2012
    
  • Maple
    A087117 := proc(n)
        local d,l,zlen ;
        if n = 0 then
            return 1 ;
        end if;
        d := convert(n,base,2) ;
        for l from nops(d)-1 to 0 by -1 do
            zlen := [seq(0,i=1..l)] ;
            if verify(zlen,d,'sublist') then
                return l ;
            end if;
        end do:
        return 0 ;
    end proc; # R. J. Mathar, Nov 05 2012
  • Mathematica
    nz[n_]:=Max[Length/@Select[Split[IntegerDigits[n,2]],MemberQ[#,0]&]]; Array[nz,110,0]/.-\[Infinity]->0 (* Harvey P. Dale, Sep 05 2017 *)
  • PARI
    h(n)=if(n<2, return(0)); my(k=valuation(n,2)); if(k, max(h(n>>k), k), n++; n>>=valuation(n,2); h(n-1))
    a(n)=if(n,h(n),1) \\ Charles R Greathouse IV, Apr 06 2022

Formula

a(n) = max(A007814(n), a(A025480(n-1))) for n >= 2. - Robert Israel, Feb 19 2017
a(2n+1) = a(n) (n>=1); indeed, the binary form of 2n+1 consists of the binary form of n with an additional 1 at the end - Emeric Deutsch, Aug 18 2017
For n > 0, a(n) = A333766(n) - 1. - Gus Wiseman, Apr 09 2020

A071585 Numerator of the continued fraction expansion whose terms are the first-order differences of exponents in the binary representation of 4*n, with the exponents of 2 being listed in descending order.

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8, 6, 9, 10, 11, 9, 12, 11, 13, 6, 9, 10, 11, 9, 12, 11, 13, 7, 11, 13, 14, 13, 17, 15, 18, 11, 16, 17, 19, 14, 19, 18, 21, 7, 11, 13, 14, 13, 17, 15, 18, 11, 16, 17, 19, 14, 19, 18, 21, 8, 13, 16, 17, 17, 22, 19, 23, 16, 23, 24, 27, 19
Offset: 0

Views

Author

Paul D. Hanna, Jun 01 2002

Keywords

Comments

Thus a(n)/a(m) = d_1 + 1/(d_2 + 1/(d_3 + ... + 1/d_k)) where m = n - 2^floor(log_2(n)) + 1 and where d_j = b_j - b_(j+1) are the differences of the binary exponents b_j > b_(j+1) defined by: 4*n = 2^b_1 + 2^b_2 + 2^b_3 + ... 2^b_k.
All the rationals are uniquely represented by this sequence - compare Stern's diatomic sequence A002487.
This sequence lists the rationals >= 1 in order by the sum of the terms of their continued fraction expansions. For example, the numerators generated from partitions of 5 that do not end with 1 are listed together as 5, 7, 7, 8, 5, 7, 7, 8, since: 5/1 = [5]; 7/2 = [3;2]; 7/3 = [2;3]; 8/3 = [2;1,2]; 5/4 = [1;4]; 7/5 = [1;2,2]; 7/4 = [1;1,3]; 8/5 = [1;1,1,2].
From Yosu Yurramendi, Jun 23 2014: (Start)
If the terms (n>0) are written as an array:
1,
2,
3, 3,
4, 5, 4, 5,
5, 7, 7, 8, 5, 7, 7, 8,
6, 9,10,11, 9,12,11,13, 6, 9,10,11, 9,12,11,13,
7,11,13,14,13,17,15,18,11,16,17,19,14,19,18,21,7,11,13,14,13,17,15,18,11, ...
then the sum of the k-th row is 2*3^(k-2) for k>1, each column is an arithmetic progression. The differences of the arithmetic sequences give the sequence A071585 itself: a(2^(p+1)+k) - a(2^p+k) = a(k). A002487 and A007306 also have these properties. The first terms of columns, excluding a(0), give A086593.
If the rows (n>0) are written on right:
1;
2;
3, 3;
4, 5, 4, 5;
5, 7, 7, 8, 5, 7, 7, 8;
6, 9, 10, 11, 9, 12, 11, 13, 6, 9, 10, 11, 9, 12, 11, 13;
then each column is a Fibonacci sequence: a(2^(p+2)+k) = a(2^(p+1)+k) + a(2^p+k). The first terms of columns, excluding a(0), give A086593. (End)
n>1 occurs in this sequence phi(n) = A000010(n) times, as it occurs in A007306 (Franklin T. Adams-Watters's comment), that is the sequence obtained by adding numerator and denominator in the Calkin-Wilf enumeration system of positive rationals. A229742(n)/A071766(n) is also an enumeration system of all positive rationals (HCS system), and in each level m >= 0 (ranks between 2^m and 2^(m+1)-1) rationals are the same in both systems. Thus a(n) (A229742(n)+A071766(n)) has the same terms in each level as A007306. The same property occurs in all numerator+denominator sequences of enumeration systems of positive rationals, as, for example, A007306 (A007305+A047679), A086592 (A020650+A020651), and A268087 (A162909+A162910). - Yosu Yurramendi, Apr 06 2016
a(n) = A086592(A059893(n)), a(A059893(n)) = A086592(n), n > 0. - Yosu Yurramendi, May 30 2017

Examples

			a(37)=17 as it is the numerator of 17/5 = 3 + 1/(2 + 1/2), which is a continued fraction that can be derived from the binary expansion of 4*37 = 2^7 + 2^4 + 2^2; the binary exponents are {7, 4, 2}, thus the differences of these exponents are {3, 2, 2}; giving the continued fraction expansion of 17/5=[3,2,2].
Illustration of Sum_{m=0..2^(k-1)-1} a(2^k + m) = 3^k:
k=2: 3^2 = a(2^2) + a(2^2 + 1) = 4 + 5;
k=3: 3^3 = a(2^3) + a(2^3 + 1) + a(2^3 + 2) + a(2^3 + 3) = 5 + 7 + 7 + 8;
k=4: 3^4 = a(2^4) + a(2^4+1) + a(2^4+2) + a(2^4+3) + a(2^4+4) + a(2^4+5) + a(2^4+6) + a(2^4+7) = 6 + 9 + 10 + 11 + 9 + 12 + 11 + 13.
1, 2, 3, 3/2, 4, 5/2, 4/3, 5/3, 5, 7/2, 7/3, 8/3, 5/4, 7/5, 7/4, 8/5, 6, ...
From _Yosu Yurramendi_, Jun 27 2014: (Start)
a(0) =             = 1;
a(1) = a(0) + a(0) = 2;
a(2) = a(0) + a(1) = 3;
a(3) = a(1) + a(0) = 3;
a(4) = a(0) + a(2) = 4;
a(5) = a(1) + a(3) = 5;
a(6) = a(2) + a(0) = 4;
a(7) = a(3) + a(1) = 5;
a(8) = a(0) + a(4) = 5;
a(9) = a(1) + a(5) = 7;
a(10) = a(2) + a(6) = 7;
a(11) = a(3) + a(7) = 8;
a(12) = a(4) + a(0) = 5;
a(13) = a(5) + a(1) = 7;
a(14) = a(6) + a(2) = 7;
a(15) = a(7) + a(3) = 8. (End)
		

Crossrefs

Cf. A071766.

Programs

  • Mathematica
    ncf[n_]:=Module[{br=Reverse[Flatten[Position[Reverse[IntegerDigits[4 n,2]],1]-1]]}, Numerator[FromContinuedFraction[Flatten[Join[{Abs[ Differences[ br]],Last[br]}]]]]]; Join[{1},Array[ncf,80]] (* Harvey P. Dale, Jul 01 2012 *)
    {1}~Join~Table[Numerator@ FromContinuedFraction@ Append[Abs@ Differences@ #, Last@ #] &@ Log2[NumberExpand[4 n, 2] /. 0 -> Nothing], {n, 120}] (* Version 11, or *)
    {1}~Join~Table[Numerator@ FromContinuedFraction@ Append[Abs@ Differences@ #, Last@ #] &@ Log2@ DeleteCases[# Reverse[2^Range[0, Length@ # - 1]] &@ IntegerDigits[4 n, 2], k_ /; k == 0], {n, 120}] (* Michael De Vlieger, Aug 15 2016 *)
  • R
    blocklevel <- 7  # arbitrary
    a <- c(1,2)
    for(k in 1:blocklevel)
    a <- c(a, a + c(a[((length(a)/2)+1):length(a)],a[1:(length(a)/2)]))
    a
    # Yosu Yurramendi, Jun 26 2014
    
  • R
    blocklevel <- 7  # arbitrary
    a <- c(1,2)
    for(p in 0:blocklevel)
      for(k in 1:2^(p+1)){
        if (k <=  2^p) a[k + 2^(p+1)] = a[k] + a[k + 2^p]
        else           a[k + 2^(p+1)] = a[k] + a[k - 2^p]
    }
    a
    # Yosu Yurramendi, Jun 27 2014

Formula

a(2^k + 2^j + m) = (k-j)*a(2^j + m) + a(m) when 2^k > 2^j > m >= 0.
a(0) = 1, a(2^k) = k + 2,
a(2^k + 1) = 2*k + 1 (k>0),
a(2^k + 2) = 3*k - 2 (k>1),
a(2^k + 3) = 3*k - 1 (k>1),
a(2^k + 4) = 4*k - 7 (k>2).
a(2^k - 1) = Fibonacci(k+2) = A000045(k+2).
Sum_{m=0..2^(k-1)-1} a(2^k + m) = 3^k (k>0).
From Yosu Yurramendi, Jun 27 2014: (Start)
Write n = k + 2^(m+1), k = 0,1,2,...,2^(m+1)-1, m = 0,1,2,...
if 0 <= k < 2^m, a(k + 2^(m+1)) = a(k) + a(k + 2^m).
if 2^m <= k < 2^(m+1), a(k + 2^(m+1)) = a(k) + a(k - 2^m).
with a(0)=1, a(1)=2. (End)
a(n) = A059893(A086592(n)), n>0. - Yosu Yurramendi, Apr 09 2016
a(n) = A093873(n) + A093875(n), n > 0. - Yosu Yurramendi, Jul 22 2016
a(n) = A093873(2n) + A093873(2n+1), n > 0; a(n) = A093875(2n) = A093875(2n+1), n > 0. - Yosu Yurramendi, Jul 25 2016
a(n) = sqrt(A071766(2^(m+1)+n)*A229742(2^(m+1)+n) - A071766(2^m+n)*A229742(2^m+n)), for n > 0, where m = floor(log_2(n)+1). - Yosu Yurramendi, Jun 10 2019
a(n) = A007306(A059893(A233279(n))), n > 0. - Yosu Yurramendi, Aug 07 2021
a(n) = A007306(A059894(A006068(n))), n > 0. - Yosu Yurramendi, Sep 29 2021
Conjecture: a(n) = a(floor(n/2)) + Sum_{k=1..A000120(n)} a(b(n, k))*(-1)^(k-1) for n > 0 with a(0) = 1 where b(n, k) = A025480(b(n, k-1) - 1) for n > 0, k > 0 with b(n, 0) = n. - Mikhail Kurkov, Feb 20 2023

A111528 Square table, read by antidiagonals, where the g.f. for row n+1 is generated by: x*R_{n+1}(x) = (1+n*x - 1/R_n(x))/(n+1) with R_0(x) = Sum_{n>=0} n!*x^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 6, 1, 1, 4, 13, 24, 1, 1, 5, 22, 71, 120, 1, 1, 6, 33, 148, 461, 720, 1, 1, 7, 46, 261, 1156, 3447, 5040, 1, 1, 8, 61, 416, 2361, 10192, 29093, 40320, 1, 1, 9, 78, 619, 4256, 23805, 99688, 273343, 362880, 1, 1, 10, 97, 876, 7045, 48096, 263313
Offset: 0

Views

Author

Paul D. Hanna, Aug 06 2005

Keywords

Examples

			Table begins:
  1, 1,  2,   6,   24,   120,    720,    5040,     40320, ...
  1, 1,  3,  13,   71,   461,   3447,   29093,    273343, ...
  1, 1,  4,  22,  148,  1156,  10192,   99688,   1069168, ...
  1, 1,  5,  33,  261,  2361,  23805,  263313,   3161781, ...
  1, 1,  6,  46,  416,  4256,  48096,  591536,   7840576, ...
  1, 1,  7,  61,  619,  7045,  87955, 1187845,  17192275, ...
  1, 1,  8,  78,  876, 10956, 149472, 2195208,  34398288, ...
  1, 1,  9,  97, 1193, 16241, 240057, 3804353,  64092553, ...
  1, 1, 10, 118, 1576, 23176, 368560, 6262768, 112784896, ...
Rows are generated by logarithms of factorial series:
log(1 + x + 2*x^2 + 6*x^3 + 24*x^4 + ... n!*x^n + ...) = x + (3/2)*x^2 + (13/3)*x^3 + (71/4)*x^4 + (461/5)*x^5 + ...
(1/2)*log(1 + 2*x + 6*x^2 + ... + ((n+1)!/1!)*x^n + ...) = x + (4/2)*x^2 + (22/3)*x^3 + (148/4)*x^4 + (1156/5)*x^5 + ...
(1/3)*log(1 + 3*x + 12*x^2 + 60*x^3 + ... + ((n+2)!/2!)*x^n + ...) = x + (5/2)*x^2 + (33/3)*x^3 + (261/4)*x^4 + (2361/5)*x^5 +...
G.f. of row n may be expressed by the continued fraction:
R_n(x) = 1/(1+n*x - (n+1)*x/(1+(n+1)*x - (n+2)*x/(1+(n+2)*x -...
or recursively by: R_n(x) = 1/(1+n*x - (n+1)*x*R_{n+1}(x)).
		

Crossrefs

Cf: A003319 (row 1), A111529 (row 2), A111530 (row 3), A111531 (row 4), A111532 (row 5), A111533 (row 6), A111534 (diagonal).
Similar recurrences: A124758, A243499, A284005, A329369, A341392.

Programs

  • Maple
    T := (n, k) -> coeff(series(hypergeom([n+1, 1], [], x)/hypergeom([n, 1], [], x), x, 21), x, k):
    #display as a sequence
    seq(seq(T(n-k, k), k = 0..n), n = 0..10);
    # display as a square array
    seq(print(seq(T(n, k), k = 0..10)), n = 0..10); # Peter Bala, Jul 16 2022
  • Mathematica
    T[n_, k_] := T[n, k] = Which[n < 0 || k < 0, 0, k == 0 || k == 1, 1, n == 0, k!, True, (T[n - 1, k + 1] - T[n - 1, k])/n - Sum[T[n, j]*T[n - 1, k - j], {j, 1, k - 1}]]; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 18 2018 *)
  • PARI
    {T(n,k)=if(n<0||k<0,0,if(k==0||k==1,1,if(n==0,k!, (T(n-1,k+1)-T(n-1,k))/n-sum(j=1,k-1,T(n,j)*T(n-1,k-j)))))}
    for(n=0,10,for(k=0,10,print1(T(n,k),", ")); print(""))
    
  • PARI
    {T(n,k)=if(n<0||k<0,0,if(k==0,1,if(n==0,k!, k/n*polcoeff(log(sum(m=0,k,(n-1+m)!/(n-1)!*x^m)),k))))}
    for(n=0,10,for(k=0,10,print1(T(n,k),", ")); print(""))

Formula

T(n, 0) = 1, T(0, k) = k!, otherwise for n>=1 and k>=1:
T(n, k) = (T(n-1, k+1) - T(n-1, k))/n - Sum_{j=1..k-1} T(n, j)*T(n-1, k-j).
T(n, k) = (k/n)*[x^k] log(Sum_{m=0..k} (n-1+m)!/(n-1)!*x^m).
T(n, k) = Sum_{j = 0..k} A089949(k, j)*n^(k-j). - Philippe Deléham, Aug 08 2005
R_n(x) = -((n-1)!/n)/Sum_{i>=1} (i+n-2)!*x^i, n > 0. - Vladeta Jovovic, May 06 2006
G.f. of row R may be expressed by the continued fraction: W(0), where W(k) = 1 - x*(k+1)/( x*(k+1) - 1/(1 - x*(k+1+R)/( x*(k+1+R) - 1/W(k+1) ))). - Sergei N. Gladkovskii, Aug 26 2013
Conjecture: T(n, k) = b(2^(k-1) - 1, n) for k > 0 with T(n, 0) = 1 where b(n, m) = b(floor(n/2), m) + b(floor((2n - 2^A007814(n))/2), m) + m*b(A025480(n-1), m) for n > 0 with b(0, m) = 1. - Mikhail Kurkov, Dec 16 2021
From Peter Bala, Jul 11 2022: (Start)
O.g.f. for row n, n >= 1: R(n,x) = ( Sum_{k >= 0} (n+k)!/n!*x^k )/( Sum_{k >= 0} (n-1+k)!/(n-1)!*x^k ).
R(n,x)/(1 - n*x*R(n,x)) = Sum_{k >= 0} (n+k)!/n!*x^k.
For n >= 0, R(n,x) satisfies the Riccati equation x^2*d/dx(R(n,x)) + n*x*R(n,x)^2 - (1 + (n-1)*x)*R(n,x) + 1 = 0 with R(n,0) = 1.
Apply Stokes 1982 to find that for n >= 0, R(n,x) = 1/(1 - x/(1 - (n+1)*x/(1 - 2*x/(1 - (n+2)*x/(1 - 3*x/(1 - (n+3)*x/(1 - 4*x/(1 - (n+4)*x/(1 - ...))))))))), a continued fraction of Stieltjes type. (End)

A011965 Second differences of Bell numbers.

Original entry on oeis.org

1, 2, 7, 27, 114, 523, 2589, 13744, 77821, 467767, 2972432, 19895813, 139824045, 1028804338, 7905124379, 63287544055, 526827208698, 4551453462543, 40740750631417, 377254241891064, 3608700264369193, 35613444194346451, 362161573323083920, 3790824599495473121
Offset: 0

Views

Author

Keywords

Comments

Number of partitions of n+3 with at least one singleton and with the smallest element in a singleton equal to 3. Alternatively, number of partitions of n+3 with at least one singleton and with the largest element in a singleton equal to n+1. - Olivier GERARD, Oct 29 2007
Out of the A005493(n) set partitions with a specific two elements clustered separately, number that have a different set of two elements clustered separately. - Andrey Goder (andy.goder(AT)gmail.com), Dec 17 2007

References

  • Olivier Gérard and Karol A. Penson, A budget of set partition statistics, in preparation, unpublished as of Sep 22 2011.

Crossrefs

Programs

  • Magma
    [Bell(n+2) -2*Bell(n+1) + Bell(n): n in [0..40]]; // G. C. Greubel, Jan 07 2025
    
  • Maple
    a:= n-> add((-1)^k*binomial(2,k)*combinat['bell'](n+k), k=0..2): seq(a(n), n=0..20);  # Alois P. Heinz, Sep 05 2008
  • Mathematica
    Differences[BellB[Range[0, 30]], 2] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
  • Python
    # requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
    from itertools import accumulate
    A011965_list, blist, b = [1], [1, 2], 2
    for _ in range(1000):
        blist = list(accumulate([b]+blist))
        b = blist[-1]
        A011965_list.append(blist[-3])
    # Chai Wah Wu, Sep 02 2014
    
  • Python
    # or Sagemath
    b=bell_number
    print([b(n+2) -2*b(n+1) +b(n) for n in range(41)]) # G. C. Greubel, Jan 07 2025

Formula

a(n) = A005493(n) - A005493(n-1).
E.g.f.: exp(exp(x)-1)*(exp(2*x)-exp(x)+1). - Vladeta Jovovic, Feb 11 2003
a(n) = A000110(n) - 2*A000110(n-1) + A000110(n-2). - Andrey Goder (andy.goder(AT)gmail.com), Dec 17 2007
G.f.: G(0) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k+2*x-1) - x*(2*k+1)*(2*k+3)*(2*x*k+2*x-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+3*x-1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 19 2012
G.f.: 1 - G(0) where G(k) = 1 - 1/(1-k*x-2*x)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 17 2013
G.f.: 1 - 1/x + (1-x)^2/x/(G(0)-x) where G(k) = 1 - x*(k+1)/(1 - x/G(k+1) ); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 26 2013
G.f.: G(0)*(1-1/x) where G(k) = 1 - 1/(1-x*(k+1))/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 07 2013
a(n) ~ n^2 * Bell(n) / LambertW(n)^2 * (1 - 2*LambertW(n)/n). - Vaclav Kotesovec, Jul 28 2021
Conjecture: a(n) = Sum_{k=0..2^n - 1} b(k) for n >= 0 where b(2n+1) = b(n) + b(A025480(n-1)), b(2n) = b(n - 2^f(n)) + b(2n - 2^f(n)) + b(A025480(n-1)) for n > 0 with b(0) = b(1) = 1 and where f(n) = A007814(n). Also b((4^n - 1)/3) = A141154(n+1). - Mikhail Kurkov, Jan 27 2022

A075300 Array A read by antidiagonals upwards: A(n, k) = array A054582(n,k) - 1 = 2^n*(2*k+1) - 1 with n,k >= 0.

Original entry on oeis.org

0, 1, 2, 3, 5, 4, 7, 11, 9, 6, 15, 23, 19, 13, 8, 31, 47, 39, 27, 17, 10, 63, 95, 79, 55, 35, 21, 12, 127, 191, 159, 111, 71, 43, 25, 14, 255, 383, 319, 223, 143, 87, 51, 29, 16, 511, 767, 639, 447, 287, 175, 103, 59, 33, 18, 1023, 1535, 1279, 895, 575, 351, 207, 119
Offset: 0

Views

Author

Antti Karttunen, Sep 12 2002

Keywords

Comments

From Philippe Deléham, Feb 19 2014: (Start)
A(0,k) = 2*k = A005843(k),
A(1,k) = 4*k + 1 = A016813(k),
A(2,k) = 8*k + 3 = A017101(k),
A(n,0) = A000225(n),
A(n,1) = A153893(n),
A(n,2) = A153894(n),
A(n,3) = A086224(n),
A(n,4) = A052996(n+2),
A(n,5) = A086225(n),
A(n,6) = A198274(n),
A(n,7) = A238087(n),
A(n,8) = A198275(n),
A(n,9) = A198276(n),
A(n,10) = A171389(n). (End)
A permutation of the nonnegative integers. - Alzhekeyev Ascar M, Jun 05 2016
The values in array row n, when expressed in binary, have n trailing 1-bits. - Ruud H.G. van Tol, Mar 18 2025

Examples

			The array A begins:
   0    2    4    6    8   10   12   14   16   18 ...
   1    5    9   13   17   21   25   29   33   37 ...
   3   11   19   27   35   43   51   59   67   75 ...
   7   23   39   55   71   87  103  119  135  151 ...
  15   47   79  111  143  175  207  239  271  303 ...
  31   95  159  223  287  351  415  479  543  607 ...
  ... - _Philippe Deléham_, Feb 19 2014
From _Wolfdieter Lang_, Jan 31 2019: (Start)
The triangle T begins:
   n\k   0    1    2   3   4   5   6   7  8  9 10 ...
   0:    0
   1:    1    2
   2:    3    5    4
   3:    7   11    9   6
   4:   15   23   19  13   8
   5    31   47   39  27  17  10
   6:   63   95   79  55  35  21  12
   7:  127  191  159 111  71  43  25  14
   8:  255  383  319 223 143  87  51  29 16
   9:  511  767  639 447 287 175 103  59 33 18
  10: 1023 1535 1279 895 575 351 207 119 67 37 20
  ...
T(3, 1) = 2^2*(2*1+1) - 1 = 12 - 1 = 11.  (End)
		

Crossrefs

Inverse permutation: A075301. Transpose: A075302. The X-projection is given by A007814(n+1) and the Y-projection A025480.

Programs

  • Maple
    A075300bi := (x,y) -> (2^x * (2*y + 1))-1;
    A075300 := n -> A075300bi(A025581(n), A002262(n));
    A002262 := n -> n - binomial(floor((1/2)+sqrt(2*(1+n))),2);
    A025581 := n -> binomial(1+floor((1/2)+sqrt(2*(1+n))),2) - (n+1);
  • Mathematica
    Table[(2^# (2 k + 1)) - 1 &[m - k], {m, 0, 10}, {k, 0, m}] (* Michael De Vlieger, Jun 05 2016 *)

Formula

From Wolfdieter Lang, Jan 31 2019: (Start)
Array A(n, k) = 2^n*(2*k+1) - 1, for n >= 0 and m >= 0.
The triangle is T(n, k) = A(n-k, k) = 2^(n-k)*(2*k+1) - 1, n >= 0, k=0..n.
See also A054582 after subtracting 1. (End)
From Ruud H.G. van Tol, Mar 17 2025: (Start)
A(0, k) is even. For n > 0, A(n, k) is odd and (3 * A(n, k) + 1) / 2 = A(n-1, 3*k+1).
A(n, k) = 2^n - 1 (mod 2^(n+1)) (equivalent to the comment about trailing 1-bits). (End)

A173732 a(n) = (A016957(n)/2^A007814(A016957(n)) - 1)/2, with A016957(n) = 6*n+4 and A007814(n) the 2-adic valuation of n.

Original entry on oeis.org

0, 2, 0, 5, 3, 8, 2, 11, 6, 14, 0, 17, 9, 20, 5, 23, 12, 26, 3, 29, 15, 32, 8, 35, 18, 38, 2, 41, 21, 44, 11, 47, 24, 50, 6, 53, 27, 56, 14, 59, 30, 62, 0, 65, 33, 68, 17, 71, 36, 74, 9, 77, 39, 80, 20, 83, 42, 86, 5, 89, 45, 92, 23, 95, 48, 98, 12, 101, 51, 104, 26, 107, 54, 110, 3
Offset: 0

Views

Author

Howard A. Landman, Feb 22 2010

Keywords

Comments

All positive integers eventually reach 1 in the Collatz problem iff all nonnegative integers eventually reach 0 with repeated application of this map, i.e., if for all n, the sequence n, a(n), a(a(n)), a(a(a(n))), ... eventually hits 0.
0 <= a(n) <= (3n+1)/2, with the upper bound being achieved for all odd n.
The positions of the zeros are given by A020988 = (2/3)*(4^n-1). This is because if n = (2/3)*(4^k-1), then m = 2n+1 = (1/3)*(4^(k+1)-1), and 3m+1 = 4^(k+1) is a power of 4. - Howard A. Landman, Mar 14 2010
Subsequence of A025480, a(n) = A025480(3n+1), i.e., A025480 = 0,[0],1,0,[2],1,3,[0],4,2,[5],1,6,[3],7,0,[8],4,9,[2],10,5,[11],1,12,[6],13,3,[14],... with elements of A173732 in brackets. - Paul Tarau, Mar 21 2010
A204418(a(n)) = 1. - Reinhard Zumkeller, Apr 29 2012
Original name: "A compression of the Collatz (or 3x+1) sequence considered as a map from odd numbers to odd numbers." - Michael De Vlieger, Oct 07 2019

Examples

			a(0) = 0 because 2n+1 = 1 (the first odd number), 3*1 + 1 = 4, dividing all powers of 2 out of 4 leaves 1, and (1-1)/2 = 0.
a(1) = 2 because 2n+1 = 3, 3*3 + 1 = 10, dividing all powers of 2 out of 10 leaves 5, and (5-1)/2 = 2.
		

Crossrefs

Programs

  • C
    #include  main() { int k,m,n; for (k = 0; ; k++) { n = 2*k + 1 ; m = 3*n + 1 ; while (!(m & 1)) { m >>= 1 ; } printf("%d,",((m - 1) >> 1)); } }
    
  • Haskell
    a173732 n = a173732_list !! n
    a173732_list = f $ tail a025480_list where f (x :  :  : xs) = x : f xs
    -- Reinhard Zumkeller, Apr 29 2012
    
  • Mathematica
    Array[(#/2^IntegerExponent[#, 2] - 1)/2 &[6 # + 4] &, 75, 0] (* Michael De Vlieger, Oct 06 2019 *)
  • PARI
    odd(n) = n >> valuation(n, 2);
    a(n) = (odd(6*n+4) - 1)/2; \\ Amiram Eldar, Aug 26 2024

Formula

From Amiram Eldar, Aug 26 2024: (Start)
a(n) = (A075677(n+1) - 1)/2.
Sum_{k=1..n} a(k) ~ n^2 / 2. (End)

Extensions

Name changed by Michael De Vlieger, Oct 07 2019

A090365 Shifts 1 place left under the INVERT transform of the BINOMIAL transform of this sequence.

Original entry on oeis.org

1, 1, 3, 11, 47, 225, 1177, 6625, 39723, 251939, 1681535, 11764185, 86002177, 655305697, 5193232611, 42726002123, 364338045647, 3215471252769, 29331858429241, 276224445794785, 2682395337435723, 26832698102762435, 276221586866499839, 2923468922184615897
Offset: 0

Views

Author

Paul D. Hanna, Nov 26 2003

Keywords

Comments

The Hankel transform of this sequence is A000178(n+1); example: det([1,1,3; 1,3,11; 3,11,47]) = 12. - Philippe Deléham, Mar 02 2005
a(n) appears to be the number of indecomposable permutations (A003319) of [n+1] that avoid both of the dashed patterns 32-41 and 41-32. - David Callan, Aug 27 2014
This is true: A nonempty permutation avoids 32-41 and 41-32 if and only if all its components do so. So if A(x) denotes the g.f. for indecomposable {32-41,41-32}-avoiders, then F(x):=1/(1-A(x)) is the g.f. for all {32-41,41-32}-avoiders. From A074664, F(x)=1/x(1-1/B(x)) where B(x) is the o.g.f. for the Bell numbers. Solve for A(x). - David Callan, Jul 21 2017
The Hankel transform of this sequence without the a(0)=1 term is also A000178(n+1). - Michael Somos, Oct 02 2024

Crossrefs

Programs

  • Maple
    bintr:= proc(p) proc(n) add(p(k) *binomial(n,k), k=0..n) end end:
    invtr:= proc(p) local b;
               b:= proc(n) option remember; local i;
                    `if`(n<1, 1, add(b(n-i) *p(i-1), i=1..n+1))
                   end;
            end:
    b:= invtr(bintr(a)):
    a:= n-> `if`(n<0, 0, b(n-1)):
    seq(a(n), n=0..30);  # Alois P. Heinz, Jun 28 2012
  • Mathematica
    a[n_] := Module[{A, B}, A = 1+x; For[k=1, k <= n, k++, B = (A /. x -> x/(1 - x))/(1-x) + O[x]^n // Normal; A = 1 + x*A*B]; SeriesCoefficient[A, {x, 0, n}]]; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Oct 23 2016, adapted from PARI *)
  • PARI
    {a(n)=local(A); if(n<0,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A,x, x/(1-x))/(1-x)+x*O(x^n); A=1+x*A*B);polcoeff(A,n,x))}

Formula

G.f.: A(x) satisfies A(x) = 1/(1 - A(x/(1-x))*x/(1-x) ).
a(n) = Sum_{k = 0..n} A085838(n, k). - Philippe Deléham, Jun 04 2004
G.f.: 1/x-1-1/(B(x)-1) where B(x) = g.f. for A000110 the Bell numbers. - Vladeta Jovovic, Aug 08 2004
a(n) = Sum_{k=0..n} A094456(n,k). - Philippe Deléham, Nov 07 2007
G.f.: 1/(1-x/(1-2x/(1-x/(1-3x/(1-x/(1-4x/(1-x/(1-5x/(1-... (continued fraction). - Paul Barry, Feb 25 2010
From Sergei N. Gladkovskii, Jan 06 2012 - May 12 2013: (Start)
Continued fractions:
G.f.: 1 - x/(G(0)+x); G(k) = x - 1 + x*k + x*(x-1+x*k)/G(k+1).
G.f.: 1/x - 1/2 + (x^2-4)/(4*U(0)-2*x^2+8) where U(k) = k*(2*k+3)*x^2 + x - 2 - (2-x+2*k*x)*(2+3*x+2*k*x)*(k+1)*x^2/U(k+1).
G.f.: 1/x+1/(U(0)-1) where U(k) = -x*k + 1 - x - x^2*(k+1)/U(k+1).
G.f.: (1 - U(0))/x - 1 where U(k) = 1 - x*(k+2) - x^2*(k+1)/U(k+1).
G.f.: (1 - U(0))/x where U(k) = 1 - x*(k+1)/(1-x/U(k+1)).
G.f.: 1/x + 1/( G(0)-1) where G(k) = 1 - x/(1 - x*(2*k+1)/(1 - x/(1 - x*(2*k+2)/ G(k+1) ))).
G.f.:1/x + 1/( G(0) - 1 ) where G(k) = 1 - x/(1 - x*(k+1)/G(k+1) ).
G.f.: (1 - Q(0))/x where Q(k) = 1 + x/(x*k - 1 )/Q(k+1).
G.f.: 1/x - 1/x/Q(0), where Q(k) = 1 + x/(1 - x + x*(k+1)/(x - 1/Q(k+1))).
(End)
Conjecture: a(n) = b(2^(n-1) - 1) for n > 0 with a(0) = 1 where b(n) = b((n - 2^f(n))/2) + b(floor((2n - 2^f(n))/2)) + b(A025480(n-1)) for n > 0 with b(0) = 1 and where f(n) = A007814(n). - Mikhail Kurkov, Jan 11 2022

A101279 a(1) = 1; a(2k) = a(k), a(2k+1) = k.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 1, 4, 2, 5, 1, 6, 3, 7, 1, 8, 4, 9, 2, 10, 5, 11, 1, 12, 6, 13, 3, 14, 7, 15, 1, 16, 8, 17, 4, 18, 9, 19, 2, 20, 10, 21, 5, 22, 11, 23, 1, 24, 12, 25, 6, 26, 13, 27, 3, 28, 14, 29, 7, 30, 15, 31, 1, 32, 16, 33, 8, 34, 17, 35, 4, 36, 18, 37, 9, 38, 19, 39, 2, 40, 20, 41, 10
Offset: 1

Views

Author

N. J. A. Sloane, May 22 2006; definition corrected May 23 2006

Keywords

Comments

From Jeremy Gardiner, Mar 22 2015: (Start)
For n > 2 write n, n-1 in binary, then align bits from the left and take contiguous matching bits as a binary number.
For example:
n = 19 10011
n-1 = 18 10010
a(n) = 9 1001
Also arrange the positive integers as a binary tree rooted at 1 as shown:
1
|
2................../ \..................3
| |
4......../ \........5 6......../ \........7
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
8 9 10 11 12 13 14 15
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
Each branch doubles the number above at the left fork or doubles and adds 1 at the right fork. Then for n > 2, a(n) is the greatest common ancestor of n and n-1, a(n) = gca(n,n-1).
(End)
From David James Sycamore, Mar 07 2023: (Start)
The following identical sequences, {b(n)} and {c(n)}, are the same as a(n+1) for n >= 1.
b(1) = 1, then reverse the conditions in Name: b(2k) = k, b(2k+1) = b(k).
c(1) = 1, then if c(n) is a first occurrence, c(n+1) = c(c(n)), else if c(n) has occurred previously, c(n+1) = n - c(n-1).
These are fractal sequences (b(2m+1) = c(2m+1), m >= 1, recovers the originals). Also {b(n)} and {c(n)} interleave A000027 with the present sequence.
(End)

Examples

			If n is a power of 2 then k=1.
		

Crossrefs

Programs

  • Maple
    a:=array(0..200); a[1]:=1; M:=200; for n from 2 to M do if n mod 2 = 1 then a[n]:=(n-1)/2; else a[n]:=a[n/2]; fi; od: [seq(a[n],n=1..M)];
  • Mathematica
    a[1] = 1; a[n_] := a[n] = If[OddQ@n, (n - 1)/2, a[n/2]]; Array[a, 84] (* Robert G. Wilson v, May 23 2006 *)
  • PARI
    a(n)=(n/2^valuation(n,2)-1)/2+if(n==2^valuation(n,2),1,0) /* Ralf Stephan, Aug 21 2013 */

Formula

a((n+1)/2) = A028310(n) if n is odd and a(n/2) = a(n) if n is even; thus this is a fractal sequence. - Robert G. Wilson v, May 23 2006; corrected by Clark Kimberling, Jul 07 2007
a(n) = A025480(n) + A036987(n) = (n/2^A007814(n) - 1)/2 + (n == 2^A007814(n)). - Ralf Stephan, Aug 21 2013
If n is a power of 2, A070939(a(n)) = 1, otherwise A070939(a(n)) = A119387(n-1).
Numbers m for which a(m) = 1 are A000079(m) and A007283(m), a(2^m + 1) = 2^(m-1); m >= 1. - David James Sycamore, Mar 07 2023
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