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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A277903 a(n) = the least k such that A000123(k) >= n.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12
Offset: 1

Views

Author

Antti Karttunen, Nov 14 2016

Keywords

Comments

Greatest monotonic left inverse of A000123.
Each n occurs A018819(n) times.

Crossrefs

Programs

Formula

a(n) = A048675(A277905(n)).
a(A000123(n)) = n.

A132880 a(n) = A000123( A000975(n-1) ) for n>=1 with a(0)=1, where A000123(n) = number of partitions of 2n into powers of 2 and A000975(n) = n-th number without consecutive equal binary digits.

Original entry on oeis.org

1, 1, 2, 4, 14, 60, 450, 4964, 95982, 3037948, 170005730, 16522010532, 2882717916878, 902450057292988, 514768747418386946, 537142988843543963620, 1033976171696917695108270, 3688322935382700002945333884, 24514290054855626308893309022498
Offset: 0

Views

Author

Paul D. Hanna, Sep 11 2007

Keywords

Examples

			Let b(n) = A000123(n) = number of partitions of 2n into powers of 2, then the initial terms of this sequence begin:
b(0), b(0), b(1), b(2), b(5), b(10), b(21), b(42), b(85), b(170),...
OSCILLATING PARITY TREE.
a(n) = number of nodes in generation n of the following tree.
Start at generation 0 with a single root node labeled [1].
From then on, each parent node [k] is attached k child nodes with labels having opposite parity to k within the range {1..2k}.
The initial generations 0..5 of the tree begin as follows; the path from the root node is given, followed by child nodes in [].
GEN.0: [1];
GEN.1: 1->[2];
GEN.2: 1-2->[1,3];
GEN.3:
1-2-1->[2]
1-2-3->[2,4,6]
GEN.4:
1-2-1-2->[1,3]
1-2-3-2->[1,3]
1-2-3-4->[1,3,5,7]
1-2-3-6->[1,3,5,7,9,11];
GEN.5:
1-2-1-2-1->[2]
1-2-1-2-3->[2,4,6]
1-2-3-2-1->[2]
1-2-3-2-3->[2,4,6]
1-2-3-4-1->[2]
1-2-3-4-3->[2,4,6]
1-2-3-4-5->[2,4,6,8,10]
1-2-3-4-7->[2,4,6,8,10,12,14]
1-2-3-6-1->[2]
1-2-3-6-3->[2,4,6]
1-2-3-6-5->[2,4,6,8,10]
1-2-3-6-7->[2,4,6,8,10,12,14]
1-2-3-6-9->[2,4,6,8,10,12,14,16,18]
1-2-3-6-11->[2,4,6,8,10,12,14,16,18,20,22] .
Note: largest node label in generation n is A000975(n) + 1.
		

Crossrefs

Programs

  • Maple
    g:= proc(b, n) option remember; local t; if b<0 then 0 elif b=0 or n=0 then 1 elif b>=n then add (g(b-t, n) *binomial (n+1, t) *(-1)^(t+1), t=1..n+1) else g(b-1, n) +g(2*b, n-1) fi end: a:= proc(n) local m, t; m:= ceil (2 *(2^(n-1) -1) /3); t:= ilog2(2*m+1); g(m /2^(t-1), t) end: seq (a(n), n=0..30);  # Alois P. Heinz, Apr 17 2009
  • Mathematica
    g[b_, n_] := g[b, n] = Module[{t}, Which[b<0, 0, b == 0 || n == 0, 1, b >= n, Sum[g[b-t, n]*Binomial[n+1, t]*(-1)^(t+1), {t, 1, n+1}], True, g[b-1, n] + g[2*b, n-1]]]; a[n_] := Module[{m, t}, m = Ceiling[2*(2^(n-1)-1) /3]; t = Log[2, 2*m+1] // Floor; g[m /2^(t-1), t]]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Nov 11 2015, after Alois P. Heinz *)
  • PARI
    {A000123(n) = if(n<1, n==0, A000123(n\2) + A000123(n-1))}
    {a(n) = A000123( (2^(n+1) + (-1)^n - 3)/6 )}

Formula

a(n) = A000123( (2^(n+1) + (-1)^n - 3)/6 ) for n>=0.
a(n) = a(n-1) + A000123( A000975(n-1) - 1) for n>0.

A268128 a(n) = (A000123(n) - A001316)/2.

Original entry on oeis.org

0, 0, 1, 1, 4, 5, 8, 9, 17, 21, 28, 33, 45, 53, 66, 75, 100, 117, 140, 161, 193, 221, 258, 291, 344, 389, 446, 499, 573, 639, 722, 797, 913, 1013, 1132, 1249, 1393, 1533, 1698, 1859, 2060, 2253, 2478, 2699, 2965, 3223, 3522, 3813, 4173, 4517, 4910, 5299, 5753
Offset: 0

Views

Author

Tom Edgar, Jan 26 2016

Keywords

Crossrefs

Programs

  • Mathematica
    b[0] = 1; b[n_] := b[n] = b[Floor[n/2]] + b[n - 1];
    c[n_] := Sum[Mod[Binomial[n, k], 2], {k, 0, n}];
    a[n_] := (b[n] - c[n])/2;
    Table[a[n], {n, 0, 52}] (* Jean-François Alcover, Dec 12 2018 *)
  • Sage
    def b(n):
        A=[1]
        for i in [1..n]:
            A.append(A[i-1] + A[floor(i/2)])
        return A[n]
    [(b(n)-prod(x+1 for x in n.digits(2)))/2 for n in [0..60]]

Formula

Let b(0) = 1 and b(n) = b(n-1) + b(floor(n/2)) and let c(n) = Product_{i=0..k}(n_i+1) where n = Sum_{i=0..k}n_i*2^i is the binary representation of n. Then a(n) = (1/2)*(b(n) - c(n)).

A278163 a(n) = the least k such that A131205(1+k) >= n; each n occurs A000123(n) times.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8
Offset: 1

Views

Author

Antti Karttunen, Nov 15 2016

Keywords

Comments

Greatest monotonic left inverse of A131205(1+n).

Crossrefs

Programs

Formula

a(A131205(1+n)) = n.

A278164 Irregular triangle read by rows: row n (n >= 0) is obtained by listing numbers 0 .. A000123(n)-1.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Nov 15 2016

Keywords

Examples

			A000123 -> range -> terms on row n
   1       [0,0]:    0;
   2       [0,1]:    0, 1;
   4       [0,3]:    0, 1, 2, 3;
   6       [0,5]:    0, 1, 2, 3, 4, 5;
  10       [0,9]:    0, 1, 2, 3, 4, 5, 6, 7, 8, 9;
etc.
		

Crossrefs

Obtained by taking the every second row of similar table A277904.

Programs

Formula

a(1) = 0; for n > 1, a(n) = n - A131205(A278163(n)) - 1.

A111414 f(f(n+1))-f(f(n)), where f(m) = binary partition(m) = A000123(m).

Original entry on oeis.org

2, 6, 10, 40, 80, 250, 510, 1890, 4270, 13738, 30630, 101960, 234864, 705046, 1580578, 5136840, 11991928, 36095322, 82990606, 255579576, 595920680, 1719038038, 3926998938, 11665004238, 26978093154, 75938910058, 172678519950, 489262646480, 1114189211016
Offset: 0

Views

Author

N. J. A. Sloane, Nov 12 2005

Keywords

Crossrefs

Cf. A000123.

Programs

  • Maple
    b:= proc(n, j) option remember; local nn, r; if n<0 then 0 elif j=0 then 1 elif j=1 then n+1 elif n f(f(n+1)) -f(f(n)): seq (a(n), n=0..50);  # Alois P. Heinz, Sep 28 2011
  • Mathematica
    b[n_, j_] := b[n, j] = Module[{nn, r}, Which[n<0, 0, j == 0, 1, j == 1, n+1, nJean-François Alcover, Mar 19 2014, after Alois P. Heinz *)

A238392 Triangle read by rows: each row is an initial segment of the terms of A000123 followed by its reflection.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 4, 2, 1, 1, 2, 4, 4, 2, 1, 1, 2, 4, 6, 4, 2, 1, 1, 2, 4, 6, 6, 4, 2, 1, 1, 2, 4, 6, 10, 6, 4, 2, 1, 1, 2, 4, 6, 10, 10, 6, 4, 2, 1, 1, 2, 4, 6, 10, 14, 10, 6, 4, 2, 1, 1, 2, 4, 6, 10, 14, 14, 10, 6, 4, 2, 1, 1, 2, 4, 6, 10, 14, 20, 14, 10, 6, 4, 2, 1
Offset: 0

Views

Author

Philippe Deléham, Feb 26 2014

Keywords

Comments

Triangle read by rows: each row is an initial segment of the terms of A000123 followed by its reflection.

Examples

			Triangle begins:
  1;
  1, 1;
  1, 2, 1;
  1, 2, 2, 1;
  1, 2, 4, 2,  1;
  1, 2, 4, 4,  2,  1;
  1, 2, 4, 6,  4,  2,  1;
  1, 2, 4, 6,  6,  4,  2,  1;
  1, 2, 4, 6, 10,  6,  4,  2,  1;
  1, 2, 4, 6, 10, 10,  6,  4,  2,  1;
  1, 2, 4, 6, 10, 14, 10,  6,  4,  2, 1;
  1, 2, 4, 6, 10, 14, 14, 10,  6,  4, 2, 1;
  1, 2, 4, 6, 10, 14, 20, 14, 10,  6, 4, 2, 1;
  1, 2, 4, 6, 10, 14, 20, 20, 14, 10, 6, 4, 2, 1;
		

Crossrefs

Programs

  • Mathematica
    a[n_] := If[n==0,  1, a[n - 1] + a[Floor[n/2]]]; Flatten[Table[a[Min[k, n - k]], {n, 0, 13}, {k, 0, n}]] (* Indranil Ghosh, Mar 14 2017 *)
  • PARI
    a(n) = if(n==0, 1, a(n-1) + a(floor(n/2)));
    tabl(nn) = {for(n=0, nn, for(k=0, n, print1(a(min(k, n - k)),", ");); print(););};
    tabl(13); \\ Indranil Ghosh, Mar 14 2017
    
  • Python
    def a(n): return 1 if n==0 else a(n - 1) + a(n//2)
    i=0
    for n in range(0, 126):
        for k in range(0, n+1):
            print(str(i)+" "+str(a(min(k, n - k))))
            i+=1 # Indranil Ghosh, Mar 14 2017

Formula

T(n,k) = A000123(min(k,n-k)).
Sum_{k=0..n} T(n,k) = A000123(n).
T(2*n,n) = A000123(n).

A278249 Least number with the prime signature of A000123(n), the number of partitions of 2n into powers of 2.

Original entry on oeis.org

1, 2, 4, 6, 6, 6, 12, 6, 36, 6, 60, 6, 6, 30, 60, 6, 6, 30, 12, 210, 210, 180, 12, 30, 12, 30, 900, 60, 6, 6, 60, 30, 12, 210, 720, 30, 420, 30, 60, 6, 12, 30, 60, 6, 6, 60, 60, 30, 60, 210, 2520, 210, 210, 30, 180, 210, 60, 120, 60, 210, 6, 30, 60, 30, 6, 30, 60, 30, 6, 30, 12, 30, 60, 30, 420, 210, 60, 30, 420, 60, 6, 30, 2520, 30, 30, 210, 12, 210, 60, 210
Offset: 0

Views

Author

Antti Karttunen, Nov 22 2016

Keywords

Crossrefs

Cf. also A278241.

Programs

Formula

a(n) = A046523(A000123(n)).

A000265 Remove all factors of 2 from n; or largest odd divisor of n; or odd part of n.

Original entry on oeis.org

1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15, 1, 17, 9, 19, 5, 21, 11, 23, 3, 25, 13, 27, 7, 29, 15, 31, 1, 33, 17, 35, 9, 37, 19, 39, 5, 41, 21, 43, 11, 45, 23, 47, 3, 49, 25, 51, 13, 53, 27, 55, 7, 57, 29, 59, 15, 61, 31, 63, 1, 65, 33, 67, 17, 69, 35, 71, 9, 73, 37, 75, 19, 77
Offset: 1

Views

Author

Keywords

Comments

When n > 0 is written as k*2^j with k odd then k = A000265(n) and j = A007814(n), so: when n is written as k*2^j - 1 with k odd then k = A000265(n+1) and j = A007814(n+1), when n > 1 is written as k*2^j + 1 with k odd then k = A000265(n-1) and j = A007814(n-1).
Also denominator of 2^n/n (numerator is A075101(n)). - Reinhard Zumkeller, Sep 01 2002
Slope of line connecting (o, a(o)) where o = (2^k)(n-1) + 1 is 2^k and (by design) starts at (1, 1). - Josh Locker (joshlocker(AT)macfora.com), Apr 17 2004
Numerator of n/2^(n-1). - Alexander Adamchuk, Feb 11 2005
From Marco Matosic, Jun 29 2005: (Start)
"The sequence can be arranged in a table:
1
1 3 1
1 5 3 7 1
1 9 5 11 3 13 7 15 1
1 17 9 19 5 21 11 23 3 25 13 27 7 29 15 31 1
Every new row is the previous row interspaced with the continuation of the odd numbers.
Except for the ones; the terms (t) in each column are t+t+/-s = t_+1. Starting from the center column of threes and working to the left the values of s are given by A000265 and working to the right by A000265." (End)
This is a fractal sequence. The odd-numbered elements give the odd natural numbers. If these elements are removed, the original sequence is recovered. - Kerry Mitchell, Dec 07 2005
2k + 1 is the k-th and largest of the subsequence of k terms separating two successive equal entries in a(n). - Lekraj Beedassy, Dec 30 2005
It's not difficult to show that the sum of the first 2^n terms is (4^n + 2)/3. - Nick Hobson, Jan 14 2005
In the table, for each row, (sum of terms between 3 and 1) - (sum of terms between 1 and 3) = A020988. - Eric Desbiaux, May 27 2009
This sequence appears in the analysis of A160469 and A156769, which resemble the numerator and denominator of the Taylor series for tan(x). - Johannes W. Meijer, May 24 2009
Indices n such that a(n) divides 2^n - 1 are listed in A068563. - Max Alekseyev, Aug 25 2013
From Alexander R. Povolotsky, Dec 17 2014: (Start)
With regard to the tabular presentation described in the comment by Marco Matosic: in his drawing, starting with the 3rd row, the first term in the row, which is equal to 1 (or, alternatively the last term in the row, which is also equal to 1), is not in the actual sequence and is added to the drawing as a fictitious term (for the sake of symmetry); an actual A000265(n) could be considered to be a(j,k) (where j >= 1 is the row number and k>=1 is the column subscript), such that a(j,1) = 1:
1
1 3
1 5 3 7
1 9 5 11 3 13 7 15
1 17 9 19 5 21 11 23 3 25 13 27 7 29 15 31
and so on ... .
The relationship between k and j for each row is 1 <= k <= 2^(j-1). In this corrected tabular representation, Marco's notion that "every new row is the previous row interspaced with the continuation of the odd numbers" remains true. (End)
Partitions natural numbers to the same equivalence classes as A064989. That is, for all i, j: a(i) = a(j) <=> A064989(i) = A064989(j). There are dozens of other such sequences (like A003602) for which this also holds: In general, all sequences for which a(2n) = a(n) and the odd bisection is injective. - Antti Karttunen, Apr 15 2017
From Paul Curtz, Feb 19 2019: (Start)
This sequence is the truncated triangle:
1, 1;
3, 1, 5;
3, 7, 1, 9;
5, 11, 3, 13, 7;
15, 1, 17, 9, 19, 5;
21, 11, 23, 3, 25, 13, 27;
7, 29, 15, 31, 1, 33, 17, 35;
...
The first column is A069834. The second column is A213671. The main diagonal is A236999. The first upper diagonal is A125650 without 0.
c(n) = ((n*(n+1)/2))/A069834 = 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 1, 1, 8, 8, 1, 1, ... for n > 0. n*(n+1)/2 is the rank of A069834. (End)
As well as being multiplicative, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 27 2019
a(n) is also the map n -> A026741(n) applied at least A007814(n) times. - Federico Provvedi, Dec 14 2021

Examples

			G.f. = x + x^2 + 3*x^3 + x^4 + 5*x^5 + 3*x^6 + 7*x^7 + x^8 + 9*x^9 + 5*x^10 + 11*x^11 + ...
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A049606 (partial products), A135013 (partial sums), A099545 (mod 4), A326937 (Dirichlet inverse).
Cf. A026741 (map), A001511 (converging steps), A038550 (prime index).
Cf. A195056 (Dgf at s=3).

Programs

  • Haskell
    a000265 = until odd (`div` 2)
    -- Reinhard Zumkeller, Jan 08 2013, Apr 08 2011, Oct 14 2010
    
  • Java
    int A000265(n){
        while(n%2==0) n>>=1;
        return n;
    }
    /* Aidan Simmons, Feb 24 2019 */
    
  • Julia
    using IntegerSequences
    [OddPart(n) for n in 1:77] |> println  # Peter Luschny, Sep 25 2021
    
  • Magma
    A000265:= func< n | n/2^Valuation(n,2) >;
    [A000265(n): n in [1..120]]; // G. C. Greubel, Jul 31 2024
    
  • Maple
    A000265:=proc(n) local t1,d; t1:=1; for d from 1 by 2 to n do if n mod d = 0 then t1:=d; fi; od; t1; end: seq(A000265(n), n=1..77);
    A000265 := n -> n/2^padic[ordp](n,2): seq(A000265(n), n=1..77); # Peter Luschny, Nov 26 2010
  • Mathematica
    a[n_Integer /; n > 0] := n/2^IntegerExponent[n, 2]; Array[a, 77] (* Josh Locker *)
    a[ n_] := If[ n == 0, 0, n / 2^IntegerExponent[ n, 2]]; (* Michael Somos, Dec 17 2014 *)
  • PARI
    {a(n) = n >> valuation(n, 2)}; /* Michael Somos, Aug 09 2006, edited by M. F. Hasler, Dec 18 2014 */
    
  • Python
    from _future_ import division
    def A000265(n):
        while not n % 2:
            n //= 2
        return n # Chai Wah Wu, Mar 25 2018
    
  • Python
    def a(n):
        while not n&1: n >>= 1
        return n
    print([a(n) for n in range(1, 78)]) # Michael S. Branicky, Jun 26 2025
    
  • SageMath
    def A000265(n): return n//2^valuation(n,2)
    [A000265(n) for n in (1..121)] # G. C. Greubel, Jul 31 2024
  • Scheme
    (define (A000265 n) (let loop ((n n)) (if (odd? n) n (loop (/ n 2))))) ;; Antti Karttunen, Apr 15 2017
    

Formula

a(n) = if n is odd then n, otherwise a(n/2). - Reinhard Zumkeller, Sep 01 2002
a(n) = n/A006519(n) = 2*A025480(n-1) + 1.
Multiplicative with a(p^e) = 1 if p = 2, p^e if p > 2. - David W. Wilson, Aug 01 2001
a(n) = Sum_{d divides n and d is odd} phi(d). - Vladeta Jovovic, Dec 04 2002
G.f.: -x/(1 - x) + Sum_{k>=0} (2*x^(2^k)/(1 - 2*x^(2^(k+1)) + x^(2^(k+2)))). - Ralf Stephan, Sep 05 2003
(a(k), a(2k), a(3k), ...) = a(k)*(a(1), a(2), a(3), ...) In general, a(n*m) = a(n)*a(m). - Josh Locker (jlocker(AT)mail.rochester.edu), Oct 04 2005
a(n) = Sum_{k=0..n} A127793(n,k)*floor((k+2)/2) (conjecture). - Paul Barry, Jan 29 2007
Dirichlet g.f.: zeta(s-1)*(2^s - 2)/(2^s - 1). - Ralf Stephan, Jun 18 2007
a(A132739(n)) = A132739(a(n)) = A132740(n). - Reinhard Zumkeller, Aug 27 2007
a(n) = 2*A003602(n) - 1. - Franklin T. Adams-Watters, Jul 02 2009
a(n) = n/gcd(2^n,n). (This also shows that the true offset is 0 and a(0) = 0.) - Peter Luschny, Nov 14 2009
a(-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2011
From Reinhard Zumkeller, May 01 2012: (Start)
A182469(n, k) = A027750(a(n), k), k = 1..A001227(n).
a(n) = A182469(n, A001227(n)). (End)
a((2*n-1)*2^p) = 2*n - 1, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 05 2013
G.f.: G(0)/(1 - 2*x^2 + x^4) - 1/(1 - x), where G(k) = 1 + 1/(1 - x^(2^k)*(1 - 2*x^(2^(k+1)) + x^(2^(k+2)))/(x^(2^k)*(1 - 2*x^(2^(k+1)) + x^(2^(k+2))) + (1 - 2*x^(2^(k+2)) + x^(2^(k+3)))/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 06 2013
a(n) = A003961(A064989(n)). - Antti Karttunen, Apr 15 2017
Completely multiplicative with a(2) = 1 and a(p) = p for prime p > 2, i.e., the sequence b(n) = a(n) * A008683(n) for n > 0 is the Dirichlet inverse of a(n). - Werner Schulte, Jul 08 2018
From Peter Bala, Feb 27 2019: (Start)
O.g.f.: F(x) - F(x^2) - F(x^4) - F(x^8) - ..., where F(x) = x/(1 - x)^2 is the generating function for the positive integers.
O.g.f. for reciprocals: Sum_{n >= 1} x^n/a(n) = L(x) + (1/2)*L(x^2) + (1/2)*L(x^4) + (1/2)*L(x^8) + ..., where L(x) = log(1/(1 - x)).
Sum_{n >= 1} x^n/a(n) = 1/2*log(G(x)), where G(x) = 1 + 2*x + 4*x^2 + 6*x^3 + 10*x^4 + ... is the o.g.f. of A000123. (End)
O.g.f.: Sum_{n >= 1} phi(2*n-1)*x^(2*n-1)/(1 - x^(2*n-1)), where phi(n) is the Euler totient function A000010. - Peter Bala, Mar 22 2019
a(n) = A049606(n) / A049606(n-1). - Flávio V. Fernandes, Dec 08 2020
a(n) = numerator of n/2^(floor(n/2)). - Federico Provvedi, Dec 14 2021
a(n) = Sum_{d divides n} (-1)^(d+1)*phi(2*n/d). - Peter Bala, Jan 14 2024
a(n) = A030101(A030101(n)). - Darío Clavijo, Sep 19 2024

Extensions

Additional comments from Henry Bottomley, Mar 02 2000
More terms from Larry Reeves (larryr(AT)acm.org), Mar 14 2000
Name clarified by David A. Corneth, Apr 15 2017

A002487 Stern's diatomic series (or Stern-Brocot sequence): a(0) = 0, a(1) = 1; for n > 0: a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1).

Original entry on oeis.org

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

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Keywords

Comments

Also called fusc(n) [Dijkstra].
a(n)/a(n+1) runs through all the reduced nonnegative rationals exactly once [Stern; Calkin and Wilf].
If the terms are written as an array:
column 0 1 2 3 4 5 6 7 8 9 ...
row 0: 0
row 1: 1
row 2: 1,2
row 3: 1,3,2,3
row 4: 1,4,3,5,2,5,3,4
row 5: 1,5,4,7,3,8,5,7,2,7,5,8,3,7,4,5
row 6: 1,6,5,9,4,11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,...
...
then (ignoring row 0) the sum of the k-th row is 3^(k-1), each column is an arithmetic progression and the steps are nothing but the original sequence. - Takashi Tokita (butaneko(AT)fa2.so-net.ne.jp), Mar 08 2003
From N. J. A. Sloane, Oct 15 2017: (Start)
The above observation can be made more precise. Let A(n,k), n >= 0, 0 <= k <= 2^(n-1)-1 for k > 0, denote the entry in row n and column k of the left-justified array above.
The equations for columns 0,1,2,3,4,... are successively (ignoring row 0):
1 (n >= 1),
n (n >= 2),
n-1 (n >= 3),
2n-3 (n >= 3),
n-2 (n >= 4),
3n-7 (n >= 4),
...
and in general column k > 0 is given by
A(n,k) = a(k)*n - A156140(k) for n >= ceiling(log_2(k+1))+1, and 0 otherwise.
(End)
a(n) is the number of odd Stirling numbers S_2(n+1, 2r+1) [Carlitz].
Moshe Newman proved that the fraction a(n+1)/a(n+2) can be generated from the previous fraction a(n)/a(n+1) = x by 1/(2*floor(x) + 1 - x). The successor function f(x) = 1/(floor(x) + 1 - frac(x)) can also be used.
a(n+1) = number of alternating bit sets in n [Finch].
If f(x) = 1/(1 + floor(x) - frac(x)) then f(a(n-1)/a(n)) = a(n)/a(n+1) for n >= 1. If T(x) = -1/x and f(x) = y, then f(T(y)) = T(x) for x > 0. - Michael Somos, Sep 03 2006
a(n+1) is the number of ways of writing n as a sum of powers of 2, each power being used at most twice (the number of hyperbinary representations of n) [Carlitz; Lind].
a(n+1) is the number of partitions of the n-th integer expressible as the sum of distinct even-subscripted Fibonacci numbers (= A054204(n)), into sums of distinct Fibonacci numbers [Bicknell-Johnson, theorem 2.1].
a(n+1) is the number of odd binomial(n-k, k), 0 <= 2*k <= n. [Carlitz], corrected by Alessandro De Luca, Jun 11 2014
a(2^k) = 1. a(3*2^k) = a(2^(k+1) + 2^k) = 2. Sequences of terms between a(2^k) = 1 and a(2^(k+1)) = 1 are palindromes of length 2^k-1 with a(2^k + 2^(k-1)) = 2 in the middle. a(2^(k-1) + 1) = a(2^k - 1) = k+1 for k > 1. - Alexander Adamchuk, Oct 10 2006
The coefficients of the inverse of the g.f. of this sequence form A073469 and are related to binary partitions A000123. - Philippe Flajolet, Sep 06 2008
It appears that the terms of this sequence are the number of odd entries in the diagonals of Pascal's triangle at 45 degrees slope. - Javier Torres (adaycalledzero(AT)hotmail.com), Aug 06 2009
Let M be an infinite lower triangular matrix with (1, 1, 1, 0, 0, 0, ...) in every column shifted down twice:
1;
1, 0;
1, 1, 0;
0, 1, 0, 0;
0, 1, 1, 0, 0;
0, 0, 1, 0, 0, 0;
0, 0, 1, 1, 0, 0, 0;
...
Then this sequence A002487 (without initial 0) is the first column of lim_{n->oo} M^n. (Cf. A026741.) - Gary W. Adamson, Dec 11 2009 [Edited by M. F. Hasler, Feb 12 2017]
Member of the infinite family of sequences of the form a(n) = a(2*n); a(2*n+1) = r*a(n) + a(n+1), r = 1 for A002487 = row 1 in the array of A178239. - Gary W. Adamson, May 23 2010
Equals row 1 in an infinite array shown in A178568, sequences of the form
a(2*n) = r*a(n), a(2*n+1) = a(n) + a(n+1); r = 1. - Gary W. Adamson, May 29 2010
Row sums of A125184, the Stern polynomials. Equivalently, B(n,1), the n-th Stern polynomial evaluated at x = 1. - T. D. Noe, Feb 28 2011
The Kn1y and Kn2y triangle sums, see A180662 for their definitions, of A047999 lead to the sequence given above, e.g., Kn11(n) = A002487(n+1) - A000004(n), Kn12(n) = A002487(n+3) - A000012(n), Kn13(n) = A002487(n+5) - A000034(n+1) and Kn14(n) = A002487(n+7) - A157810(n+1). For the general case of the knight triangle sums see the Stern-Sierpiński triangle A191372. This triangle not only leads to Stern's diatomic series but also to snippets of this sequence and, quite surprisingly, their reverse. - Johannes W. Meijer, Jun 05 2011
Maximum of terms between a(2^k) = 1 and a(2^(k+1)) = 1 is the Fibonacci number F(k+2). - Leonid Bedratyuk, Jul 04 2012
Probably the number of different entries per antidiagonal of A223541. That would mean there are exactly a(n+1) numbers that can be expressed as a nim-product 2^x*2^y with x + y = n. - Tilman Piesk, Mar 27 2013
Let f(m,n) be the frequency of the integer n in the interval [a(2^(m-1)), a(2^m-1)]. Let phi(n) be Euler's totient function (A000010). Conjecture: for all integers m,n n<=m f(m,n) = phi(n). - Yosu Yurramendi, Sep 08 2014
Back in May 1995, it was proved that A000360 is the modulo 3 mapping, (+1,-1,+0)/2, of this sequence A002487 (without initial 0). - M. Jeremie Lafitte (Levitas), Apr 24 2017
Define a sequence chf(n) of Christoffel words over an alphabet {-,+}: chf(1) = '-'; chf(2*n+0) = negate(chf(n)); chf(2*n+1) = negate(concatenate(chf(n),chf(n+1))). Then the length of the chf(n) word is fusc(n) = a(n); the number of '-'-signs in the chf(n) word is c-fusc(n) = A287729(n); the number of '+'-signs in the chf(n) word is s-fusc(n) = A287730(n). See examples below. - I. V. Serov, Jun 01 2017
The sequence can be extended so that a(n) = a(-n), a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1) for all n in Z. - Michael Somos, Jun 25 2019
Named after the German mathematician Moritz Abraham Stern (1807-1894), and sometimes also after the French clockmaker and amateur mathematician Achille Brocot (1817-1878). - Amiram Eldar, Jun 06 2021
It appears that a(n) is equal to the multiplicative inverse of A007305(n+1) mod A007306(n+1). For example, a(12) is 2, the multiplicative inverse of A007305(13) mod A007306(13), where A007305(13) is 4 and A007306(13) is 7. - Gary W. Adamson, Dec 18 2023

Examples

			Stern's diatomic array begins:
  1,1,
  1,2,1,
  1,3,2,3,1,
  1,4,3,5,2,5,3,4,1,
  1,5,4,7,3,8,5,7,2,7,5,8,3,7,4,5,1,
  1,6,5,9,4,11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,7,11,4,9,...
  ...
a(91) = 19, because 91_10 = 1011011_2; b_6=b_4=b_3=b_1=b_0=1, b_5=b_2=0;  L=5; m_1=0, m_2=1, m_3=3, m_4=4, m_5=6; c_1=2, c_2=3, c_3=2, c_4=3; f(1)=1, f(2)=2, f(3)=5, f(4)=8, f(5)=19. - _Yosu Yurramendi_, Jul 13 2016
From _I. V. Serov_, Jun 01 2017: (Start)
a(n) is the length of the Christoffel word chf(n):
n  chf(n) A070939(n)   a(n)
1   '-'       1          1
2   '+'       2          1
3   '+-'      2          2
4   '-'       3          1
5   '--+'     3          3
6   '-+'      3          2
... (End)
G.f. = x + x^2 + 2*x^3 + x^4 + 3*x^5 + 2*x^6 + 3*x^7 + x^8 + ... - _Michael Somos_, Jun 25 2019
		

References

  • M. Aigner and G. M. Ziegler, Proofs from THE BOOK, 3rd ed., Berlin, Heidelberg, New York: Springer-Verlag, 2004, p. 97.
  • Elwyn R. Berlekamp, John H. Conway and Richard K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 114.
  • Krishna Dasaratha, Laure Flapan, Chansoo Lee, Cornelia Mihaila, Nicholas Neumann-Chun, Sarah Peluse and Matthew Stroegeny, A family of multi-dimensional continued fraction Stern sequences, Abtracts Amer. Math. Soc., Vol. 33 (#1, 2012), #1077-05-2543.
  • Edsger W. Dijkstra, Selected Writings on Computing, Springer, 1982, p. 232 (sequence is called fusc).
  • F. G. M. Eisenstein, Eine neue Gattung zahlentheoretischer Funktionen, welche von zwei Elementen abhaengen und durch gewisse lineare Funktional-Gleichungen definirt werden, Verhandlungen der Koenigl. Preuss. Akademie der Wiss. Berlin (1850), pp. 36-42, Feb 18, 1850. Werke, II, pp. 705-711.
  • Graham Everest, Alf van der Poorten, Igor Shparlinski and Thomas Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 2.16.3; pp. 148-149.
  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 117.
  • Thomas Koshy, Fibonacci and Lucas numbers with applications, Wiley, 2001, p. 98.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Record values are in A212289.
If the 1's are replaced by pairs of 1's we obtain A049456.
Inverse: A020946.
Cf. a(A001045(n)) = A000045(n). a(A062092(n)) = A000032(n+1).
Cf. A064881-A064886 (Stern-Brocot subtrees).
A column of A072170.
Cf. A049455 for the 0,1 version of Stern's diatomic array.
Cf. A000119, A262097 for analogous sequences in other bases and A277189, A277315, A277328 for related sequences with similar graphs.
Cf. A086592 and references therein to other sequences related to Kepler's tree of fractions.

Programs

  • Haskell
    a002487 n = a002487_list !! n
    a002487_list = 0 : 1 : stern [1] where
       stern fuscs = fuscs' ++ stern fuscs' where
         fuscs' = interleave fuscs $ zipWith (+) fuscs $ (tail fuscs) ++ [1]
       interleave []     ys = ys
       interleave (x:xs) ys = x : interleave ys xs
    -- Reinhard Zumkeller, Aug 23 2011
    
  • Julia
    using Nemo
    function A002487List(len)
        a, A = QQ(0), [0,1]
        for n in 1:len
            a = next_calkin_wilf(a)
            push!(A, denominator(a))
        end
    A end
    A002487List(91) |> println # Peter Luschny, Mar 13 2018
    
  • Magma
    [&+[(Binomial(k, n-k-1) mod 2): k in [0..n]]: n in [0..100]]; // Vincenzo Librandi, Jun 18 2019
    
  • Maple
    A002487 := proc(n) option remember; if n <= 1 then n elif n mod 2 = 0 then procname(n/2); else procname((n-1)/2)+procname((n+1)/2); fi; end: seq(A002487(n),n=0..91);
    A002487 := proc(m) local a,b,n; a := 1; b := 0; n := m; while n>0 do if type(n,odd) then b := a+b else a := a+b end if; n := floor(n/2); end do; b; end proc: seq(A002487(n),n=0..91); # Program adapted from E. Dijkstra, Selected Writings on Computing, Springer, 1982, p. 232. - Igor Urbiha (urbiha(AT)math.hr), Oct 28 2002. Since A007306(n) = a(2*n+1), this program can be adapted for A007306 by replacing b := 0 by b := 1.
    A002487 := proc(n::integer) local k; option remember; if n = 0 then 0 elif n=1 then 1 else add(K(k,n-1-k)*procname(n - k), k = 1 .. n) end if end proc:
    K := proc(n::integer, k::integer) local KC; if 0 <= k and k <= n and n-k <= 2 then KC:=1; else KC:= 0; end if; end proc: seq(A002487(n),n=0..91); # Thomas Wieder, Jan 13 2008
    # next Maple program:
    a:= proc(n) option remember; `if`(n<2, n,
          (q-> a(q)+(n-2*q)*a(n-q))(iquo(n, 2)))
        end:
    seq(a(n), n=0..100);  # Alois P. Heinz, Feb 11 2021
    fusc := proc(n) local a, b, c; a := 1; b := 0;
        for c in convert(n, base, 2) do
            if c = 0 then a := a + b else b := a + b fi od;
        b end:
    seq(fusc(n), n = 0..91); # Peter Luschny, Nov 09 2022
    Stern := proc(n, u) local k, j, b;
        b := j -> nops({seq(Bits:-Xor(k, j-k), k = 0..j)}):
        ifelse(n=0, 1-u, seq(b(j), j = 2^(n-1)-1..2^n-1-u)) end:
    seq(print([n], Stern(n, 1)), n = 0..5); # As shown in the comments.
    seq(print([n], Stern(n, 0)), n = 0..5); # As shown in the examples. # Peter Luschny, Sep 29 2024
  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := If[ EvenQ[n], a[n/2], a[(n-1)/2] + a[(n+1)/2]]; Table[ a[n], {n, 0, 100}] (* end of program *)
    Onemore[l_] := Transpose[{l, l + RotateLeft[l]}] // Flatten;
    NestList[Onemore, {1}, 5] // Flatten  (*gives [a(1), ...]*) (* Takashi Tokita, Mar 09 2003 *)
    ToBi[l_] := Table[2^(n - 1), {n, Length[l]}].Reverse[l]; Map[Length,
    Split[Sort[Map[ToBi, Table[IntegerDigits[n - 1, 3], {n, 500}]]]]]  (*give [a(1), ...]*) (* Takashi Tokita, Mar 10 2003 *)
    A002487[m_] := Module[{a = 1, b = 0, n = m}, While[n > 0, If[OddQ[n], b = a+b, a = a+b]; n = Floor[n/2]]; b]; Table[A002487[n], {n, 0, 100}] (* Jean-François Alcover, Sep 06 2013, translated from 2nd Maple program *)
    a[0] = 0; a[1] = 1;
    Flatten[Table[{a[2*n] = a[n], a[2*n + 1] = a[n] + a[n + 1]}, {n, 0, 50}]] (* Horst H. Manninger, Jun 09 2021 *)
    nmax = 100; CoefficientList[Series[x*Product[(1 + x^(2^k) + x^(2^(k+1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 08 2022 *)
  • PARI
    {a(n) = n=abs(n); if( n<2, n>0, a(n\2) + if( n%2, a(n\2 + 1)))};
    
  • PARI
    fusc(n)=local(a=1,b=0);while(n>0,if(bitand(n,1),b+=a,a+=b);n>>=1);b \\ Charles R Greathouse IV, Oct 05 2008
    
  • PARI
    A002487(n,a=1,b=0)=for(i=0,logint(n,2),if(bittest(n,i),b+=a,a+=b));b \\ M. F. Hasler, Feb 12 2017, updated Feb 14 2019
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def a(n): return n if n<2 else a(n//2) if n%2==0 else a((n - 1)//2) + a((n + 1)//2) # Indranil Ghosh, Jun 08 2017; corrected by Reza K Ghazi, Dec 27 2021
    
  • Python
    def a(n):
        a, b = 1, 0
        while n > 0:
            if n & 1:
                b += a
            else:
                a += b
            n >>= 1
        return b
    # Reza K Ghazi, Dec 29 2021
    
  • Python
    def A002487(n): return sum(int(not (n-k-1) & ~k) for k in range(n)) # Chai Wah Wu, Jun 19 2022
    
  • Python
    # (fast way for big vectors)
    from math import log, ceil
    import numpy
    how_many_terms = 2**20  # (Powers of 2 recommended but other integers are also possible.)
    A002487, A002487[1]  = numpy.zeros(2**(ce:=ceil(log(how_many_terms,2))), dtype=object), 1
    for exponent in range(1,ce):
        L, L2 = 2**exponent, 2**(exponent+1)
        A002487[L2 - 1] = exponent + 1
        A002487[L:L2][::2] = A002487[L >> 1: L]
        A002487[L + 1:L2 - 2][::2] = A002487[L:L2 - 3][::2]  +  A002487[L + 2:L2 - 1][::2]
    print(list(A002487[0:100])) # Karl-Heinz Hofmann, Jul 22 2025
  • R
    N <- 50 # arbitrary
    a <- 1
    for (n in 1:N)
    {
      a[2*n    ] = a[n]
      a[2*n + 1] = a[n] + a[n+1]
      a
    }
    a
    # Yosu Yurramendi, Oct 04 2014
    
  • R
    # Given n, compute a(n) by taking into account the binary representation of n
    a <- function(n){
      b <- as.numeric(intToBits(n))
      l <- sum(b)
      m <- which(b == 1)-1
      d <- 1
      if(l > 1) for(j in 1:(l-1)) d[j] <- m[j+1]-m[j]+1
      f <- c(0,1)
      if(l > 1) for(j in 3:(l+1)) f[j] <- d[j-2]*f[j-1]-f[j-2]
      return(f[l+1])
    } # Yosu Yurramendi, Dec 13 2016
    
  • R
    # computes the sequence as a vector A, rather than function a() as above.
    A <- c(1,1)
    maxlevel <- 5 # by choice
    for(m in 1:maxlevel) {
      A[2^(m+1)] <- 1
      for(k in 1:(2^m-1)) {
        r <- m - floor(log2(k)) - 1
        A[2^r*(2*k+1)] <- A[2^r*(2*k)] + A[2^r*(2*k+2)]
    }}
    A # Yosu Yurramendi, May 08 2018
    
  • Sage
    def A002487(n):
        M = [1, 0]
        for b in n.bits():
            M[b] = M[0] + M[1]
        return M[1]
    print([A002487(n) for n in (0..91)])
    # For a dual see A174980. Peter Luschny, Nov 28 2017
    
  • Scheme
    ;; An implementation of memoization-macro definec can be found for example in: http://oeis.org/wiki/Memoization
    (definec (A002487 n) (cond ((<= n 1) n) ((even? n) (A002487 (/ n 2))) (else (+ (A002487 (/ (- n 1) 2)) (A002487 (/ (+ n 1) 2))))))
    ;; Antti Karttunen, Nov 05 2016
    

Formula

a(n+1) = (2*k+1)*a(n) - a(n-1) where k = floor(a(n-1)/a(n)). - David S. Newman, Mar 04 2001
Let e(n) = A007814(n) = exponent of highest power of 2 dividing n. Then a(n+1) = (2k+1)*a(n)-a(n-1), n > 0, where k = e(n). Moreover, floor(a(n-1)/a(n)) = e(n), in agreement with D. Newman's formula. - Dragutin Svrtan (dsvrtan(AT)math.hr) and Igor Urbiha (urbiha(AT)math.hr), Jan 10 2002
Calkin and Wilf showed 0.9588 <= limsup a(n)/n^(log(phi)/log(2)) <= 1.1709 where phi is the golden mean. Does this supremum limit = 1? - Benoit Cloitre, Jan 18 2004. Coons and Tyler show the limit is A246765 = 0.9588... - Kevin Ryde, Jan 09 2021
a(n) = Sum_{k=0..floor((n-1)/2)} (binomial(n-k-1, k) mod 2). - Paul Barry, Sep 13 2004
a(n) = Sum_{k=0..n-1} (binomial(k, n-k-1) mod 2). - Paul Barry, Mar 26 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = v^3 + 2*u*v*w - u^2*w. - Michael Somos, May 02 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^3), A(x^6)) where f(u1, u2, u3, u6) = u1^3*u6 - 3*u1^2*u2*u6 + 3*u2^3*u6 - u2^3*u3. - Michael Somos, May 02 2005
G.f.: x * Product_{k>=0} (1 + x^(2^k) + x^(2^(k+1))) [Carlitz].
a(n) = a(n-2) + a(n-1) - 2*(a(n-2) mod a(n-1)). - Mike Stay, Nov 06 2006
A079978(n) = (1 + e^(i*Pi*A002487(n)))/2, i=sqrt(-1). - Paul Barry, Jan 14 2005
a(n) = Sum_{k=1..n} K(k, n-k)*a(n - k), where K(n,k) = 1 if 0 <= k AND k <= n AND n-k <= 2 and K(n,k) = 0 else. (When using such a K-coefficient, several different arguments to K or several different definitions of K may lead to the same integer sequence. For example, if we drop the condition k <= n in the above definition, then we arrive at A002083 = Narayana-Zidek-Capell numbers.) - Thomas Wieder, Jan 13 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1; a(2^n - k) + a(k) = a(2^(n+1) + k). Both formulas hold for 0 <= k <= 2^n - 1. G.f.: G(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ... Define f(z) = (1 + z + z^2), then G(z) = lim f(z)*f(z^2)*f(z^4)* ... *f(z^(2^n))*... = (1 + z + z^2)*G(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 11 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1 (0 <= k <= 2^n - 1). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
a(2^n + k) = a(2^n - k) + a(k) (0 <= k <= 2^n). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
Let g(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ..., f(z) = 1 + z + z^2. Then g(z) = lim_{n->infinity} f(z)*f(z^2)*f(z^4)*...*f(z^(2^n)), g(z) = f(z)*g(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
For 0 <= k <= 2^n - 1, write k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1) where b(0), b(1), etc. are 0 or 1. Define a 2 X 2 matrix X(m) with entries x(1,1) = x(2,2) = 1, x(1,2) = 1 - b(m), x(2,1) = b(m). Let P(n)= X(0)*X(1)* ... *X(n-1). The entries of the matrix P are members of the sequence: p(1,1) = a(k+1), p(1,2) = a(2^n - (k+1)), p(2,1) = a(k), p(2,2) = a(2^n - k). - Arie Werksma (werksma(AT)tiscali.nl), Apr 20 2008
Let f(x) = A030101(x); if 2^n + 1 <= x <= 2^(n + 1) and y = 2^(n + 1) - f(x - 1) then a(x) = a(y). - Arie Werksma (Werksma(AT)Tiscali.nl), Jul 11 2008
a(n) = A126606(n + 1) / 2. - Reikku Kulon, Oct 05 2008
Equals infinite convolution product of [1,1,1,0,0,0,0,0,0] aerated A000079 - 1 times, i.e., [1,1,1,0,0,0,0,0,0] * [1,0,1,0,1,0,0,0,0] * [1,0,0,0,1,0,0,0,1]. - Mats Granvik and Gary W. Adamson, Oct 02 2009; corrected by Mats Granvik, Oct 10 2009
a(2^(p+2)*n+2^(p+1)-1) - a(2^(p+1)*n+2^p-1) = A007306(n+1), p >= 0 and n >= 0. - Johannes W. Meijer, Feb 07 2013
a(2*n-1) = A007306(n), n > 0. - Yosu Yurramendi, Jun 23 2014
a(n*2^m) = a(n), m>0, n > 0. - Yosu Yurramendi, Jul 03 2014
a(k+1)*a(2^m+k) - a(k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Nov 07 2014
a(2^(m+1)+(k+1))*a(2^m+k) - a(2^(m+1)+k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Nov 07 2014
a(5*2^k) = 3. a(7*2^k) = 3. a(9*2^k) = 4. a(11*2^k) = 5. a(13*2^k) = 5. a(15*2^k) = 4. In general: a((2j-1)*2^k) = A007306(j), j > 0, k >= 0 (see Adamchuk's comment). - Yosu Yurramendi, Mar 05 2016
a(2^m+2^m'+k') = a(2^m'+k')*(m-m'+1) - a(k'), m >= 0, m' <= m-1, 0 <= k' < 2^m'. - Yosu Yurramendi, Jul 13 2016
From Yosu Yurramendi, Jul 13 2016: (Start)
Let n be a natural number and [b_m b_(m-1) ... b_1 b_0] its binary expansion with b_m=1.
Let L = Sum_{i=0..m} b_i be the number of binary digits equal to 1 (L >= 1).
Let {m_j: j=1..L} be the set of subindices such that b_m_j = 1, j=1..L, and 0 <= m_1 <= m_2 <= ... <= m_L = m.
If L = 1 then c_1 = 1, otherwise let {c_j: j=1..(L-1)} be the set of coefficients such that c_(j) = m_(j+1) - m_j + 1, 1 <= j <= L-1.
Let f be a function defined on {1..L+1} such that f(1) = 0, f(2) = 1, f(j) = c_(j-2)*f(j-1) - f(j-2), 3 <= j <= L+1.
Then a(n) = f(L+1) (see example). (End)
a(n) = A001222(A260443(n)) = A000120(A277020(n)). Also a(n) = A000120(A101624(n-1)) for n >= 1. - Antti Karttunen, Nov 05 2016
(a(n-1) + a(n+1))/a(n) = A037227(n) for n >= 1. - Peter Bala, Feb 07 2017
a(0) = 0; a(3n) = 2*A000360(3n-1); a(3n+1) = 2*A000360(3n) - 1; a(3n+2) = 2*A000360(3n+1) + 1. - M. Jeremie Lafitte (Levitas), Apr 24 2017
From I. V. Serov, Jun 14 2017: (Start)
a(n) = A287896(n-1) - 1*A288002(n-1) for n > 1;
a(n) = A007306(n-1) - 2*A288002(n-1) for n > 1. (End)
From Yosu Yurramendi, Feb 14 2018: (Start)
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1) + 2^m + k) = 2*a(k), m >= 0, 0 <= k < 2^m.
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1) + k) = a(2^m + k), m >= 0, 0 <= k < 2^m.
a(2^m + k) = a(k)*(m - floor(log_2(k)) - 1) + a(2^(floor(log_2(k))+1) + k), m >= 0, 0 < k < 2^m, a(2^m) = 1, a(0) = 0. (End)
From Yosu Yurramendi, May 08 2018: (Start)
a(2^m) = 1, m >= 0.
a(2^r*(2*k+1)) = a(2^r*(2*k)) + a(2^r*(2*k+2)), r < - m - floor(log_2(k)) - 1, m > 0, 1 <= k < 2^m. (End)
Trow(n) = [card({k XOR (j-k): k=0..j}) for j = 2^(n-1)-1..2^n-2] when regarded as an irregular table (n >= 1). - Peter Luschny, Sep 29 2024
a(n) = A000120(A168081(n)). - Karl-Heinz Hofmann, Jun 16 2025

Extensions

Additional references and comments from Len Smiley, Joshua Zucker, Rick L. Shepherd and Herbert S. Wilf
Typo in definition corrected by Reinhard Zumkeller, Aug 23 2011
Incorrect formula deleted and text edited by Johannes W. Meijer, Feb 07 2013
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