cp's OEIS Frontend

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

Showing 1-3 of 3 results.

A023531 a(n) = 1 if n is of the form m(m+3)/2, otherwise 0.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Jun 14 1998

Keywords

Comments

Can be read as table: a(n,m) = 1 if n = m >= 0, else 0 (unit matrix).
a(n) = number of 1's between successive 0's (see also A005614, A003589 and A007538). - Eric Angelini, Jul 06 2005
Triangle T(n,k), 0 <= k <= n, read by rows, given by A000004 DELTA A000007 where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 03 2009
Sequence B is called a reverse reluctant sequence of sequence A, if B is triangle array read by rows: row number k lists first k elements of the sequence A in reverse order.
A023531 is reverse reluctant sequence of sequence A000007. - Boris Putievskiy, Jan 11 2013
Also the Bell transform (and the inverse Bell transform) of 0^n (A000007). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 19 2016
This is the turn sequence of the triangle spiral. To form the spiral: go a unit step forward, turn left a(n)*120 degrees, and repeat. The triangle sides are the runs of a(n)=0 (no turn). The sequence can be generated by a morphism with a special symbol S for the start of the sequence: S -> S,1; 1 -> 0,1; 0->0. The expansion lengthens each existing side and inserts a new unit side at the start. See the Fractint L-system in the links to draw the spiral this way. - Kevin Ryde, Dec 06 2019

Examples

			As a triangle:
       1
      0 1
     0 0 1
    0 0 0 1
   0 0 0 0 1
  0 0 0 0 0 1
G.f. = 1 + x^2 + x^5 + x^9 + x^14 + x^20 + x^27 + x^35 + x^44 + x^54 + ...
From _Kevin Ryde_, Dec 06 2019: (Start)
.
              1            Triangular spiral: start at S;
             / \             go a unit step forward,
            0   0   .        turn left a(n)*120 degrees,
           /     \   .       repeat.
          0   1   0   .
         /   / \   \   \   Each side's length is 1 greater
        0   0   0   0   0    than that of the previous side.
       /   /     \   \   \
      0   0   S---1   0   0
     /   /             \   \
    0   1---0---0---0---1   0
   /                         \
  1---0---0---0---0---0---0---1
(End)
		

Crossrefs

Programs

  • Haskell
    a023531 n = a023531_list !! n
    a023531_list = concat $ iterate ([0,1] *) [1]
    instance Num a => Num [a] where
       fromInteger k = [fromInteger k]
       (p:ps) + (q:qs) = p + q : ps + qs
       ps + qs         = ps ++ qs
       (p:ps) * qs'@(q:qs) = p * q : ps * qs' + [p] * qs
        *                = []
    -- Reinhard Zumkeller, Apr 02 2011
    
  • Maple
    seq(op([0$m,1]),m=0..10); # Robert Israel, Jan 18 2015
    # alternative
    A023531 := proc(n)
        option remember ;
        local m,t ;
        for m from 0 do
            t := m*(m+3)/2 ;
            if t > n then
                return 0 ;
            elif t = n then
                return 1 ;
            end if;
        end do:
    end proc:
    seq(A023531(n),n=0..40) ; # R. J. Mathar, May 15 2025
  • Mathematica
    If[IntegerQ[(Sqrt[9+8#]-3)/2],1,0]&/@Range[0,100] (* Harvey P. Dale, Jul 27 2011 *)
    a[ n_] := If[ n < 0, 0, Boole @ IntegerQ @ Sqrt[ 8 n + 9]]; (* Michael Somos, May 17 2014 *)
    a[ n_] := SeriesCoefficient[ (EllipticTheta[ 2, 0, x^(1/2)] / (2 x^(1/8)) - 1) / x, {x, 0, n}]; (* Michael Somos, May 17 2014 *)
  • PARI
    {a(n) = if( n<0, 0, issquare(8*n + 9))}; /* Michael Somos, May 17 2014 */
    
  • PARI
    A023531(n)=issquare(8*n+9) \\ M. F. Hasler, Apr 12 2018
    
  • Python
    from math import isqrt
    def A023531(n): return int((k:=n+1<<1)==(m:=isqrt(k))*(m+1)) # Chai Wah Wu, Nov 09 2024
  • Sage
    def A023531_row(n) :
        if n == 0: return [1]
        return [0] + A023531_row(n-1)
    for n in (0..9): print(A023531_row(n))  # Peter Luschny, Jul 22 2012
    

Formula

If (floor(sqrt(2*n))-(2*n/(floor(sqrt(2*n)))) = -1, 1, 0). - Gerald Hillier, Sep 11 2005
a(n) = 1 - A023532(n); a(n) = 1 - mod(floor(((10^(n+2) - 10)/9)10^(n+1 - binomial(floor((1+sqrt(9+8n))/2), 2) - (1+floor(log((10^(n+2) - 10)/9, 10))))), 10). - Paul Barry, May 25 2004
a(n) = floor((sqrt(9+8n)-1)/2) - floor((sqrt(1+8n)-1)/2). - Paul Barry, May 25 2004
a(n) = round(sqrt(2n+3)) - round(sqrt(2n+2)). - Hieronymus Fischer, Aug 06 2007
a(n) = ceiling(2*sqrt(2n+3)) - floor(2*sqrt(2n+2)) - 1. - Hieronymus Fischer, Aug 06 2007
From Franklin T. Adams-Watters, Jun 29 2009: (Start)
G.f.: (1/2 x^{-1/8}theta_2(0,x^{1/2}) - 1)/x, where theta_2 is a Jacobi theta function.
G.f. for triangle: Sum T(n,k) x^n y^k = 1/(1-x*y). Sum T(n,k) x^n y^k / n! = Sum T(n,k) x^n y^k / k! = exp(x*y). Sum T(n,k) x^n y^k / (n! k!) = I_0(2*sqrt(x*y)), where I is the modified Bessel function of the first kind. (End)
a(n) = A000007(m), where m=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Jan 11 2013
The row polynomials are p(n,x) = x^n = (-1)^n n!Lag(n,-n,x), the normalized, associated Laguerre polynomials of order -n. As the prototypical Appell sequence with e.g.f. exp(x*y), its raising operator is R = x and lowering operator, L = d/dx, i.e., R p(n,x) = p(n+1,x), and L p(n,x) = n * p(n-1,x). - Tom Copeland, May 10 2014
a(n) = A010054(n+1) if n >= 0. - Michael Somos, May 17 2014
a(n) = floor(sqrt(2*(n+1)+1/2)-1/2) - floor(sqrt(2*n+1/2)-1/2). - Mikael Aaltonen, Jan 18 2015
a(n) = A003057(n+3) - A003057(n+2). - Robert Israel, Jan 18 2015
a(A000096(n)) = 1; a(A007701(n)) = 0. - Reinhard Zumkeller, Feb 14 2015
Characteristic function of A000096. - M. F. Hasler, Apr 12 2018
Sum_{k=1..n} a(k) ~ sqrt(2*n). - Amiram Eldar, Jan 13 2024

A005614 The binary complement of the infinite Fibonacci word A003849. Start with 1, apply 0->1, 1->10, iterate, take limit.

Original entry on oeis.org

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

Views

Author

Keywords

Comments

Previous name was: The infinite Fibonacci word (start with 1, apply 0->1, 1->10, iterate, take limit).
Characteristic function of A022342. - Philippe Deléham, May 03 2004
a(n) = number of 0's between successive 1's (see also A003589 and A007538). - Eric Angelini, Jul 06 2005
With offset 1 this is the characteristic sequence for Wythoff A-numbers A000201=[1,3,4,6,...].
Eric Angelini's comment made me think that if 1 is defined to be the number of 0's between successive 1's in a string of 0's and 1's, then this string is 101. Applying the same operation to the digits of 101 leads to 101101, the iteration leads to successive palindromes of lengths given by A001911, up to a(n). - Rémi Schulz, Jul 06 2010
For generalized Fibonacci words see A221150, A221151, A221152, ... - Peter Bala, Nov 11 2013
The limiting mean of the first n terms is phi - 1; the limiting variance is phi (A001622). - Clark Kimberling, Mar 12 2014
Apply the difference operator to every column of the Wythoff difference array, A080164, to get an array of Fibonacci numbers, F(h). Replace each F(h) with h, and apply the difference operator to every column. In the resulting array, every column is A005614. - Clark Kimberling, Mar 02 2015
Binary expansion of the rabbit constant A014565. - M. F. Hasler, Nov 10 2018

Examples

			The infinite word is 101101011011010110101101101011...
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • G. Melançon, Factorizing infinite words using Maple, MapleTech journal, vol. 4, no. 1, 1997, pp. 34-42, esp. p. 36.

Crossrefs

Binary complement of A003849, which is the standard form of this sequence.
Two other essentially identical sequences are A096270, A114986.
Subwords: A178992, A171676.
Cf. A000045 (Fibonacci numbers), A001468, A001911, A005206 (partial sums), A014565, A014675, A022342, A036299, A044432, A221150, A221151, A221152.
Cf. A339051 (odd bisection), A339052 (even bisection).
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a005614 n = a005614_list !! n
    a005614_list = map (1 -) a003849_list
    -- Reinhard Zumkeller, Apr 07 2012
    
  • Magma
    [Floor((n+1)*(-1+Sqrt(5))/2)-Floor(n*(-1+Sqrt(5))/2): n in [1..100]]; // Vincenzo Librandi, Jan 17 2019
    
  • Maple
    Digits := 50; u := evalf((1-sqrt(5))/2); A005614 := n->floor((n+1)*u)-floor(n*u);
  • Mathematica
    Nest[ Flatten[ # /. {0 -> {1}, 1 -> {1, 0}}] &, {1}, 10] (* Robert G. Wilson v, Jan 30 2005 *)
    Flatten[Nest[{#, #[[1]]} &, {1, 0}, 9]] (* IWABUCHI Yu(u)ki, Oct 23 2013 *)
    SubstitutionSystem[{0 -> {1}, 1 -> {1, 0}}, {1, 0}, 9] // Last (* Jean-François Alcover, Feb 06 2020 *)
  • PARI
    a(n,w1,s0,s1)=local(w2); for(i=2,n,w2=[ ]; for(k=1,length(w1),w2=concat(w2, if(w1[ k ],s1,s0))); w1=w2); w2
    for(n=2,10,print(n" "a(n,[ 0 ],[ 1 ],[ 1,0 ]))) \\ Gives successive convergents to sequence
    
  • PARI
    /* for m>=1 compute exactly A183136(m+1)+1 terms of the sequence */
    r=(1+sqrt(5))/2;v=[1,0];for(n=2,m,v=concat(v,vector(floor((n+1)/r),i,v[i]));a(n)=v[n];) /* Benoit Cloitre, Jan 16 2013 */
    
  • Python
    from math import isqrt
    def A005614(n): return (n+isqrt(m:=5*(n+2)**2)>>1)-(n+1+isqrt(m-10*n-15)>>1) # Chai Wah Wu, Aug 17 2022

Formula

Define strings S(0)=1, S(1)=10, thereafter S(n)=S(n-1)S(n-2); iterate. Sequence is S(oo). The individual S(n)'s are given in A036299.
a(n) = floor((n+2)*u) - floor((n+1)*u), where u = (-1 + sqrt(5))/2.
Sum_{n>=0} a(n)/2^(n+1) = A014565. - R. J. Mathar, Jul 19 2013
From Peter Bala, Nov 11 2013: (Start)
If we read the present sequence as the digits of a decimal constant c = 0.101101011011010 ... then we have the series representation c = Sum_{n >= 1} 1/10^floor(n*phi). An alternative representation is c = Sum_{n >= 1} 1/10^floor(n/phi) - 10/9.
The constant 9*c has the simple continued fraction representation [0; 1, 10, 10, 100, 1000, ..., 10^Fibonacci(n), ...]. See A010100.
Using this result we can find the alternating series representation c = 1/9 - 9*Sum_{n >= 1} (-1)^(n+1)*(1 + 10^Fibonacci(3*n+1))/( (10^(Fibonacci(3*n - 1)) - 1)*(10^(Fibonacci(3*n + 2)) - 1) ). The series converges very rapidly: for example, the first 10 terms of the series give a value for c accurate to more than 5.7 million decimal places. Cf. A014565. (End)
a(n) = A005206(n+1) - A005206(n). a(2*n) = A339052(n); a(2*n+1) = A339051(n+1). - Peter Bala, Aug 09 2022

Extensions

Corrected by Clark Kimberling, Oct 04 2000
Name corrected by Michel Dekking, Apr 02 2019

A188590 [(n+1)*r] - [n*r], where r = 3/2 + sqrt(13)/2 and [...] denotes the floor function.

Original entry on oeis.org

3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3
Offset: 1

Views

Author

John W. Layman, Apr 04 2011

Keywords

Comments

It appears that this sequence is a fixed-pt of the morphism 3 -> 334, 4 -> 3343, starting with 3. The orbit of 3 under the indicated morphism is 3, 334, 3343343343, 334334334333433433433343343343334, ...
The sequence of the lengths of the words in this orbit appears to be A006190 = {1,3,10,33,109,...}, a solution of the difference equation a(n) = 3*a(n-1) + a(n-2). A root of the auxiliary equation r^2 - 3r -1 = 0 of this difference equation is 3/2 + sqrt(13)/2, the value of r used in the definition of {a(n)}.
See A003849 for the infinite Fibonacci word (start with 0, apply 0->01, 1->0, take limit).
It appears that {a(n)-1} = {2,2,3,2,2,3,2,2,3,2,2,2,3,...} is the same as A003589 (the number of 2's between consecutive 3's in A003589 gives the original sequence). This has been verified up to 2000 terms.

Crossrefs

Programs

  • Mathematica
    r = 3/2 + Sqrt[13]/2; Table[Floor[(n + 1)r] - Floor[n * r], {n, 100}] (* Alonso del Arte, Apr 04 2011 *)

Formula

a(n) = [(n+1)*r] - [n*r], where r = 3/2 + sqrt(13)/2 and [...] denotes the floor function.
Showing 1-3 of 3 results.