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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A119972 a(n) = n * A034947(n).

Original entry on oeis.org

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, 32, 33, 34, -35, 36, 37, -38, -39, 40, 41, 42, -43, -44, 45, -46, -47, -48, 49, 50, -51, 52, 53, -54, -55, -56, 57, 58, -59, -60, 61, -62, -63, 64, 65, 66, -67, 68, 69, -70, -71, 72, 73, 74, -75
Offset: 1

Views

Author

Alford Arnold, Jun 01 2006

Keywords

Comments

Previous name was: Flag n when the first difference of the decimal encoding of the Gray code is negative. (With "flag" meaning negate n when the difference is negative.)
Merge A091072 with minus A091067 maintaining increasing absolute value.

Examples

			A003188 begins 0  1  3  2  6  7  5  4 12  13  15  14  10  11  9 ... so
A055975 begins   1  2 -1  4  1 -2 -1  8  1   2  -1  -4   1  -2  ...
Sequence         1, 2,-3, 4, 5,-6,-7, 8, 9, 10,-11,-12, 13,-14, ...
Negative terms are at positions 3,6,7,11,12,14,..., = A091067.
Positive terms are the complement, which is A091072.
		

Crossrefs

Programs

  • Maple
    isA091067 := proc(n) option remember ; if n mod 4 = 3 then RETURN(true) ; else if n mod 2 = 0 then if isA091067(n/2) then RETURN(true) ; fi ; fi ; RETURN(false) ; fi ; end: A119972 := proc(n) if isA091067(n) then -n ; else n ; fi ; end: for n from 1 to 180 do printf("%d, ",A119972(n)) ; od ; # R. J. Mathar, May 14 2007
    # second Maple program:
    a:= n-> numtheory[jacobi](-1, n)*n:
    seq(a(n), n=1..75);  # Alois P. Heinz, Jan 19 2023
  • Mathematica
    a[n_] := n KroneckerSymbol[-1, n];
    Array[a, 75] (* Jean-François Alcover, Apr 09 2020 *)
  • PARI
    a(n) = n*kronecker(-1, n); \\ Andrew Howroyd, Aug 06 2018

Formula

a(n) = n*Kronecker(-1, n) = n * A034947(n). - Andrew Howroyd, Aug 06 2018

Extensions

More terms from R. J. Mathar, May 14 2007
Keyword:mult added by Andrew Howroyd, Aug 06 2018
New name using existing formula from Joerg Arndt, Jan 19 2023

A381929 Ending positions of runs in the regular paperfolding sequence A034947.

Original entry on oeis.org

2, 3, 5, 7, 10, 12, 13, 15, 18, 19, 21, 24, 26, 28, 29, 31, 34, 35, 37, 39, 42, 44, 45, 48, 50, 51, 53, 56, 58, 60, 61, 63, 66, 67, 69, 71, 74, 76, 77, 79, 82, 83, 85, 88, 90, 92, 93, 96, 98, 99, 101, 103, 106, 108, 109, 112, 114, 115, 117, 120, 122, 124, 125
Offset: 1

Views

Author

Jeffrey Shallit, Mar 10 2025

Keywords

Comments

A "run" is a maximal block of consecutive identical terms.

Examples

			The first few terms of A034947 are 1,1,-1,1,1,-1,-1,1,1,1,-1, and the runs end at positions 2,3,5,7,10,... .
		

Crossrefs

Cf. A034947. A371594 gives the starting positions of the runs, and A088431 gives the length of the runs.

A000364 Euler (or secant or "Zig") numbers: e.g.f. (even powers only) sec(x) = 1/cos(x).

Original entry on oeis.org

1, 1, 5, 61, 1385, 50521, 2702765, 199360981, 19391512145, 2404879675441, 370371188237525, 69348874393137901, 15514534163557086905, 4087072509293123892361, 1252259641403629865468285, 441543893249023104553682821, 177519391579539289436664789665
Offset: 0

Views

Author

Keywords

Comments

Inverse Gudermannian gd^(-1)(x) = log(sec(x) + tan(x)) = log(tan(Pi/4 + x/2)) = arctanh(sin(x)) = 2 * arctanh(tan(x/2)) = 2 * arctanh(csc(x) - cot(x)). - Michael Somos, Mar 19 2011
a(n) is the number of downup permutations of [2n]. Example: a(2)=5 counts 4231, 4132, 3241, 3142, 2143. - David Callan, Nov 21 2011
a(n) is the number of increasing full binary trees on vertices {0,1,2,...,2n} for which the leftmost leaf is labeled 2n. - David Callan, Nov 21 2011
a(n) is the number of unordered increasing trees of size 2n+1 with only even degrees allowed and degree-weight generating function given by cosh(t). - Markus Kuba, Sep 13 2014
a(n) is the number of standard Young tableaux of skew shape (n+1,n,n-1,...,3,2)/(n-1,n-2,...2,1). - Ran Pan, Apr 10 2015
Since cos(z) has a root at z = Pi/2 and no other root in C with a smaller |z|, the radius of convergence of the e.g.f. (intended complex-valued) is Pi/2 = A019669 (see also A028296). - Stanislav Sykora, Oct 07 2016
All terms are odd. - Alois P. Heinz, Jul 22 2018
The sequence starting with a(1) is periodic modulo any odd prime p. The minimal period is (p-1)/2 if p == 1 mod 4 and p-1 if p == 3 mod 4 [Knuth & Buckholtz, 1967, Theorem 2]. - Allen Stenger, Aug 03 2020
Conjecture: taking the sequence [a(n) : n >= 1] modulo an integer k gives a purely periodic sequence with period dividing phi(k). For example, the sequence taken modulo 21 begins [1, 5, 19, 20, 16, 2, 1, 5, 19, 20, 16, 2, 1, 5, 19, 20, 16, 2, 1, 5, 19, ...] with an apparent period of 6 = phi(21)/2. - Peter Bala, May 08 2023

Examples

			G.f. = 1 + x + 5*x^2 + 61*x^3 + 1385*x^4 + 50521*x^5 + 2702765*x^6 + 199360981*x^7 + ...
sec(x) = 1 + 1/2*x^2 + 5/24*x^4 + 61/720*x^6 + ...
From _Gary W. Adamson_, Jul 18 2011: (Start)
The first few rows of matrix M are:
   1,  1,  0,  0,  0, ...
   4,  4,  4,  0,  0, ...
   9,  9,  9,  9,  0, ...
  16, 16, 16, 16, 16, ... (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 810; gives a version with signs: E_{2n} = (-1)^n*a(n) (this is A028296).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
  • J. M. Borwein and D. M. Bailey, Mathematics by Experiment, Peters, Boston, 2004; p. 49
  • J. M. Borwein, D. H. Bailey, and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 141.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 420.
  • G. Chrystal, Algebra, Vol. II, p. 342.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 110.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 69.
  • L. Euler, Inst. Calc. Diff., Section 224.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 444.
  • L. Seidel, Über eine einfache Entstehungsweise der Bernoulli'schen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187.
  • 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).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapters 5 and 33, pages 41, 314.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 269.

Crossrefs

Essentially same as A028296 and A122045.
First column of triangle A060074.
Two main diagonals of triangle A060058 (as iterated sums of squares).
Absolute values of row sums of A160485. - Johannes W. Meijer, Jul 06 2009
Left edge of triangle A210108, see also A125053, A076552. Cf. A255881.
Bisection (even part) of A317139.
The sequences [(-k^2)^n*Euler(2*n, 1/k), n = 0, 1, ...] are: A000007 (k=1), A000364 (k=2), |A210657| (k=3), A000281 (k=4), A272158 (k=5), A002438 (k=6), A273031 (k=7).

Programs

  • Maple
    series(sec(x),x,40): SERIESTOSERIESMULT(%): subs(x=sqrt(y),%): seriestolist(%);
    # end of program
    A000364_list := proc(n) local S,k,j; S[0] := 1;
    for k from 1 to n do S[k] := k*S[k-1] od;
    for k from  1 to n do
        for j from k to n do
            S[j] := (j-k)*S[j-1]+(j-k+1)*S[j] od od;
    seq(S[j], j=1..n)  end:
    A000364_list(16);  # Peter Luschny, Apr 02 2012
    A000364 := proc(n)
        abs(euler(2*n)) ;
    end proc: # R. J. Mathar, Mar 14 2013
  • Mathematica
    Take[ Range[0, 32]! * CoefficientList[ Series[ Sec[x], {x, 0, 32}], x], {1, 32, 2}] (* Robert G. Wilson v, Apr 23 2006 *)
    Table[Abs[EulerE[2n]], {n, 0, 30}] (* Ray Chandler, Mar 20 2007 *)
    a[ n_] := If[ n < 0, 0, With[{m = 2 n}, m! SeriesCoefficient[ Sec[ x], {x, 0, m}]]]; (* Michael Somos, Nov 22 2013 *)
    a[ n_] := If[ n < 0, 0, With[{m = 2 n + 1}, m! SeriesCoefficient[ InverseGudermannian[ x], {x, 0, m}]]]; (* Michael Somos, Nov 22 2013 *)
    a[n_] := Sum[Sum[Binomial[k, m] (-1)^(n+k)/(2^(m-1)) Sum[Binomial[m, j]* (2j-m)^(2n), {j, 0, m/2}] (-1)^(k-m), {m, 0, k}], {k, 1, 2n}]; Table[ a[n], {n, 0, 16}] (* Jean-François Alcover, Jun 26 2019, after Vladimir Kruchinin *)
    a[0] := 1; a[n_] := a[n] = -Sum[a[n - k]/(2 k)!, {k, 1, n}]; Map[(-1)^# (2 #)! a[#] &, Range[0, 16]] (* Oliver Seipel, May 18 2024 *)
  • Maxima
    a(n):=sum(sum(binomial(k,m)*(-1)^(n+k)/(2^(m-1))*sum(binomial(m,j)*(2*j-m)^(2*n),j,0,m/2)*(-1)^(k-m),m,0,k),k,1,2*n); /* Vladimir Kruchinin, Aug 05 2010 */
    
  • Maxima
    a[n]:=if n=0 then 1 else sum(sum((i-k)^(2*n)*binomial(2*k, i)*(-1)^(i+k+n), i, 0, k-1)/ (2^(k-1)), k, 1, 2*n); makelist(a[n], n, 0, 16); /* Vladimir Kruchinin, Oct 05 2012 */
    
  • PARI
    {a(n)=local(CF=1+x*O(x^n));if(n<0,return(0), for(k=1,n,CF=1/(1-(n-k+1)^2*x*CF));return(Vec(CF)[n+1]))} \\ Paul D. Hanna Oct 07 2005
    
  • PARI
    {a(n) = if( n<0, 0, (2*n)! * polcoeff( 1 / cos(x + O(x^(2*n + 1))), 2*n))}; /* Michael Somos, Jun 18 2002 */
    
  • PARI
    {a(n) = my(A); if( n<0, 0, n = 2*n+1 ; A = x * O(x^n); n! * polcoeff( log(1 / cos(x + A) + tan(x + A)), n))}; /* Michael Somos, Aug 15 2007 */
    
  • PARI
    {a(n)=polcoeff(sum(m=0, n, (2*m)!/2^m * x^m/prod(k=1, m, 1+k^2*x+x*O(x^n))), n)} \\ Paul D. Hanna, Sep 20 2012
    
  • PARI
    list(n)=my(v=Vec(1/cos(x+O(x^(2*n+1)))));vector(n,i,v[2*i-1]*(2*i-2)!) \\ Charles R Greathouse IV, Oct 16 2012
    
  • PARI
    a(n)=subst(bernpol(2*n+1),'x,1/4)*4^(2*n+1)*(-1)^(n+1)/(2*n+1) \\ Charles R Greathouse IV, Dec 10 2014
    
  • PARI
    a(n)=abs(eulerfrac(2*n)) \\ Charles R Greathouse IV, Mar 23 2022
    
  • PARI
    \\ Based on an algorithm of Peter Bala, cf. link in A110501.
    upto(n) = my(v1, v2, v3); v1 = vector(n+1, i, 0); v1[1] = 1; v2 = vector(n, i, i^2); v3 = v1; for(i=2, n+1, for(j=2, i-1, v1[j] += v2[i-j+1]*v1[j-1]); v1[i] = v1[i-1]; v3[i] = v1[i]); v3 \\ Mikhail Kurkov, Aug 30 2025
    
  • Python
    from functools import lru_cache
    from math import comb
    @lru_cache(maxsize=None)
    def A000364(n): return 1 if n == 0 else (1 if n % 2 else -1)*sum((-1 if i % 2 else 1)*A000364(i)*comb(2*n,2*i) for i in range(n)) # Chai Wah Wu, Jan 14 2022
    
  • Python
    # after Mikhail Kurkov, based on an algorithm of Peter Bala, cf. link in A110501.
    def euler_list(len: int) -> list[int]:
        if len == 0: return []
        v1 = [1] + [0] * (len - 1)
        v2 = [i**2 for i in range(1, len + 1)]
        result = [0] * len
        result[0] = 1
        for i in range(1, len):
            for j in range(1, i):
                v1[j] += v2[i - j] * v1[j - 1]
            v1[i] = v1[i - 1]
            result[i] = v1[i]
        return result
    print(euler_list(1000))  # Peter Luschny, Aug 30 2025
  • Sage
    # Algorithm of L. Seidel (1877)
    # n -> [a(0), a(1), ..., a(n-1)] for n > 0.
    def A000364_list(len) :
        R = []; A = {-1:0, 0:1}; k = 0; e = 1
        for i in (0..2*len-1) :
            Am = 0; A[k + e] = 0; e = -e
            for j in (0..i) : Am += A[k]; A[k] = Am; k += e
            if e < 0 : R.append(A[-i//2])
        return R
    A000364_list(17) # Peter Luschny, Mar 31 2012
    

Formula

E.g.f.: Sum_{n >= 0} a(n) * x^(2*n) / (2*n)! = sec(x). - Michael Somos, Aug 15 2007
E.g.f.: Sum_{n >= 0} a(n) * x^(2*n+1) / (2*n+1)! = gd^(-1)(x). - Michael Somos, Aug 15 2007
E.g.f.: Sum_{n >= 0} a(n)*x^(2*n+1)/(2*n+1)! = 2*arctanh(cosec(x)-cotan(x)). - Ralf Stephan, Dec 16 2004
Pi/4 - [Sum_{k=0..n-1} (-1)^k/(2*k+1)] ~ (1/2)*[Sum_{k>=0} (-1)^k*E(k)/(2*n)^(2k+1)] for positive even n. [Borwein, Borwein, and Dilcher]
Also, for positive odd n, log(2) - Sum_{k = 1..(n-1)/2} (-1)^(k-1)/k ~ (-1)^((n-1)/2) * Sum_{k >= 0} (-1)^k*E(k)/n^(2*k+1), where E(k) is the k-th Euler number, by Borwein, Borwein, and Dilcher, Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then replace x with (n-1)/2. - Peter Bala, Oct 29 2016
Let M_n be the n X n matrix M_n(i, j) = binomial(2*i, 2*(j-1)) = A086645(i, j-1); then for n>0, a(n) = det(M_n); example: det([1, 1, 0, 0; 1, 6, 1, 0; 1, 15, 15, 1; 1, 28, 70, 28 ]) = 1385. - Philippe Deléham, Sep 04 2005
This sequence is also (-1)^n*EulerE(2*n) or abs(EulerE(2*n)). - Paul Abbott (paul(AT)physics.uwa.edu.au), Apr 14 2006
a(n) = 2^n * E_n(1/2), where E_n(x) is an Euler polynomial.
a(k) = a(j) (mod 2^n) if and only if k == j (mod 2^n) (k and j are even). [Stern; see also Wagstaff and Sun]
E_k(3^(k+1)+1)/4 = (3^k/2)*Sum_{j=0..2^n-1} (-1)^(j-1)*(2j+1)^k*[(3j+1)/2^n] (mod 2^n) where k is even and [x] is the greatest integer function. [Sun]
a(n) ~ 2^(2*n+2)*(2*n)!/Pi^(2*n+1) as n -> infinity. [corrected by Vaclav Kotesovec, Jul 10 2021]
a(n) = Sum_{k=0..n} A094665(n, k)*2^(n-k). - Philippe Deléham, Jun 10 2004
Recurrence: a(n) = -(-1)^n*Sum_{i=0..n-1} (-1)^i*a(i)*binomial(2*n, 2*i). - Ralf Stephan, Feb 24 2005
O.g.f.: 1/(1-x/(1-4*x/(1-9*x/(1-16*x/(...-n^2*x/(1-...)))))) (continued fraction due to T. J. Stieltjes). - Paul D. Hanna, Oct 07 2005
a(n) = (Integral_{t=0..Pi} log(tan(t/2)^2)^(2n)dt)/Pi^(2n+1). - Logan Kleinwaks (kleinwaks(AT)alumni.princeton.edu), Mar 15 2007
From Peter Bala, Mar 24 2009: (Start)
Basic hypergeometric generating function: 2*exp(-t)*Sum {n >= 0} Product_{k = 1..n} (1-exp(-(4*k-2)*t))*exp(-2*n*t)/Product_{k = 1..n+1} (1+exp(-(4*k-2)*t)) = 1 + t + 5*t^2/2! + 61*t^3/3! + .... For other sequences with generating functions of a similar type see A000464, A002105, A002439, A079144 and A158690.
a(n) = 2*(-1)^n*L(-2*n), where L(s) is the Dirichlet L-function L(s) = 1 - 1/3^s + 1/5^s - + .... (End)
Sum_{n>=0} a(n)*z^(2*n)/(4*n)!! = Beta(1/2-z/(2*Pi),1/2+z/(2*Pi))/Beta(1/2,1/2) with Beta(z,w) the Beta function. - Johannes W. Meijer, Jul 06 2009
a(n) = Sum_(Sum_(binomial(k,m)*(-1)^(n+k)/(2^(m-1))*Sum_(binomial(m,j)*(2*j-m)^(2*n),j,0,m/2)*(-1)^(k-m),m,0,k),k,1,2*n), n>0. - Vladimir Kruchinin, Aug 05 2010
If n is prime, then a(n)==1 (mod 2*n). - Vladimir Shevelev, Sep 04 2010
From Peter Bala, Jan 21 2011: (Start)
(1)... a(n) = (-1/4)^n*B(2*n,-1),
where {B(n,x)}n>=1 = [1, 1+x, 1+6*x+x^2, 1+23*x+23*x^2+x^3, ...] is the sequence of Eulerian polynomials of type B - see A060187. Equivalently,
(2)... a(n) = Sum_{k = 0..2*n} Sum_{j = 0..k} (-1)^(n-j) *binomial(2*n+1,k-j)*(j+1/2)^(2*n).
We also have
(3)... a(n) = 2*A(2*n,i)/(1+i)^(2*n+1),
where i = sqrt(-1) and where {A(n,x)}n>=1 = [x, x + x^2, x + 4*x^2 + x^3, ...] denotes the sequence of Eulerian polynomials - see A008292. Equivalently,
(4)... a(n) = i*Sum_{k = 1..2*n} (-1)^(n+k)*k!*Stirling2(2*n,k) *((1+i)/2)^(k-1)
= i*Sum_{k = 1..2*n} (-1)^(n+k)*((1+i)/2)^(k-1) Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*j^(2*n).
Either this explicit formula for a(n) or (2) above may be used to obtain congruence results for a(n). For example, for prime p
(5a)... a(p) = 1 (mod p)
(5b)... a(2*p) = 5 (mod p)
and for odd prime p
(6a)... a((p+1)/2) = (-1)^((p-1)/2) (mod p)
(6b)... a((p-1)/2) = -1 + (-1)^((p-1)/2) (mod p).
(End)
a(n) = (-1)^n*2^(4*n+1)*(zeta(-2*n,1/4) - zeta(-2*n,3/4)). - Gerry Martens, May 27 2011
a(n) may be expressed as a sum of multinomials taken over all compositions of 2*n into even parts (Vella 2008): a(n) = Sum_{compositions 2*i_1 + ... + 2*i_k = 2*n} (-1)^(n+k)* multinomial(2*n, 2*i_1, ..., 2*i_k). For example, there are 4 compositions of the number 6 into even parts, namely 6, 4+2, 2+4 and 2+2+2, and hence a(3) = 6!/6! - 6!/(4!*2!) - 6!/(2!*4!) + 6!/(2!*2!*2!) = 61. A companion formula expressing a(n) as a sum of multinomials taken over the compositions of 2*n-1 into odd parts has been given by Malenfant 2011. - Peter Bala, Jul 07 2011
a(n) = the upper left term in M^n, where M is an infinite square production matrix; M[i,j] = A000290(i) = i^2, i >= 1 and 1 <= j <= i+1, and M[i,j] = 0, i >= 1 and j >= i+2 (see examples). - Gary W. Adamson, Jul 18 2011
E.g.f. A'(x) satisfies the differential equation A'(x)=cos(A(x)). - Vladimir Kruchinin, Nov 03 2011
From Peter Bala, Nov 28 2011: (Start)
a(n) = D^(2*n)(cosh(x)) evaluated at x = 0, where D is the operator cosh(x)*d/dx. a(n) = D^(2*n-1)(f(x)) evaluated at x = 0, where f(x) = 1+x+x^2/2! and D is the operator f(x)*d/dx.
Other generating functions: cosh(Integral_{t = 0..x} 1/cos(t)) dt = 1 + x^2/2! + 5*x^4/4! + 61*x^6/6! + 1385*x^8/8! + .... Cf. A012131.
A(x) := arcsinh(tan(x)) = log( sec(x) + tan(x) ) = x + x^3/3! + 5*x^5/5! + 61*x^7/7! + 1385*x^9/9! + .... A(x) satisfies A'(x) = cosh(A(x)).
B(x) := Series reversion( log(sec(x) + tan(x)) ) = x - x^3/3! + 5*x^5/5! - 61*x^7/7! + 1385*x^9/9! - ... = arctan(sinh(x)). B(x) satisfies B'(x) = cos(B(x)). (End)
HANKEL transform is A097476. PSUM transform is A173226. - Michael Somos, May 12 2012
a(n+1) - a(n) = A006212(2*n). - Michael Somos, May 12 2012
a(0) = 1 and, for n > 0, a(n) = (-1)^n*((4*n+1)/(2*n+1) - Sum_{k = 1..n} (4^(2*k)/2*k)*binomial(2*n,2*k-1)*A000367(k)/A002445(k)); see the Bucur et al. link. - L. Edson Jeffery, Sep 17 2012
O.g.f.: Sum_{n>=0} (2*n)!/2^n * x^n / Product_{k=1..n} (1 + k^2*x). - Paul D. Hanna, Sep 20 2012
From Sergei N. Gladkovskii, Oct 31 2011 to Oct 11 2013: (Start)
Continued fractions:
E.g.f.: (sec(x)) = 1+x^2/T(0), T(k) = 2(k+1)(2k+1) - x^2 + x^2*(2k+1)(2k+2)/T(k+1).
E.g.f.: 2/Q(0) where Q(k) = 1 + 1/(1 - x^2/(x^2 - 2*(k+1)*(2*k+1)/Q(k+1))).
G.f.: 1/Q(0) where Q(k) = 1 + x*k*(3*k-1) - x*(k+1)*(2*k+1)*(x*k^2+1)/Q(k+1).
E.g.f.: (2 + x^2 + 2*U(0))/(2 + (2 - x^2)*U(0)) where U(k)= 4*k + 4 + 1/( 1 + x^2/(2 - x^2 + (2*k+3)*(2*k+4)/U(k+1))).
E.g.f.: 1/cos(x) = 8*(x^2+1)/(4*x^2 + 8 - x^4*U(0)) where U(k) = 1 + 4*(k+1)*(k+2)/(2*k+3 - x^2*(2*k+3)/(x^2 - 8*(k+1)*(k+2)*(k+3)/U(k+1))).
G.f.: 1/U(0) where U(k) = 1 + x - x*(2*k+1)*(2*k+2)/(1 - x*(2*k+1)*(2*k+2)/U(k+1)).
G.f.: 1 + x/G(0) where G(k) = 1 + x - x*(2*k+2)*(2*k+3)/(1 - x*(2*k+2)*(2*k+3)/G(k+1)).
Let F(x) = sec(x^(1/2)) = Sum_{n>=0} a(n)*x^n/(2*n)!, then F(x)=2/(Q(0) + 1) where Q(k)= 1 - x/(2*k+1)/(2*k+2)/(1 - 1/(1 + 1/Q(k+1))).
G.f.: Q(0), where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/Q(k+1)).
E.g.f.: 1/cos(x) = 1 + x^2/(2-x^2)*Q(0), where Q(k) = 1 - 2*x^2*(k+1)*(2*k+1)/( 2*x^2*(k+1)*(2*k+1)+ (12-x^2 + 14*k + 4*k^2)*(2-x^2 + 6*k + 4*k^2)/Q(k+1)). (End)
a(n) = Sum_{k=1..2*n} (Sum_{i=0..k-1} (i-k)^(2*n)*binomial(2*k,i)*(-1)^(i+k+n)) / 2^(k-1) for n>0, a(0)=1. - Vladimir Kruchinin, Oct 05 2012
It appears that a(n) = 3*A076552(n -1) + 2*(-1)^n for n >= 1. Conjectural congruences: a(2*n) == 5 (mod 60) for n >= 1 and a(2*n+1) == 1 (mod 60) for n >= 0. - Peter Bala, Jul 26 2013
From Peter Bala, Mar 09 2015: (Start)
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 - sqrt(-x)*(2*k + 1)) = Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 + x*(2*k + 1)^2).
O.g.f. is 1 + x*d/dx(log(F(x))), where F(x) = 1 + x + 3*x^2 + 23*x^3 + 371*x^4 + ... is the o.g.f. for A255881. (End)
Sum_(n >= 1, A034947(n)/n^(2d+1)) = a(d)*Pi^(2d+1)/(2^(2d+2)-2)(2d)! for d >= 0; see Allouche and Sondow, 2015. - Jonathan Sondow, Mar 21 2015
Asymptotic expansion: 4*(4*n/(Pi*e))^(2*n+1/2)*exp(1/2+1/(24*n)-1/(2880*n^3) +1/(40320*n^5)-...). (See the Luschny link.) - Peter Luschny, Jul 14 2015
a(n) = 2*(-1)^n*Im(Li_{-2n}(i)), where Li_n(x) is polylogarithm, i=sqrt(-1). - Vladimir Reshetnikov, Oct 22 2015
Limit_{n->infinity} ((2*n)!/a(n))^(1/(2*n)) = Pi/2. - Stanislav Sykora, Oct 07 2016
O.g.f.: 1/(1 + x - 2*x/(1 - 2*x/(1 + x - 12*x/(1 - 12*x/(1 + x - 30*x/(1 - 30*x/(1 + x - ... - (2*n - 1)*(2*n)*x/(1 - (2*n - 1)*(2*n)*x/(1 + x - ... ))))))))). - Peter Bala, Nov 09 2017
For n>0, a(n) = (-PolyGamma(2*n, 1/4) / 2^(2*n - 1) - (2^(2*n + 2) - 2) * Gamma(2*n + 1) * zeta(2*n + 1)) / Pi^(2*n + 1). - Vaclav Kotesovec, May 04 2020
a(n) ~ 2^(4*n + 3) * n^(2*n + 1/2) / (Pi^(2*n + 1/2) * exp(2*n)) * exp(Sum_{k>=1} bernoulli(k+1) / (k*(k+1)*2^k*n^k)). - Vaclav Kotesovec, Mar 05 2021
Peter Bala's conjectured congruences, a(2n) == 5 (mod 60) for n >= 1 and a(2n + 1) == 1 (mod 60), hold due to the results of Stern (mod 4) and Knuth & Buckholtz (mod 3 and 5). - Charles R Greathouse IV, Mar 23 2022

Extensions

Typo in name corrected by Anders Claesson, Dec 01 2015

A005811 Number of runs in binary expansion of n (n>0); number of 1's in Gray code for n.

Original entry on oeis.org

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

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Author

Keywords

Comments

Starting with a(1) = 0 mirror all initial 2^k segments and increase by one.
a(n) gives the net rotation (measured in right angles) after taking n steps along a dragon curve. - Christopher Hendrie (hendrie(AT)acm.org), Sep 11 2002
This sequence generates A082410: (0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...) and A014577; identical to the latter except starting 1, 1, 0, ...; by writing a "1" if a(n+1) > a(n); if not, write "0". E.g., A014577(2) = 0, since a(3) < a(2), or 1 < 2. - Gary W. Adamson, Sep 20 2003
Starting with 1 = partial sums of A034947: (1, 1, -1, 1, 1, -1, -1, 1, 1, 1, ...). - Gary W. Adamson, Jul 23 2008
The composer Per Nørgård's name is also written in the OEIS as Per Noergaard.
Can be used as a binomial transform operator: Let a(n) = the n-th term in any S(n); then extract 2^k strings, adding the terms. This results in the binomial transform of S(n). Say S(n) = 1, 3, 5, ...; then we obtain the strings: (1), (3, 1), (3, 5, 3, 1), (3, 5, 7, 5, 3, 5, 3, 1), ...; = the binomial transform of (1, 3, 5, ...) = (1, 4, 12, 32, 80, ...). Example: the 8-bit string has a sum of 32 with a distribution of (1, 3, 3, 1) or one 1, three 3's, three 5's, and one 7; as expected. - Gary W. Adamson, Jun 21 2012
Considers all positive odd numbers as nodes of a graph. Two nodes are connected if and only if the sum of the two corresponding odd numbers is a power of 2. Then a(n) is the distance between 2n + 1 and 1. - Jianing Song, Apr 20 2019

Examples

			Considered as a triangle with 2^k terms per row, the first few rows are:
  1
  2, 1
  2, 3, 2, 1
  2, 3, 4, 3, 2, 3, 2, 1
  ...
The n-th row becomes right half of next row; left half is mirrored terms of n-th row increased by one. - _Gary W. Adamson_, Jun 20 2012
		

References

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

Crossrefs

Cf. A037834 (-1), A088748 (+1), A246960 (mod 4), A034947 (first differences), A000975 (indices of record highs), A173318 (partial sums).
Partial sums of A112347. Recursion depth of A035327.

Programs

  • Haskell
    import Data.List (group)
    a005811 0 = 0
    a005811 n = length $ group $ a030308_row n
    a005811_list = 0 : f [1] where
       f (x:xs) = x : f (xs ++ [x + x `mod` 2, x + 1 - x `mod` 2])
    -- Reinhard Zumkeller, Feb 16 2013, Mar 07 2011
    
  • Maple
    A005811 := proc(n)
        local i, b, ans;
        if n = 0 then
            return 0 ;
        end if;
        ans := 1;
        b := convert(n, base, 2);
        for i from nops(b)-1 to 1 by -1 do
            if b[ i+1 ]<>b[ i ] then
                ans := ans+1
            fi
        od;
        return ans ;
    end proc:
    seq(A005811(i), i=1..50) ;
    # second Maple program:
    a:= n-> add(i, i=Bits[Split](Bits[Xor](n, iquo(n, 2)))):
    seq(a(n), n=0..100);  # Alois P. Heinz, Feb 01 2023
  • Mathematica
    Table[ Length[ Length/@Split[ IntegerDigits[ n, 2 ] ] ], {n, 1, 255} ]
    a[n_] := DigitCount[BitXor[n, Floor[n/2]], 2, 1]; Array[a, 100, 0] (* Amiram Eldar, Jul 11 2024 *)
  • PARI
    a(n)=sum(k=1,n,(-1)^((k/2^valuation(k,2)-1)/2))
    
  • PARI
    a(n)=if(n<1,0,a(n\2)+(a(n\2)+n)%2) \\ Benoit Cloitre, Jan 20 2014
    
  • PARI
    a(n) = hammingweight(bitxor(n, n>>1));  \\ Gheorghe Coserea, Sep 03 2015
    
  • Python
    def a(n): return bin(n^(n>>1))[2:].count("1") # Indranil Ghosh, Apr 29 2017

Formula

a(2^k + i) = a(2^k - i + 1) + 1 for k >= 0 and 0 < i <= 2^k. - Reinhard Zumkeller, Aug 14 2001
a(2n+1) = 2a(n) - a(2n) + 1, a(4n) = a(2n), a(4n+2) = 1 + a(2n+1).
a(j+1) = a(j) + (-1)^A014707(j). - Christopher Hendrie (hendrie(AT)acm.org), Sep 11 2002
G.f.: (1/(1-x)) * Sum_{k>=0} x^2^k/(1+x^2^(k+1)). - Ralf Stephan, May 02 2003
Delete the 0, make subsets of 2^n terms; and reverse the terms in each subset to generate A088696. - Gary W. Adamson, Oct 19 2003
a(0) = 0, a(2n) = a(n) + [n odd], a(2n+1) = a(n) + [n even]. - Ralf Stephan, Oct 20 2003
a(n) = Sum_{k=1..n} (-1)^((k/2^A007814(k)-1)/2) = Sum_{k=1..n} (-1)^A025480(k-1). - Ralf Stephan, Oct 29 2003
a(n) = A069010(n) + A033264(n). - Ralf Stephan, Oct 29 2003
a(0) = 0 then a(n) = a(floor(n/2)) + (a(floor(n/2)) + n) mod 2. - Benoit Cloitre, Jan 20 2014
a(n) = A037834(n) + 1.
a(n) = A000120(A003188(n)). - Amiram Eldar, Jul 11 2024

Extensions

Additional description from Wouter Meeussen

A056594 Period 4: repeat [1,0,-1,0]; expansion of 1/(1 + x^2).

Original entry on oeis.org

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

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Author

Wolfdieter Lang, Aug 04 2000

Keywords

Comments

G.f. is inverse of cyclotomic(4,x). Unsigned: A000035(n+1).
Real part of i^n and imaginary part of i^(n+1), i=sqrt(-1). - Reinhard Zumkeller, Jul 22 2007
The BINOMIAL transform generates A009116(n); the inverse BINOMIAL transform generates (-1)^n*A009116(n). - R. J. Mathar, Apr 07 2008
a(n-1), n >= 1, is the nontrivial Dirichlet character modulo 4, called Chi_2(4;n) (the trivial one is Chi_1(4;n) given by periodic(1,0) = A000035(n)). See the Apostol reference, p. 139, the k = 4, phi(k) = 2 table. - Wolfdieter Lang, Jun 21 2011
a(n-1), n >= 1, is the character of the Dirichlet beta function. - Daniel Forgues, Sep 15 2012
a(n-1), n >= 1, is also the (strongly) multiplicative function h(n) of Theorem 5.12, p. 150, of the Niven-Zuckerman reference. See the formula section. This function h(n) can be employed to count the integer solutions to n = x^2 + y^2. See A002654 for a comment with the formula. - Wolfdieter Lang, Apr 19 2013
This sequence is duplicated in A101455 but with offset 1. - Gary Detlefs, Oct 04 2013
For n >= 2 this gives the determinant of the bipartite graph with 2*n nodes and the adjacency matrix A(n) with elements A(n;1,2) = 1 = A(n;n,n-1), and for 1 < i < n A(n;i,i+1) = 1 = A(n;i,i-1), otherwise 0. - Wolfdieter Lang, Jun 25 2023

Examples

			With a(n-1) = h(n) of Niven-Zuckerman: a(62) = h(63) = h(3^2*7^1) = (-1)^(2*1)*(-1)^(1*3) = -1 = h(3)^2*h(7) = a(2)^2*a(6) = (-1)^2*(-1) = -1. - _Wolfdieter Lang_, Apr 19 2013
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1986.
  • I. S. Gradstein and I. M. Ryshik, Tables of series, products, and integrals, Volume 1, Verlag Harri Deutsch, 1981.
  • Ivan Niven and Herbert S. Zuckerman, An Introduction to the Theory of Numbers, New York: John Wiley (1980), p. 150.
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 32, equation 32:6:1 at page 300.

Crossrefs

Cf. A049310, A074661, A131852, A002654, A146559 (binomial transform).

Programs

  • Magma
    &cat[ [1, 0, -1, 0]: n in [0..23] ]; // Bruno Berselli, Feb 08 2011
    
  • Maple
    A056594 := n->(1-irem(n,2))*(-1)^iquo(n,2); # Peter Luschny, Jul 27 2011
  • Mathematica
    CoefficientList[Series[1/(1 + x^2), {x, 0, 50}], x]
    a[n_]:= KroneckerSymbol[-4,n+1];Table[a[n],{n,0,93}] (* Thanks to Jean-François Alcover. - Wolfdieter Lang, May 31 2013 *)
    CoefficientList[Series[1/Cyclotomic[4, x], {x, 0, 100}], x] (* Vincenzo Librandi, Apr 03 2014 *)
  • Maxima
    A056594(n) := block(
            [1,0,-1,0][1+mod(n,4)]
    )$ /* R. J. Mathar, Mar 19 2012 */
    
  • PARI
    {a(n) = real( I^n )}
    
  • PARI
    {a(n) = kronecker(-4, n+1) }
    
  • Python
    def A056594(n): return (1,0,-1,0)[n&3] # Chai Wah Wu, Sep 23 2023

Formula

G.f.: 1/(1+x^2).
E.g.f.: cos(x).
a(n) = (1/2)*((-i)^n + i^n), where i = sqrt(-1). - Mitch Harris, Apr 19 2005
a(n) = (1/2)*((-1)^(n+floor(n/2)) + (-1)^floor(n/2)).
Recurrence: a(n)=a(n-4), a(0)=1, a(1)=0, a(2)=-1, a(3)=0.
a(n) = T(n, 0) = A053120(n, 0); T(n, x) Chebyshev polynomials of the first kind. - Wolfdieter Lang, Aug 21 2009
a(n) = S(n, 0) = A049310(n, 0); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind.
Sum_{k>=0} a(k)/(k+1) = Pi/4. - Jaume Oliver Lafont, Mar 30 2010
a(n) = Sum_{k=0..n} A101950(n,k)*(-1)^k. - Philippe Deléham, Feb 10 2012
a(n) = (1/2)*(1 + (-1)^n)*(-1)^(n/2). - Bruno Berselli, Mar 13 2012
a(0) = 1, a(n-1) = 0 if n is even, a(n-1) = Product_{j=1..m} (-1)^(e_j*(p_j-1)/2) if the odd n-1 = p_1^(e_1)*p_2^(e_2)*...*p_m^(e_m) with distinct odd primes p_j, j=1..m. See the function h(n) of Theorem 5.12 of the Niven-Zuckerman reference. - Wolfdieter Lang, Apr 19 2013
a(n) = (-4/(n+1)), n >= 0, where (k/n) is the Kronecker symbol. See the Eric Weisstein and Wikipedia links. Thanks to Wesley Ivan Hurt. - Wolfdieter Lang, May 31 2013
a(n) = R(n,0)/2 with the row polynomials R of A127672. This follows from the product of the zeros of R, and the formula Product_{k=0..n-1} 2*cos((2*k+1)*Pi/(2*n)) = (1 + (-1)^n)*(-1)^(n/2), n >= 1 (see the Gradstein and Ryshik reference, p. 63, 1.396 4., with x = sqrt(-1)). - Wolfdieter Lang, Oct 21 2013
a(n) = Sum_{k=0..n} i^(k*(k+1)), where i=sqrt(-1). - Bruno Berselli, Mar 11 2015
Dirichlet g.f. of a(n) shifted right: L(chi_2(4),s) = beta(s) = (1-2^(-s))*(d.g.f. of A034947), see comments by Lang and Forgues. - Ralf Stephan, Mar 27 2015
a(n) = cos(n*Pi/2). - Ridouane Oudra, Sep 29 2024

A014577 The regular paper-folding sequence (or dragon curve sequence). Alphabet {1,0}.

Original entry on oeis.org

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

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Author

Keywords

Comments

a(n) is the complement of the bit to the left of the least significant "1" in the binary expansion of n+1. E.g., n = 3, n+1 = 4 = 100_2, so a(3) = (complement of bit to left of 1) = 1. - Robert L. Brown, Nov 28 2001 [Adjusted to match offset by N. J. A. Sloane, Apr 15 2021]
To construct the sequence: start from 1,(..),0,(..),1,(..),0,(..),1,(..),0,(..),1,(..),0,... and fill undefined places with the sequence itself. - Benoit Cloitre, Jul 08 2007
This is the characteristic function of A091072 - 1. - Gary W. Adamson, Apr 11 2010
Turns (by 90 degrees) of the Heighway dragon which can be rendered as follows: [Init] Set n=0 and direction=0. [Draw] Draw a unit line (in the current direction). Turn left/right if a(n) is zero/nonzero respectively. [Next] Set n=n+1 and goto (draw). See fxtbook link below. - Joerg Arndt, Apr 15 2010
Sequence can be obtained by the L-system with rules L->L1R, R->L0R, 1->1, 0->0, starting with L, and then dropping all L's and R's (see example). - Joerg Arndt, Aug 28 2011
From Gary W. Adamson, Jun 20 2012: (Start)
One half of the infinite Farey tree can be mapped one-to-one onto A014577 since both sequences can be derived directly from the binary. First few terms are
1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...
1/2 2/3 1/3 3/4 3/5 2/5 1/4 4/5 5/7 5/8, ...
Infinite Farey tree fractions can be derived from the binary by appending a repeat of rightmost binary term to the right, then recording the number of runs to obtain the continued fraction representation. Example: 9 = 1001 which becomes 10011 which becomes [1,2,2] = 5/7. (End)
The sequence can be considered as a binomial transform operator for a target sequence S(n). Replace the first 1 in A014577 with the first term in S(n), then for successive "1" term in A014577, map the next higher term in S(n). If "0" in A014577, map the next lower term in S(n). Using the sequence S(n) = (1, 3, 5, 7, ...), we obtain (1), (3, 1), (3, 5, 3, 1), (3, 5, 7, 5, 3, 5, 3, 1), .... Then parse the terms into subsequences of 2^k terms, adding the terms in each string. We obtain (1, 4, 12, 32, 80, ...), the binomial transform of (1, 3, 5, 7, ...). The 8-bit string has one 1, three 5's, three 7's and one 1) as expected, or (1, 3, 3, 1) dot (1, 3, 5, 7). - Gary W. Adamson, Jun 24 2012
From Gary W. Adamson, May 29 2013: (Start)
The sequence can be generated directly from the lengths of continued fraction representations of fractions in one half of the Stern-Brocot tree (fractions between 0 and 1):
1/2
1/3 2/3
1/4 2/5 3/5 3/4
1/5 2/7 3/8 3/7 4/7 5/8 5/7 4/5
...
and their corresponding continued fraction representations are:
[2]
[3] [1,2]
[4] [2,2] [1,1,2] [1,3]
[5] [3,2] [2,1,2] [2,3] [1,1,3] [1,1,1,2] [1,2,2] [1,4]
...
Record the lengths by rows then reverse rows, getting:
1,
2, 1,
2, 3, 2, 1,
2, 3, 4, 3, 2, 3, 2, 1,
...
start with "1" and if the next term is greater than the current term, record a 1, otherwise 0; getting the present sequence, the Harter-Heighway dragon curve. (End)
The paper-folding word "110110011100100111011000..." can be created by concatenating the terms of a fixed point of the morphism or string substitution rule: 00 -> 1000, 01 -> 1001, 10 -> 1100 & 11 -> 1101, beginning with "11". - Robert G. Wilson v, Jun 11 2015
From Gary W. Adamson, Jun 04 2021: (Start)
Since the Heighway dragon is composed of right angles, it can be mapped with unit trajectories (Right = 1, Left = (-1), Up = i and Down = -i) on the complex plane where i = sqrt(-1). The initial (0th) iterate is chosen in this case to be the unit line from (0,0) to (-1,0). Then follow the directions below as indicated, resulting in a reflected variant of the dragon curve shown at the Eric Weisstein link. The conjectured system of complex plane trajectories is:
0 -1
1 -1, i
2 -1, i, 1, i
3 -1, i, 1, i, 1, -i, 1, i
4 -1, i, 1, i, 1, -i, 1, i, 1, -i, -1, -i, 1, -i, 1, i
...
The conjecture succeeds through the 4th iterate. It appears that to generate the (n+1)-th row, bring down the n-th row as the left half of row (n+1). For the right half of row (n+1), bring down the n-th row but change the signs of the first half of row n. For example, to get the complex plane instructions for the third iterate of the dragon curve, bring down (-1, i, 1, i) as the left half, and the right half is (1, -i, 1, i). (End)
From Gary W. Adamson, Jun 09 2021: (Start)
Partial sums of the iterate trajectories produce a sequence of complex addresses for unit segments. Partial sums of row 4 are: -1, (-1+i), i, 2i, (1+2i), (1+i), (2+i), (2+2i), (3+2i), (3+i), (2+i), 2, 3, (3-i), (4-i), 4. (zeros are omitted with terms of the form a+0i). The reflected variant of the dragon curve has the 0th iterate from (0,0) to (1,0), and the respective addresses simply change the signs of the real terms. (End)

Examples

			1 + x + x^3 + x^4 + x^7 + x^8 + x^9 + x^12 + x^15 + x^16 + x^17 + x^19 + ...
From _Joerg Arndt_, Aug 28 2011: (Start)
Generation via string substitution:
Start: L
Rules:
  L --> L1R
  R --> L0R
  0 --> 0
  1 --> 1
-------------
0:   (#=1)
  L
1:   (#=3)
  L1R
2:   (#=7)
  L1R1L0R
3:   (#=15)
  L1R1L0R1L1R0L0R
4:   (#=31)
  L1R1L0R1L1R0L0R1L1R1L0R0L1R0L0R
5:   (#=63)
  L1R1L0R1L1R0L0R1L1R1L0R0L1R0L0R1L1R1L0R1L1R0L0R0L1R1L0R0L1R0L0R
Drop all L and R to obtain 1101100111001001110110001100100
(End)
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, pp. 155, 182.
  • Chandler Davis and Donald E. Knuth, Number Representations and Dragon Curves -- I and II, Journal of Recreational Mathematics, volume 3, number 2, April 1970, pages 66-81, and number 3, July 1970, pages 133-149. Reprinted in Donald E. Knuth, Selected Papers on Fun and Games, CSLI Publications, 2010, pages 571-614.
  • Michel Dekking, 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).
  • Martin Gardner, Mathematical Magic Show, New York: Vintage, pp. 207-209 and 215-220, 1978.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Crossrefs

Cf. A059125, A065339, A005811, A220466, A088748, A091072, A343173 (first differences), A343174 (partial sums).
The two bisections are A000035 and the sequence itself.
See A343181 for prefixes of length 2^k-1.

Programs

  • Magma
    [(1+KroneckerSymbol(-1,n))/2: n in [1..100]]; // Vincenzo Librandi, Aug 05 2015
    
  • Magma
    [Floor(1/2*(1+(-1)^(1/2*((n+1)/2^Valuation(n+1, 2)-1)))): n in [0..100]]; // Vincenzo Librandi, Aug 05 2015
    
  • Maple
    nmax:=98: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 0 to ceil(nmax/(p+2))+1 do a((2*n+1)*2^p-1) := (n+1) mod 2 od: od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Jan 28 2013
    a014577 := proc(n) local p,s1,s2,i;
    if n=0 then return(1); fi;
    s1:=convert(n,base,2); s2:=nops(s1);
    for i from 1 to s2 do if s1[i]=1 then p:=i; break; fi; od:
    if p <= s2-1 then 1-s1[p+1]; else 1; fi; end;
    [seq(a014577(i),i=1..120)]; # N. J. A. Sloane, Apr 08 2021
    # third Maple program:
    a:= n-> 1-irem(iquo((n+1)/2^padic[ordp](n+1, 2), 2), 2):
    seq(a(n), n=0..120);  # Alois P. Heinz, Apr 08 2021
  • Mathematica
    a[n_] := Boole[EvenQ[((n+1)/2^IntegerExponent[n+1, 2]-1)/2]]; Table[a[n], {n, 0, 98}] (* Jean-François Alcover, Feb 16 2012, after Gary W. Adamson, updated Nov 21 2014 *)
    Table[1-(((Mod[#1,2^(#2+2)]/2^#2)&[n,IntegerExponent[n,2]])-1)/2,{n,1,100,1}] (* WolframAlpha compatible code; Robert L. Brown, Jan 06 2015 *)
    MapThread[(a[x_/;IntegerQ[(x-#1)/4]]:= #2)&,{{1,3},{1,0}}];a[x_/;IntegerQ[x/2]]:=a[x/2];a/@ Range[100] (* Bradley Klee, Aug 04 2015 *)
    (1 + JacobiSymbol[-1, Range[100]])/2 (* Paolo Xausa, May 22 2024 *)
    Array[Boole[BitAnd[#, BitAnd[#, -#]*2] == 0] &, 100] (* Paolo Xausa, May 22 2024, after Joerg Arndt C++ code *)
  • PARI
    {a(n) = if( n%2, a(n\2), 1 - (n/2%2))} /* Michael Somos, Feb 05 2012 */
    
  • PARI
    a(n)=1/2*(1+(-1)^(1/2*((n+1)/2^valuation(n+1,2)-1))) \\ Ralf Stephan, Sep 02 2013
    
  • PARI
    a(n)=!bittest(n+1,valuation(n+1,2)+1); \\ Robert L. Brown, Jul 07 2025
    
  • Python
    def A014577(n):
        s = bin(n+1)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return 1-int(s[m-i-2]) if m-i-2 >= 0 else 1 # Chai Wah Wu, Apr 08 2021

Formula

a(n) = (1+Jacobi(-1,n+1))/2 (cf. A034947). - N. J. A. Sloane, Jul 27 2012 [Adjusted to match offset by Peter Munn, Jul 01 2022]
Set a=1, b=0, S(0)=a, S(n+1) = S(n), a, F(S(n)), where F(x) reverses x and then interchanges a and b; sequence is limit S(infinity).
From Ralf Stephan, Jul 03 2003: (Start)
a(4*n) = 1, a(4*n+2) = 0, a(2*n+1) = a(n).
a(n) = 1 - A014707(n) = 2 - A014709(n) = A014710(n) - 1. (End)
Set a=1, b=0, S(0)=a, S(n+1)=S(n), a, M(S(n)), where M(S) is S but the bit in the middle position flipped. (Proof via isomorphism of both formulas to a modified string substitution.) - Benjamin Heiland, Dec 11 2011
a(n) = 1 if A005811(n+1) > A005811(n), otherwise a(n) = 0. - Gary W. Adamson, Jun 20 2012
a((2*n+1)*2^p-1) = (n+1) mod 2, p >= 0. - Johannes W. Meijer, Jan 28 2013
G.f. g(x) satisfies g(x) = x*g(x^2) + 1/(1-x^4). - Robert Israel, Jan 06 2015
a(n) = 1 - A065339(n+1) mod 2. - Peter Munn, Jun 29 2022
From A.H.M. Smeets, Mar 19 2023: (Start)
a(n) = 1 - A038189(n+1).
a(n) = A082410(n+2).
a(n) = 1 - A089013(n+1)
a(n) = (3 - A099545(n+1))/2.
a(n) = (A112347(n+1) + 1)/2.
a(n) = (A121238(n+1) + 1)/2.
a(n) = (A317335(n) + 2)/3.
a(n) = (A317336(n) + 10)/3. (End)
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/2. - Amiram Eldar, Sep 14 2024

Extensions

More terms from Ralf Stephan, Jul 03 2003

A091067 Numbers whose odd part is of the form 4k+3.

Original entry on oeis.org

3, 6, 7, 11, 12, 14, 15, 19, 22, 23, 24, 27, 28, 30, 31, 35, 38, 39, 43, 44, 46, 47, 48, 51, 54, 55, 56, 59, 60, 62, 63, 67, 70, 71, 75, 76, 78, 79, 83, 86, 87, 88, 91, 92, 94, 95, 96, 99, 102, 103, 107, 108, 110, 111, 112, 115, 118, 119, 120, 123, 124, 126, 127, 131
Offset: 1

Views

Author

Ralf Stephan, Feb 22 2004

Keywords

Comments

Either of form 2*a(m) or 4k+3, k >= 0, 0 < m < n.
A000265(a(n)) is an element of A004767.
a(n) such that A038189(a(n)) = 1.
Numbers n such that Kronecker(-n, m) = Kronecker(m, n) for all m. - Michael Somos, Sep 22 2005
From Antti Karttunen, Feb 20-21 2015: (Start)
Gives all n for which A005811(n) - A005811(n-1) = -1, from which follows that a(n) = the least k such that A255070(k) = n.
Gives the positions of even terms in A003602. (End)
Indices of negative terms in A164677. - M. F. Hasler, Aug 06 2015
Indices of the 0's in A014577. - Gabriele Fici, Jun 02 2016
Also indices of -1 in A034947. - Jianing Song, Apr 24 2021
Conjecture: alternate definition of same sequence is that a(1)=3 and a(n) is the smallest number > a(n-1) so that no number that is the sum of at most 2 terms in this sequence is a power of 2. - J. Lowell, Jan 20 2024
The asymptotic density of this sequence is 1/2. - Amiram Eldar, Aug 31 2024

Crossrefs

Essentially one less than A060833.
Characteristic function: A038189.
Complement of A091072.
First differences are in A106836 (from its second term onward).
Sequence A246590 gives the even terms.
Gives the positions of records (after zero) for A255070 (equally, the position of the first n there).
Cf. A106837 (gives n such that both n and n+1 are terms of this sequence).
Cf. A098502 (gives n such that both n and n+2 are, but n+1 is not in this sequence).

Programs

  • Haskell
    import Data.List (elemIndices)
    a091067 n = a091067_list !! (n-1)
    a091067_list = map (+ 1) $ elemIndices 1 a014707_list
    -- Reinhard Zumkeller, Sep 28 2011
    (Scheme, with Antti Karttunen's IntSeq-library, two versions)
    (define A091067 (MATCHING-POS 1 1 (COMPOSE even? A003602)))
    (define A091067 (NONZERO-POS 1 0 A038189))
    ;; Antti Karttunen, Feb 20 2015
  • Mathematica
    Select[Range[150], Mod[# / 2^IntegerExponent[#, 2], 4] == 3 &] (* Amiram Eldar, Aug 31 2024 *)
  • PARI
    for(n=1,200,if(((n/2^valuation(n,2)-1)/2)%2,print1(n",")))
    
  • PARI
    {a(n) = local(m, c); if( n<1, 0, c=0; m=1; while( cMichael Somos, Sep 22 2005 */
    
  • PARI
    is_A091067(n)=bittest(n,valuation(n,2)+1) \\ M. F. Hasler, Aug 06 2015
    
  • PARI
    a(n) = my(t=1); n<<=1; forstep(i=logint(n,2),0,-1, if(bittest(n,i)==t, n++;t=!t)); n; \\ Kevin Ryde, Mar 21 2021
    

Formula

a(n) = A060833(n+1) - 1. [See N. Sato's Feb 12 2013 comment in A060833.]
Other identities. For all n >= 1 it holds that:
A014707(a(n) + 1) = 1. - Reinhard Zumkeller, Sep 28 2011
A055975(a(n)) < 0. - Reinhard Zumkeller, Apr 28 2012
From Antti Karttunen, Feb 20-21 2015: (Start)
a(n) = A246590(n)/2.
A255070(a(n)) = n, or equally, A236840(a(n)) = 2n.
a(n) = 1 + A255068(n-1). (End)

A014707 a(4n) = 0, a(4n+2) = 1, a(2n+1) = a(n).

Original entry on oeis.org

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
Offset: 0

Views

Author

Keywords

Comments

The regular paper-folding (or dragon curve) sequence.
It appears that the sequence of run lengths is A088431. - Dimitri Hendriks, May 06 2010
Runs of three consecutive ones appear around positions n = 22, 46, 54, 86, 94, 118, 150, 174, 182, ..., or for n of the form 2^(k+3)*(4*t+3)-2, k >= 0, t >= 0. - Vladimir Shevelev, Mar 19 2011

References

  • Guy Melançon, Factorizing infinite words using Maple, MapleTech journal, Vol. 4, No. 1, 1997, pp. 34-42, esp. p. 36.

Crossrefs

Equals 1 - A014577, which see for further references. Also a(n) = A038189(n+1).
The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410.

Programs

  • Haskell
    a014707 n = a014707_list !! n
    a014707_list = f 0 $ cycle [0,0,1,0] where
       f i (x:_:xs) = x : a014707 i : f (i+1) xs
    -- Reinhard Zumkeller, Sep 28 2011
    
  • Maple
    nmax:=92: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 0 to ceil(nmax/(p+2))+1 do a((2*n+1)*2^p-1) := n mod 2 od: od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Jan 28 2013
    # second Maple program:
    a:= proc(n) option remember;
         `if`(n::even, irem(n/2, 2), a((n-1)/2))
        end:
    seq(a(n), n=0..92);  # Alois P. Heinz, Jun 27 2022
  • Mathematica
    a[n_ /; Mod[n, 4] == 0] = 0; a[n_ /; Mod[n, 4] == 2] = 1; a[n_ /; Mod[n, 2] == 1] := a[n] = a[(n - 1)/2]; Table[a[n],{n,0,92}] (* Jean-François Alcover, May 17 2011 *)
    (1 - JacobiSymbol[-1, Range[100]])/2 (* Paolo Xausa, May 26 2024 *)
  • PARI
    a(n)=n+=1;my(h=bitand(n,-n));n=bitand(n,h<<1);n!=0; \\ Joerg Arndt, Apr 09 2021
  • Python
    def A014707(n):
        s = bin(n+1)[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 A014707(n): n+=1; h=n&-n; n=n&(h<<1); return int(n!=0)
    print([A014707(n) for n in range(93)]) # Michael S. Branicky, Mar 29 2024 after Joerg Arndt
    

Formula

a(A091072(n)-1) = 0; a(A091067(n)-1) = 1. - Reinhard Zumkeller, Sep 28 2011 [Adjusted to match offset by Peter Munn, Jul 01 2022]
a(n) = (1-Jacobi(-1,n+1))/2 (cf. A034947). - N. J. A. Sloane, Jul 27 2012 [Adjusted to match offset by Peter Munn, Jul 01 2022]
Set a=0, b=1, S(0)=a, S(n+1) = S(n)aF(S(n)), where F(x) reverses x and then interchanges a and b; sequence is limit S(infinity).
a((2*n+1)*2^p-1) = n mod 2, p >= 0. - Johannes W. Meijer, Jan 28 2013
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/2. - Amiram Eldar, Aug 31 2024

Extensions

More terms from Scott C. Lindhurst (ScottL(AT)alumni.princeton.edu)

A091072 Positive numbers k such that the Kronecker Symbol (-1 / k) > 0.

Original entry on oeis.org

1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 21, 25, 26, 29, 32, 33, 34, 36, 37, 40, 41, 42, 45, 49, 50, 52, 53, 57, 58, 61, 64, 65, 66, 68, 69, 72, 73, 74, 77, 80, 81, 82, 84, 85, 89, 90, 93, 97, 98, 100, 101, 104, 105, 106, 109, 113, 114, 116, 117, 121, 122, 125, 128, 129
Offset: 1

Views

Author

Ralf Stephan, Feb 22 2004

Keywords

Comments

Numbers whose odd part is of the form 4k+1. The bit to the left of the least significant bit of each term is unset. Either of form 2a(m) or 4k+1, k >= 0, 0 < m < n.
A000265(a(n)) is an element of A016813.
a(n) such that A038189(a(n)) = 0.
Numbers n such that kronecker(n, m) = kronecker(m, n) for all m. - Michael Somos, Sep 24 2005
The Dragon curve A014577 (but changing the offset to 1): (1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...) = the characteristic function of A091072. - Gary W. Adamson, Apr 11 2010
Also indices of 1 in A034947. - Jianing Song, Apr 24 2021
The terms in the sequence are the same as the terms in the odd columns of the table in A135764 with headings 4k+1: (1, 5, 9, 13...). A014577(n) = 1 if n is in that set, but A014577(n) = 0 if n is in the set of even columns in the A135764 table. - Gary W. Adamson, May 29 2021
The asymptotic density of this sequence is 1/2. - Amiram Eldar, Sep 14 2024

Examples

			x + 2*x^2 + 4*x^3 + 5*x^4 + 8*x^5 + 9*x^6 + 10*x^7 + 13*x^8 + 16*x^9 + ...
		

Crossrefs

Complement of A091067.
Cf. A000265, A014577 (characteristic function), A014707, A016813, A034947, A055975, A106841 (first of triplet), A088742 (first differences), A339597.

Programs

  • Haskell
    import Data.List (elemIndices)
    a091072 n = a091072_list !! (n-1)
    a091072_list = map (+ 1) $ elemIndices 0 a014707_list
    -- Reinhard Zumkeller, Sep 28 2011
  • Maple
    KS := (n, k) -> NumberTheory:-KroneckerSymbol(n, k):
    aList := upto -> select(n -> 0 < KS(-1, n), [seq(1..upto)]):
    aList(129);  # Peter Luschny, Mar 20 2025
  • Mathematica
    Select[ Range[129], EvenQ[ (#/2^IntegerExponent[#, 2] - 1)/2 ] & ] (* Jean-François Alcover, Feb 16 2012, after Pari *)
  • PARI
    for(n=1,200,if(((n/2^valuation(n,2)-1)/2)%2==0,print1(n",")))
    
  • PARI
    {a(n) = local(m, c); if( n<1, 0, c=1; m=1; while( cMichael Somos, Sep 24 2005 */
    
  • PARI
    a(n) = if(n=2*n-2, my(t=1); forstep(i=logint(n,2),0,-1, if(bittest(n,i)==t, n--;t=!t))); n+1; \\ Kevin Ryde, Mar 21 2021
    
  • PARI
    isok(k) = kronecker(-1, k) > 0; \\ Michel Marcus, Mar 20 2025
    

Formula

A014707(a(n) + 1) = 0. - Reinhard Zumkeller, Sep 28 2011
A055975(a(n)) > 0. - Reinhard Zumkeller, Apr 28 2012

Extensions

New name from Peter Luschny, Mar 20 2025

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
Showing 1-10 of 32 results. Next