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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A089608 a(n) = ((-1)^(n+1)*A002425(n)) modulo 6.

Original entry on oeis.org

1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 5, 1
Offset: 1

Views

Author

Benoit Cloitre, Dec 30 2003

Keywords

Comments

Let S(1)={1} and S(n+1)=S(n)S'(n) where S'(n) is obtained from S(n) by changing last term using the cyclic permutation 1->5->1, then sequence is S(infinity).

Crossrefs

Programs

  • Mathematica
    a[n_] := Mod[IntegerExponent[n, 2], 2] * 4 + 1; Array[a, 100] (* Amiram Eldar, Nov 28 2022 *)
  • PARI
    a(n)=numerator(2/n*(4^n-1)*bernfrac(2*n))%6
    
  • PARI
    a(n)=valuation(n,2)%2 * 4 + 1; \\ Andrew Howroyd, Aug 01 2018
    
  • Scheme
    (define (A035263 n) (let loop ((n n) (i 1)) (cond ((odd? n) (modulo i 2)) (else (loop (/ n 2) (+ 1 i))))))
    (define (A089608 n) (- 5 (* 4 (A035263 n))))
    ;; Antti Karttunen, Sep 11 2017

Formula

a(n) = 5 - 4*A035263(n).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 7/3. - Amiram Eldar, Nov 28 2022
From Amiram Eldar, Jan 04 2023: (Start)
Multiplicative with a(2^e) = 5 if e is odd, and 1 if e is even, a(p^e) = 1 for p >= 3.
Dirichlet g.f.: zeta(s)*(2^s+5)/(2^s+1). (End)

Extensions

Keyword:mult added by Andrew Howroyd, Aug 01 2018

A089607 a(n)=((-1)^(n+1)*A002425(n)) modulo 4.

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Dec 30 2003

Keywords

Crossrefs

Cf. A056832.

Programs

  • PARI
    a(n)=numerator(2/n*(4^n-1)*bernfrac(2*n))%4

Formula

a(1)=1; for n>=2, a(n) = 2*A014577(n-1)+1

A089612 a(n) = ((-1)^(n+1)*A002425(n)) modulo 5.

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Dec 30 2003

Keywords

Crossrefs

Programs

  • Magma
    [Numerator(2/n*(4^n-1)*Bernoulli(2*n)) mod 5: n in [1..100]]; // Vincenzo Librandi, Aug 01 2018
  • Mathematica
    Table[Mod[Numerator[2 / n (4^n - 1) BernoulliB[2 n]], 5], {n, 100}] (* Vincenzo Librandi, Aug 01 2018 *)
  • PARI
    a(n)=numerator(2/n*(4^n-1)*bernfrac(2*n))%5
    
  • PARI
    a(n)=if(n%2, 1, 2*2^valuation(n,2) % 5); \\ Andrew Howroyd, Aug 01 2018
    

Formula

Let S(1) = {1, 4} and S(n+1) = S(n)*S'(n), where S'(n) is obtained from S(n) by changing last term using the cyclic permutation 4->3->1->2->4; sequence is S(infinity).
Multiplicative with a(2^e) = 2^(e + 1) mod 5, a(p^e) = 1 for odd prime p. - Andrew Howroyd, Aug 01 2018
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 21/10. - Amiram Eldar, Nov 10 2022

A089172 Values of n where numerators of series coefficients of 1/(1 + Cosh[Sqrt[x]]) differ from A002425.

Original entry on oeis.org

11, 19, 24, 27, 29, 33, 34, 35, 39, 43, 44, 48, 49, 50, 51, 53, 54, 55, 59, 61, 62, 64, 65, 67, 69, 71, 75, 76, 79, 80, 83, 84, 87, 89, 90, 91, 94, 97, 98, 99, 101, 103, 104, 105, 107, 108, 109, 111, 113, 114, 115, 116, 118, 119, 120, 123, 124, 125, 127
Offset: 1

Views

Author

Wouter Meeussen, Dec 07 2003

Keywords

Crossrefs

Cf. A002425.

Programs

  • Mathematica
    it=Table[Numerator[2*BernoulliB[2*n]*(4^n-1)/(2*n)]/ Numerator[SeriesCoefficient[Series[1/(1+Cosh[x]), {x, 0, 2n}], 2n-2]], {n, 1, 96}];Flatten[Position[it, q_Integer/;q>1]]-1

A000182 Tangent (or "Zag") numbers: e.g.f. tan(x), also (up to signs) e.g.f. tanh(x).

Original entry on oeis.org

1, 2, 16, 272, 7936, 353792, 22368256, 1903757312, 209865342976, 29088885112832, 4951498053124096, 1015423886506852352, 246921480190207983616, 70251601603943959887872, 23119184187809597841473536, 8713962757125169296170811392, 3729407703720529571097509625856
Offset: 1

Views

Author

Keywords

Comments

Number of Joyce trees with 2n-1 nodes. Number of tremolo permutations of {0,1,...,2n}. - Ralf Stephan, Mar 28 2003
The Hankel transform of this sequence is A000178(n) for n odd = 1, 12, 34560, ...; example: det([1, 2, 16; 2, 16, 272, 16, 272, 7936]) = 34560. - Philippe Deléham, Mar 07 2004
a(n) is the number of increasing labeled full binary trees with 2n-1 vertices. Full binary means every non-leaf vertex has two children, distinguished as left and right; labeled means the vertices are labeled 1,2,...,2n-1; increasing means every child has a label greater than its parent. - David Callan, Nov 29 2007
From Micha Hofri (hofri(AT)wpi.edu), May 27 2009: (Start)
a(n) was found to be the number of permutations of [2n] which when inserted in order, to form a binary search tree, yield the maximally full possible tree (with only one single-child node).
The e.g.f. is sec^2(x)=1+tan^2(x), and the same coefficients can be manufactured from the tan(x) itself, which is the e.g.f. for the number of trees as above for odd number of nodes. (End)
a(n) is the number of increasing strict binary trees with 2n-1 nodes. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014
For relations to alternating permutations, Euler and Bernoulli polynomials, zigzag numbers, trigonometric functions, Fourier transform of a square wave, quantum algebras, and integrals over and in n-dimensional hypercubes and over Green functions, see Hodges and Sukumar. For further discussion on the quantum algebra, see the later Hodges and Sukumar reference and the paper by Hetyei presenting connections to the general combinatorial theory of Viennot on orthogonal polynomials, inverse polynomials, tridiagonal matrices, and lattice paths (thereby related to continued fractions and cumulants). - Tom Copeland, Nov 30 2014
The Zigzag Hankel transform is A000178. That is, A000178(2*n - k) = det( [a(i+j - k)]{i,j = 1..n} ) for n>0 and k=0,1. - _Michael Somos, Mar 12 2015
a(n) is the number of standard Young tableaux of skew shape (n,n,n-1,n-2,...,3,2)/(n-1,n-2,n-3,...,2,1). - Ran Pan, Apr 10 2015
For relations to the Sheffer Appell operator calculus and a Riccati differential equation for generating the Meixner-Pollaczek and Krawtchouk orthogonal polynomials, see page 45 of the Feinsilver link and Rzadkowski. - Tom Copeland, Sep 28 2015
For relations to an elliptic curve, a Weierstrass elliptic function, the Lorentz formal group law, a Lie infinitesimal generator, and the Eulerian numbers A008292, see A155585. - Tom Copeland, Sep 30 2015
Absolute values of the alternating sums of the odd-numbered rows (where the single 1 at the apex of the triangle is counted as row #1) of the Eulerian triangle, A008292. The actual alternating sums alternate in sign, e.g., 1, -2, 16, -272, etc. (Even-numbered rows have alternating sums always 0.) - Gregory Gerard Wojnar, Sep 28 2018
The sequence 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 1]. - Allen Stenger, Aug 03 2020
From Peter Bala, Dec 24 2021: (Start)
Conjectures:
1) The sequence taken modulo any integer k eventually becomes periodic with period dividing phi(k).
2) The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k, except when p = 2, n = 1 and k = 1 or 2.
3) For i >= 1 define a_i(n) = a(n+i). The Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k. If true, then for each i >= 1 the expansion of exp(Sum_{n >= 1} a_i(n)*x^n/n) has integer coefficients. For an example, see A262145.(End)

Examples

			tan(x) = x + 2*x^3/3! + 16*x^5/5! + 272*x^7/7! + ... = x + 1/3*x^3 + 2/15*x^5 + 17/315*x^7 + 62/2835*x^9 + O(x^11).
tanh(x) = x - 1/3*x^3 + 2/15*x^5 - 17/315*x^7 + 62/2835*x^9 - 1382/155925*x^11 + ...
(sec x)^2 = 1 + x^2 + 2/3*x^4 + 17/45*x^6 + ...
a(3)=16 because we have: {1, 3, 2, 5, 4}, {1, 4, 2, 5, 3}, {1, 4, 3, 5, 2},
  {1, 5, 2, 4, 3}, {1, 5, 3, 4, 2}, {2, 3, 1, 5, 4}, {2, 4, 1, 5, 3},
  {2, 4, 3, 5, 1}, {2, 5, 1, 4, 3}, {2, 5, 3, 4, 1}, {3, 4, 1, 5, 2},
  {3, 4, 2, 5, 1}, {3, 5, 1, 4, 2}, {3, 5, 2, 4, 1}, {4, 5, 1, 3, 2},
  {4, 5, 2, 3, 1}. - _Geoffrey Critzer_, May 19 2013
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 111.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 69.
  • L. M. Milne-Thompson, Calculus of Finite Differences, 1951, p. 148 (the numbers |C^{2n-1}|).
  • J. W. Milnor and J. D. Stasheff, Characteristic Classes, Princeton, 1974, p. 282.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 444.
  • H. Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1, p. 20.
  • L. Seidel, Über eine einfache Entstehungsweise der Bernoullischen 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).
  • E. van Fossen Conrad, Some continued fraction expansions of elliptic functions, PhD thesis, The Ohio State University, 2002, p. 28.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, pp. 267-268.

Crossrefs

A350972 is essentially the same sequence.
a(n)=2^(n-1)*A002105(n). Apart from signs, 2^(2n-2)*A001469(n) = n*a(n).
Cf. A001469, A002430, A036279, A000364 (secant numbers), A000111 (secant-tangent numbers), A024283, A009764. First diagonal of A059419 and of A064190.
Equals A002425(n) * 2^A101921(n).
Equals leftmost column of A162005. - Johannes W. Meijer, Jun 27 2009

Programs

  • Maple
    series(tan(x),x,40);
    with(numtheory): a := n-> abs(2^(2*n)*(2^(2*n)-1)*bernoulli(2*n)/(2*n));
    A000182_list := proc(n) local T,k,j; T[1] := 1;
    for k from 2 to n do T[k] := (k-1)*T[k-1] od;
       for k from 2 to n do
           for j from k to n do
               T[j] := (j-k)*T[j-1]+(j-k+2)*T[j] od od;
    seq(T[j], j=1..n)  end:
    A000182_list(15);  # Peter Luschny, Apr 02 2012
  • Mathematica
    Table[ Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}], {n, 0, 7}] (* Victor Adamchik, Oct 05 2005 *)
    v[1] = 2; v[n_] /; n >= 2 := v[n] = Sum[ Binomial[2 n - 3, 2 k - 2] v[k] v[n - k], {k, n - 1}]; Table[ v[n]/2, {n, 15}] (* Zerinvary Lajos, Jul 08 2009 *)
    Rest@ Union[ Range[0, 29]! CoefficientList[ Series[ Tan[x], {x, 0, 30}], x]] (* Harvey P. Dale, Oct 19 2011; modified by Robert G. Wilson v, Apr 02 2012 *)
    t[1, 1] = 1; t[1, 0] = 0; t[n_ /; n > 1, m_] := t[n, m] = m*(m+1)*Sum[t[n-1, k], {k, m-1, n-1}]; a[n_] := t[n, 1]; Table[a[n], {n, 1, 15}]  (* Jean-François Alcover, Jan 02 2013, after A064190 *)
    a[ n_] := If[ n < 1, 0, With[{m = 2 n - 1}, m! SeriesCoefficient[ Tan[x], {x, 0, m}]]]; (* Michael Somos, Mar 12 2015 *)
    a[ n_] := If[ n < 1, 0, ((-16)^n - (-4)^n) Zeta[1 - 2 n]]; (* Michael Somos, Mar 12 2015 *)
    Table[2 PolyGamma[2n - 1, 1/2]/Pi^(2n), {n, 1, 10}] (* Vladimir Reshetnikov, Oct 18 2015 *)
    a[ n_] := a[n] = If[ n < 2, Boole[n == 1], Sum[Binomial[2 n - 2, 2 k - 1] a[k] a[n - k], {k, n - 1}]]; (* Michael Somos, Aug 02 2018 *)
    a[n_] := (2^(2*n)*(2^(2*n) - 1)*Abs[BernoulliB[2*n]])/(2*n); a /@  Range[20]  (* Stan Wagon, Nov 21 2022 *)
  • Maxima
    a(n):=sum(sum(binomial(k,r)*sum(sum(binomial(l,j)/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(j-2*i)^(2*n),i,0,floor((j-1)/2))*(-1)^(l-j),j,1,l)*(-1)^l*binomial(r+l-1,r-1),l,1,2*n)*(-1)^(1-r),r,1,k)/k,k,1,2*n); /* Vladimir Kruchinin, Aug 23 2010 */
    
  • Maxima
    a[n]:=if n=1 then 1 else 2*sum(sum(binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j),k,1,j),j,1,n-1);
    makelist(a[n],n,1,30); /* Tani Akinari, Sep 20 2023 */
    
  • PARI
    {a(n) = if( n<1, 0, ((-4)^n - (-16)^n) * bernfrac(2*n) / (2*n))};
    
  • PARI
    {a(n) = my(an); if( n<2, n==1, an = vector(n, m, 1); for( m=2, n, an[m] = sum( k=1, m-1, binomial(2*m - 2, 2*k - 1) * an[k] * an[m-k])); an[n])}; /* Michael Somos */
    
  • PARI
    {a(n) = if( n<1, 0, (2*n - 1)! * polcoeff( tan(x + O(x^(2*n + 2))), 2*n - 1))}; /* Michael Somos */
    
  • PARI
    {a(n) = my(X=x+x*O(x^n),Egf); Egf = x*sum(m=0,n, prod(k=1,m, tanh(2*k*X))); (n-1)!*polcoeff(Egf,n)} /* Paul D. Hanna, May 11 2010 */
    
  • PARI
    /* Continued Fraction for the e.g.f. tan(x), from Paul D. Hanna: */
    {a(n)=local(CF=1+O(x)); for(i=1, n, CF=1/(2*(n-i+1)-1-x^2*CF)); (2*n-1)!*polcoeff(x*CF, 2*n-1)}
    
  • PARI
    /* O.g.f. Sum_{n>=1} a(n)*x^n, from Paul D. Hanna Feb 05 2013: */
    {a(n)=polcoeff( x+2*x*sum(m=1, n, x^m*prod(k=1, m, (2*k-1)^2/(1+(2*k-1)^2*x +x*O(x^n))) ), n)}
    
  • Python
    # The objective of this implementation is efficiency.
    # n -> [0, a(1), a(2), ..., a(n)] for n > 0.
    def A000182_list(n):
        T = [0 for i in range(1, n+2)]
        T[1] = 1
        for k in range(2, n+1):
            T[k] = (k-1)*T[k-1]
        for k in range(2, n+1):
            for j in range(k, n+1):
                T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
        return T
    print(A000182_list(100)) # Peter Luschny, Aug 07 2011
    
  • Python
    from sympy import bernoulli
    def A000182(n): return abs(((2-(2<<(m:=n<<1)))*bernoulli(m)<Chai Wah Wu, Apr 14 2023
    
  • Sage
    # Algorithm of L. Seidel (1877)
    # n -> [a(1), ..., a(n)] for n >= 1.
    def A000182_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
    A000182_list(15) # Peter Luschny, Mar 31 2012

Formula

E.g.f.: log(sec x) = Sum_{n > 0} a(n)*x^(2*n)/(2*n)!.
E.g.f.: tan x = Sum_{n >= 0} a(n+1)*x^(2*n+1)/(2*n+1)!.
E.g.f.: (sec x)^2 = Sum_{n >= 0} a(n+1)*x^(2*n)/(2*n)!.
2/(exp(2x)+1) = 1 + Sum_{n>=1} (-1)^(n+1) a(n) x^(2n-1)/(2n-1)! = 1 - x + x^3/3 - 2*x^5/15 + 17*x^7/315 - 62*x^9/2835 + ...
a(n) = 2^(2*n) (2^(2*n) - 1) |B_(2*n)| / (2*n) where B_n are the Bernoulli numbers (A000367/A002445 or A027641/A027642).
Asymptotics: a(n) ~ 2^(2*n+1)*(2*n-1)!/Pi^(2*n).
Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}]. - Victor Adamchik, Oct 05 2005
a(n) = abs[c(2*n-1)] where c(n)= 2^(n+1) * (1-2^(n+1)) * Ber(n+1)/(n+1) = 2^(n+1) * (1-2^(n+1)) * (-1)^n * Zeta(-n) = [ -(1+EN(.))]^n = 2^n * GN(n+1)/(n+1) = 2^n * EP(n,0) = (-1)^n * E(n,-1) = (-2)^n * n! * Lag[n,-P(.,-1)/2] umbrally = (-2)^n * n! * C{T[.,P(.,-1)/2] + n, n} umbrally for the signed Euler numbers EN(n), the Bernoulli numbers Ber(n), the Genocchi numbers GN(n), the Euler polynomials EP(n,t), the Eulerian polynomials E(n,t), the Touchard / Bell polynomials T(n,t), the binomial function C(x,y) = x!/[(x-y)!*y! ] and the polynomials P(j,t) of A131758. - Tom Copeland, Oct 05 2007
a(1) = A094665(0,0)*A156919(0,0) and a(n) = Sum_{k=1..n-1} 2^(n-k-1)*A094665(n-1, k)*A156919(k,0) for n = 2, 3, .., see A162005. - Johannes W. Meijer, Jun 27 2009
G.f.: 1/(1-1*2*x/(1-2*3*x/(1-3*4*x/(1-4*5*x/(1-5*6*x/(1-... (continued fraction). - Paul Barry, Feb 24 2010
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-2x-12x^2/(1-18x-240x^2/(1-50x-1260x^2/(1-98x-4032x^2/(1-162x-9900x^2/(1-... (continued fraction);
coefficient sequences given by 4*(n+1)^2*(2n+1)*(2n+3) and 2(2n+1)^2 (see Van Fossen Conrad reference). (End)
E.g.f.: x*Sum_{n>=0} Product_{k=1..n} tanh(2*k*x) = Sum_{n>=1} a(n)*x^n/(n-1)!. - Paul D. Hanna, May 11 2010 [corrected by Paul D. Hanna, Sep 28 2023]
a(n) = (-1)^(n+1)*Sum_{j=1..2*n+1} j!*Stirling2(2*n+1,j)*2^(2*n+1-j)*(-1)^j for n >= 0. Vladimir Kruchinin, Aug 23 2010: (Start)
If n is odd such that 2*n-1 is prime, then a(n) == 1 (mod (2*n-1)); if n is even such that 2*n-1 is prime, then a(n) == -1 (mod (2*n-1)). - Vladimir Shevelev, Sep 01 2010
Recursion: a(n) = (-1)^(n-1) + Sum_{i=1..n-1} (-1)^(n-i+1)*C(2*n-1,2*i-1)* a(i). - Vladimir Shevelev, Aug 08 2011
E.g.f.: tan(x) = Sum_{n>=1} a(n)*x^(2*n-1)/(2*n-1)! = x/(1 - x^2/(3 - x^2/(5 - x^2/(7 - x^2/(9 - x^2/(11 - x^2/(13 -...))))))) (continued fraction from J. H. Lambert - 1761). - Paul D. Hanna, Sep 21 2011
From Sergei N. Gladkovskii, Oct 31 2011 to Oct 09 2013: (Start)
Continued fractions:
E.g.f.: (sec(x))^2 = 1+x^2/(x^2+U(0)) where U(k) = (k+1)*(2k+1) - 2x^2 + 2x^2*(k+1)*(2k+1)/U(k+1).
E.g.f.: tan(x) = x*T(0) where T(k) = 1-x^2/(x^2-(2k+1)*(2k+3)/T(k+1)).
E.g.f.: tan(x) = x/(G(0)+x) where G(k) = 2*k+1 - 2*x + x/(1 + x/G(k+1)).
E.g.f.: tanh(x) = x/(G(0)-x) where G(k) = k+1 + 2*x - 2*x*(k+1)/G(k+1).
E.g.f.: tan(x) = 2*x - x/W(0) where W(k) = 1 + x^2*(4*k+5)/((4*k+1)*(4*k+3)*(4*k+5) - 4*x^2*(4*k+3) + x^2*(4*k+1)/W(k+1)).
E.g.f.: tan(x) = x/T(0) where T(k) = 1 - 4*k^2 + x^2*(1 - 4*k^2)/T(k+1).
E.g.f.: tan(x) = -3*x/(T(0)+3*x^2) where T(k)= 64*k^3 + 48*k^2 - 4*k*(2*x^2 + 1) - 2*x^2 - 3 - x^4*(4*k -1)*(4*k+7)/T(k+1).
G.f.: 1/G(0) where G(k) = 1 - 2*x*(2*k+1)^2 - x^2*(2*k+1)*(2*k+2)^2*(2*k+3)/G(k+1).
G.f.: 2*Q(0) - 1 where Q(k) = 1 + x^2*(4*k + 1)^2/(x + x^2*(4*k + 1)^2 - x^2*(4*k + 3)^2*(x + x^2*(4*k + 1)^2)/(x^2*(4*k + 3)^2 + (x + x^2*(4*k + 3)^2)/Q(k+1) )).
G.f.: (1 - 1/G(0))*sqrt(-x), where G(k) = 1 + sqrt(-x) - x*(k+1)^2/G(k+1).
G.f.: Q(0), where Q(k) = 1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/Q(k+1)). (End)
O.g.f.: x + 2*x*Sum_{n>=1} x^n * Product_{k=1..n} (2*k-1)^2 / (1 + (2*k-1)^2*x). - Paul D. Hanna, Feb 05 2013
a(n) = (-4)^n*Li_{1-2*n}(-1). - Peter Luschny, Jun 28 2012
a(n) = (-4)^n*(4^n-1)*Zeta(1-2*n). - Jean-François Alcover, Dec 05 2013
Asymptotic expansion: 4*((2*(2*n-1))/(Pi*e))^(2*n-1/2)*exp(1/2+1/(12*(2*n-1))-1/(360*(2*n-1)^3)+1/(1260*(2*n-1)^5)-...). (See Luschny link.) - Peter Luschny, Jul 14 2015
From Peter Bala, Sep 11 2015: (Start)
The e.g.f. A(x) = tan(x) satisfies the differential equation A''(x) = 2*A(x)*A'(x) with A(0) = 0 and A'(0) = 1, leading to the recurrence a(0) = 0, a(1) = 1, else a(n) = 2*Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the aerated sequence [0, 1, 0, 2, 0, 16, 0, 272, ...].
Note, the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence n! and with a(0) = 1/2 and a(1) = 1 produces A080635. Cf. A002105, A234797. (End)
a(n) = 2*polygamma(2*n-1, 1/2)/Pi^(2*n). - Vladimir Reshetnikov, Oct 18 2015
a(n) = 2^(2n-2)*|p(2n-1,-1/2)|, where p_n(x) are the shifted row polynomials of A019538. E.g., a(2) = 2 = 2^2 * |1 + 6(-1/2) + 6(-1/2)^2|. - Tom Copeland, Oct 19 2016
From Peter Bala, May 05 2017: (Start)
With offset 0, the o.g.f. A(x) = 1 + 2*x + 16*x^2 + 272*x^3 + ... has the property that its 4th binomial transform 1/(1 - 4*x) A(x/(1 - 4*x)) has the S-fraction representation 1/(1 - 6*x/(1 - 2*x/(1 - 20*x/(1 - 12*x/(1 - 42*x/(1 - 30*x/(1 - ...))))))), where the coefficients [6, 2, 20, 12, 42, 30, ...] in the partial numerators of the continued fraction are obtained from the sequence [2, 6, 12, 20, ..., n*(n + 1), ...] by swapping adjacent terms. Compare with the S-fraction associated with A(x) given above by Paul Barry.
A(x) = 1/(1 + x - 3*x/(1 - 4*x/(1 + x - 15*x/(1 - 16*x/(1 + x - 35*x/(1 - 36*x/(1 + x - ...))))))), where the unsigned coefficients in the partial numerators [3, 4, 15, 16, 35, 36,...] come in pairs of the form 4*n^2 - 1, 4*n^2 for n = 1,2,.... (End)
a(n) = Sum_{k = 1..n-1} binomial(2*n-2, 2*k-1) * a(k) * a(n-k), with a(1) = 1. - Michael Somos, Aug 02 2018
a(n) = 2^(2*n-1) * |Euler(2*n-1, 0)|, where Euler(n,x) are the Euler polynomials. - Daniel Suteu, Nov 21 2018 (restatement of one of Copeland's 2007 formulas.)
x - Sum_{n >= 1} (-1)^n*a(n)*x^(2*n)/(2*n)! = x - log(cosh(x)). The series reversion of x - log(cosh(x)) is (1/2)*x - (1/2)*log(2 - exp(x)) = Sum_{n >= 0} A000670(n)*x^(n+1)/(n+1)!. - Peter Bala, Jul 11 2022
For n > 1, a(n) = 2*Sum_{j=1..n-1} Sum_{k=1..j} binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j). - Tani Akinari, Sep 20 2023
a(n) = A110501(n) * 4^(n-1) / n (Han and Liu, 2018). - Amiram Eldar, May 17 2024

A048896 a(n) = 2^(A000120(n+1) - 1), n >= 0.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 4, 1, 2, 2, 4, 2, 4, 4, 8, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 2, 4, 4
Offset: 0

Views

Author

Keywords

Comments

a(n) = 2^A048881 = 2^{maximal power of 2 dividing the n-th Catalan number (A000108)}. [Comment corrected by N. J. A. Sloane, Apr 30 2018]
Row sums of triangle A128937. - Philippe Deléham, May 02 2007
a(n) = sum of (n+1)-th row terms of triangle A167364. - Gary W. Adamson, Nov 01 2009
a(n), n >= 1: Numerators of Maclaurin series for 1 - ((sin x)/x)^2, A117972(n), n >= 2: Denominators of Maclaurin series for 1 - ((sin x)/x)^2, the correlation function in Montgomery's pair correlation conjecture. - Daniel Forgues, Oct 16 2011
For n > 0: a(n) = A007954(A007931(n)). - Reinhard Zumkeller, Oct 26 2012
a(n) = A261363(2*(n+1), n+1). - Reinhard Zumkeller, Aug 16 2015
From Gus Wiseman, Oct 30 2022: (Start)
Also the number of coarsenings of the (n+1)-th composition in standard order. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See link for sequences related to standard compositions. For example, the a(10) = 4 coarsenings of (2,1,1) are: (2,1,1), (2,2), (3,1), (4).
Also the number of times n+1 appears in A357134. For example, 11 appears at positions 11, 20, 33, and 1024, so a(10) = 4.
(End)

Examples

			From _Omar E. Pol_, Jul 21 2009: (Start)
If written as a triangle:
  1;
  1,2;
  1,2,2,4;
  1,2,2,4,2,4,4,8;
  1,2,2,4,2,4,4,8,2,4,4,8,4,8,8,16;
  1,2,2,4,2,4,4,8,2,4,4,8,4,8,8,16,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32;
  ...,
the first half-rows converge to Gould's sequence A001316.
(End)
		

Crossrefs

This is Guy Steele's sequence GS(3, 5) (see A135416).
Equals first right hand column of triangle A160468.
Equals A160469(n+1)/A002425(n+1).
Standard compositions are listed by A066099.
The opposite version (counting refinements) is A080100.
The version for Heinz numbers of partitions is A317141.

Programs

  • Haskell
    a048896 n = a048896_list !! n
    a048896_list = f [1] where f (x:xs) = x : f (xs ++ [x,2*x])
    -- Reinhard Zumkeller, Mar 07 2011
    
  • Haskell
    import Data.List (transpose)
    a048896 = a000079 . a000120
    a048896_list = 1 : concat (transpose
       [zipWith (-) (map (* 2) a048896_list) a048896_list,
        map (* 2) a048896_list])
    -- Reinhard Zumkeller, Jun 16 2013
    
  • Magma
    [Numerator(2^n / Factorial(n+1)): n in [0..100]]; // Vincenzo Librandi, Apr 12 2014
  • Maple
    a := n -> 2^(add(i,i=convert(n+1,base,2))-1): seq(a(n), n=0..97); # Peter Luschny, May 01 2009
  • Mathematica
    NestList[Flatten[#1 /. a_Integer -> {a, 2 a}] &, {1}, 4] // Flatten (* Robert G. Wilson v, Aug 01 2012 *)
    Table[Numerator[2^n / (n + 1)!], {n, 0, 200}] (* Vincenzo Librandi, Apr 12 2014 *)
    Denominator[Table[BernoulliB[2*n] / (Zeta[2*n]/Pi^(2*n)), {n, 1, 100}]] (* Terry D. Grant, May 29 2017 *)
    Table[Denominator[((2 n)!/2^(2 n + 1)) (-1)^n], {n, 1, 100}]/4 (* Terry D. Grant, May 29 2017 *)
    2^IntegerExponent[CatalanNumber[Range[0,100]],2] (* Harvey P. Dale, Apr 30 2018 *)
  • PARI
    a(n)=if(n<1,1,if(n%2,a(n/2-1/2),2*a(n-1)))
    
  • PARI
    a(n) = 1 << (hammingweight(n+1)-1); \\ Kevin Ryde, Feb 19 2022
    

Formula

a(n) = 2^A048881(n).
a(n) = 2^k if 2^k divides A000108(n) but 2^(k+1) does not divide A000108(n).
It appears that a(n) = Sum_{k=0..n} binomial(2*(n+1), k) mod 2. - Christopher Lenard (c.lenard(AT)bendigo.latrobe.edu.au), Aug 20 2001
a(0) = 1; a(2*n) = 2*a(2*n-1); a(2*n+1) = a(n).
a(n) = (1/2) * A001316(n+1). - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 26 2004
It appears that a(n) = Sum_{k=0..2n} floor(binomial(2n+2, k+1)/2)(-1)^k = 2^n - Sum_{k=0..n+1} floor(binomial(n+1, k)/2). - Paul Barry, Dec 24 2004
a(n) = Sum_{k=0..n} (T(n,k) mod 2) where T = A039598, A053121, A052179, A124575, A126075, A126093. - Philippe Deléham, May 02 2007
a(n) = numerator(b(n)), where sin(x)^2/x = Sum_{n>0} b(n)*(-1)^n x^(2*n-1). - Vladimir Kruchinin, Feb 06 2013
a((2*n+1)*2^p-1) = A001316(n), p >= 0 and n >= 0. - Johannes W. Meijer, Feb 12 2013
a(n) = numerator(2^n / (n+1)!). - Vincenzo Librandi, Apr 12 2014
a(2n) = (2n+1)!/(n!n!)/A001803(n). - Richard Turk, Aug 23 2017
a(2n-1) = (2n-1)!/(n!(n-1)!)/A001790(n). - Richard Turk, Aug 23 2017

Extensions

New definition from N. J. A. Sloane, Mar 01 2008

A198631 Numerators of the rational sequence with e.g.f. 1/(1+exp(-x)).

Original entry on oeis.org

1, 1, 0, -1, 0, 1, 0, -17, 0, 31, 0, -691, 0, 5461, 0, -929569, 0, 3202291, 0, -221930581, 0, 4722116521, 0, -968383680827, 0, 14717667114151, 0, -2093660879252671, 0, 86125672563201181, 0, -129848163681107301953, 0, 868320396104950823611, 0
Offset: 0

Views

Author

Wolfdieter Lang, Oct 31 2011

Keywords

Comments

Numerators of the row sums of the Euler triangle A060096/A060097.
The corresponding denominator sequence looks like 2*A006519(n+1) when n is odd.
Also numerator of the value at the origin of the n-th derivative of the standard logistic function. - Enrique Pérez Herrero, Feb 15 2016

Examples

			The rational sequence r(n) = a(n) / A006519(n+1) starts:
1, 1/2, 0, -1/4, 0, 1/2, 0, -17/8, 0, 31/2, 0, -691/4, 0, 5461/2, 0, -929569/16, 0, 3202291/2, 0, -221930581/4, 0, 4722116521/2, 0, -968383680827/8, 0, 14717667114151/2, 0, -2093660879252671/4, ...
		

Crossrefs

Programs

  • Maple
    seq(denom(euler(i,x))*euler(i,1),i=0..33); # Peter Luschny, Jun 16 2012
  • Mathematica
    Join[{1},Table[Numerator[EulerE[n,1]/(2^n-1)], {n, 34}]] (* Peter Luschny, Jul 14 2013 *)
  • Sage
    def A198631_list(n):
        x = var('x')
        s = (1/(1+exp(-x))).series(x,n+2)
        return [(factorial(i)*s.coefficient(x,i)).numerator() for i in (0..n)]
    A198631_list(34) # Peter Luschny, Jul 12 2012
    
  • Sage
    # Alternatively:
    def A198631_list(len):
        e, f, R, C = 2, 1, [], [1]+[0]*(len-1)
        for n in (1..len-1):
            for k in range(n, 0, -1):
                C[k] = -C[k-1] / (k+1)
            C[0] = -sum(C[k] for k in (1..n))
            R.append(numerator((e-1)*f*C[0]))
            f *= n; e <<= 1
        return R
    print(A198631_list(36)) # Peter Luschny, Feb 21 2016

Formula

a(n) = numerator(sum(E(n,m),m=0..n)), n>=0, with the Euler triangle E(n,m)=A060096(n,m)/A060097(n,m).
E.g.f.: 2/(1+exp(-x)) (see a comment in A060096).
r(n) := sum(E(n,m),m=0..n) = ((-1)^n)*sum(((-1)^m)*m!*S2(n,m)/2^m, m=0..n), n>=0, where S2 are the Stirling numbers of the second kind A048993. From the e.g.f. with y=exp(-x), dx=-y*dy, putting y=1 at the end. - Wolfdieter Lang, Nov 03 2011
a(n) = numerator(euler(n,1)/(2^n-1)) for n > 0. - Peter Luschny, Jul 14 2013
a(n) = numerator(2*(2^n-1)*B(n,1)/n) for n > 0, B(n,x) the Bernoulli polynomials. - Peter Luschny, May 24 2014
Numerators of the Taylor series coefficients 4*(2^(n+1)-1)*B(n+1)/(n+1) for n>0 of 1 + 2 * tanh(x/2) (cf. A000182 and A089171). - Tom Copeland, Oct 19 2016
a(n) = -2*zeta(-n)*A335956(n+1). - Peter Luschny, Jul 21 2020
Conjecture: r(n) = Sum_{k=0..n} A001147(k) * A039755(n, k) * (-1)^k / (k+1) where r(n) = a(n) / A006519(n+1) = (n!) * ([x^n] (2 / (1 + exp(-x)))), for n >= 0. - Werner Schulte, Feb 16 2024

Extensions

New name, a simpler standalone definition by Peter Luschny, Jul 13 2012
Second comment corrected by Robert Israel, Feb 21 2016

A056832 All a(n) = 1 or 2; a(1) = 1; get next 2^k terms by repeating first 2^k terms and changing last element so sum of first 2^(k+1) terms is odd.

Original entry on oeis.org

1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1
Offset: 1

Views

Author

Jonas Wallgren, Aug 30 2000

Keywords

Comments

Dekking (2016) calls this the Toeplitz sequence or period-doubling sequence. - N. J. A. Sloane, Nov 08 2016
Fixed point of the morphism 1->12 and 2->11 (1 -> 12 -> 1211 -> 12111212 -> ...). - Benoit Cloitre, May 31 2004
a(n) is multiplicative. - Christian G. Bower, Jun 03 2005
a(n) is the least k such that A010060(n-1+k) = 1 - A010060(n-1); the sequence {a(n+1)-1} is the characteristic sequence for A079523. - Vladimir Shevelev, Jun 22 2009
The squarefree part of the even part of n. - Peter Munn, Dec 03 2020

Examples

			1 -> 1,2 -> 1,2,1,1 -> 1,2,1,1,1,2,1,2 -> 1,2,1,1,1,2,1,2,1,2,1,1,1,2,1,1.
Here we have 1 element, then 2 elements, then 4, 8, 16, etc.
		

References

  • Manfred R. Schroeder, Fractals, Chaos, Power Laws, W. H. Freeman, NY, 1991; pp. 277-279.

Crossrefs

Cf. A197911 (partial sums).
Essentially same as first differences of Thue-Morse, A010060. - N. J. A. Sloane, Jul 02 2015
See A035263 for an equivalent version.
Limit of A317956(n) for large n.
Row/column 2 of A059895.
Positions of 1s: A003159.
Positions of 2s: A036554.
A002425, A006519, A079523, A096268, A214682, A234957 are used in a formula defining this sequence.
A059897 is used to express relationship between terms of this sequence.

Programs

  • Haskell
    a056832 n = a056832_list !! (n-1)
    a056832_list = 1 : f [1] where
       f xs = y : f (y : xs) where
              y = 1 + sum (zipWith (*) xs $ reverse xs) `mod` 2
    -- Reinhard Zumkeller, Jul 29 2014
    
  • Mathematica
    Nest[ Function[l, {Flatten[(l /. {1 -> {1, 2}, 2 -> {1, 1}})]}], {1}, 7] (* Robert G. Wilson v, Mar 03 2005 *)
    Table[Mod[-(-1)^(n + 1) (-1)^n Numerator[EulerE[2 n + 1, 1]], 3] , {n, 0, 120}] (* Michael De Vlieger, Aug 15 2016, after Jean-François Alcover at A002425 *)
  • PARI
    a(n)=numerator(2/n*(4^n-1)*bernfrac(2*n))%3
    
  • PARI
    a(n)=if(n<1, 0, valuation(n,2)%2+1) /* Michael Somos, Jun 18 2005 */
    
  • Python
    def A056832(n): return 1+((~n&n-1).bit_length()&1) # Chai Wah Wu, Jan 09 2023

Formula

a(n) = ((-1)^(n+1)*A002425(n)) modulo 3. - Benoit Cloitre, Dec 30 2003
a(1)=1, a(n) = 1 + ((Sum_{i=1..n-1} a(i)*a(n-i)) mod 2). - Benoit Cloitre, Mar 16 2004
a(n) is multiplicative with a(2^e) = 1 + (1-(-1)^e)/2, a(p^e)=1 if p > 2. - Michael Somos, Jun 18 2005
[a(2^n+1) .. a(2^(n+1)-1)] = [a(1) .. a(2^n-1)]; a(2^(n+1)) = 3 - a(2^n).
For n > 0, a(n) = 2 - A035263(n). - Benoit Cloitre, Nov 24 2002
a(n)=2 if n-1 is in A079523; a(n)=1 otherwise. - Vladimir Shevelev, Jun 22 2009
a(n) = A096268(n-1) + 1. - Reinhard Zumkeller, Jul 29 2014
From Peter Munn, Dec 03 2020: (Start)
a(n) = A007913(A006519(n)) = A006519(n)/A234957(n).
a(n) = A059895(n, 2) = n/A214682(n).
a(n*k) = (a(n) * a(k)) mod 3.
a(A059897(n, k)) = A059897(a(n), a(k)).
(End)
Asymptotic mean: lim_{m->oo} (1/m) * Sum__{k=1..m} a(k) = 4/3. - Amiram Eldar, Mar 09 2021

A089171 Numerators of series coefficients of 1/(1 + cosh(sqrt(x))).

Original entry on oeis.org

1, -1, 1, -17, 31, -691, 5461, -929569, 3202291, -221930581, 4722116521, -56963745931, 14717667114151, -2093660879252671, 86125672563201181, -129848163681107301953, 868320396104950823611, -209390615747646519456961, 14129659550745551130667441
Offset: 0

Views

Author

Wouter Meeussen, Dec 07 2003

Keywords

Comments

Unsigned version is equal to A002425 up to n=11, but differs beyond that point.
Unsigned version: numerators of series coefficients of 1/(1 + cos(sqrt(x))); see Mathematica. - Clark Kimberling, Dec 06 2016

Crossrefs

Programs

  • Maple
    with(numtheory): c := n->(2^(2*n)-1)*bernoulli(2*n)/(2*n)!; seq(numer(c(n)),n=1..20); # C. Ronaldo
  • Mathematica
    Numerator[CoefficientList[Series[1/(1+Cosh[Sqrt[x]]), {x, 0, 24}], x]]
    Numerator[CoefficientList[Series[1/(1+Cos[Sqrt[x]]), {x, 0, 30}], x]]
    (* unsigned version, Clark Kimberling, Dec 06 2016 *)

Formula

a(n) = numerator(c(n+1)) where c(n)=(2^(2*n)-1)*B(2*n)/(2*n)!, B(k) denotes the k-th Bernoulli number. - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 19 2004
Numerators of expansion of cosec(x)-cot(x) = 1/2*x+1/4*x^3/3!+1/2*x^5/5!+17/8*x^7/7!+31/2*x^9/9!+... - Ralf Stephan, Dec 21 2004 (Comment was applied to wrong entry, corrected by Alessandro Musesti (musesti(AT)gmail.com), Nov 02 2007)
E.g.f.: 1/sin(x)-cot(x). - Sergei N. Gladkovskii, Nov 22 2011
E.g.f.: x/G(0); G(k) = 4*k+2-x^2/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 22 2011
E.g.f.: (1+x/(x-2*Q(0)))/2; Q(k) = 8*k+2+x/(1+(2*k+1)*(2*k+2)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 22 2011
E.g.f.: x/(x+Q(0)); Q(k) = x+(x^2)/((4*k+1)*(4*k+2)-(4*k+1)*(4*k+2)/(1+(4*k+3)*(4*k+4)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 22 2011
E.g.f.: T(0)/2, where T(k) = 1 - x^2/(x^2 - (4*k+2)*(4*k+6)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 12 2013
Aerated, these are the numerators of the Taylor series coefficients of 2 * tanh(x/2) (cf. A000182 and A198631). - Tom Copeland, Oct 19 2016

A160469 The left hand column of the triangle A160468.

Original entry on oeis.org

1, 1, 2, 17, 62, 1382, 21844, 929569, 6404582, 443861162, 18888466084, 1936767361654, 58870668456604, 8374643517010684, 689005380505609448, 129848163681107301953, 1736640792209901647222, 418781231495293038913922
Offset: 1

Views

Author

Johannes W. Meijer, May 24 2009

Keywords

Comments

Resembles A002430, the numerators of the Taylor series for tan(x). The first difference occurs at a(12). (Its resemblance to this sequence led to the conjecture A160469(n) = A002430(n)*A089170(n-1).)

Crossrefs

Equals the first left hand column of A160468.
Equals A002430(n)*A089170(n-1).
Equals (A002430(n)/A036279(n))*(A117972(n)/A000265(n)).
Equals A048896(n-1)*A002425(n).
Cf. A156769 (which resembles the denominators of the Taylor series for tan(x)).

Formula

a(n) = A002430(n)*A089170(n-1) with A002430 (n) = numer((-1)^(n-1)*2^(2*n)*(2^(2*n)-1)* bernoulli(2*n)/(2*n)!) and A089170 (n-1) = numer(2*bernoulli(2*n)* (4^n-1)/(2*n))/ numer((4^n-1)*bernoulli(2*n)/(2*n)!) for n = 1, 2, 3, ....
Showing 1-10 of 28 results. Next