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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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

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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

A002105 Reduced tangent numbers: 2^n*(2^{2n} - 1)*|B_{2n}|/n, where B_n = Bernoulli numbers.

Original entry on oeis.org

1, 1, 4, 34, 496, 11056, 349504, 14873104, 819786496, 56814228736, 4835447317504, 495812444583424, 60283564499562496, 8575634961418940416, 1411083019275488149504, 265929039218907754399744, 56906245479134057176170496, 13722623393637762299131396096, 3704005473270641755597685653504
Offset: 1

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Author

Keywords

Comments

Comments from R. L. Graham, Apr 25 2006 and Jun 08 2006: "This sequence also gives the number of ways of arranging 2n tokens in a row, with 2 copies of each token from 1 through n, such that the first token is a 1 and between every pair of tokens labeled i (i=1..n-1) there is exactly one token labeled i+1.
"For example, for n=3, there are 4 possibilities: 123123, 121323, 132312 and 132132 and indeed a(3) = 4. This is the work of my Ph. D. student Nan Zang. See also A117513, A117514, A117515.
"Develin and Sullivant give another occurrence of this sequence and show that their numbers have the same generating function, although they were unable to find a 1-1-mapping between their problem and Poupard's."
The sequence 1,0,1,0,4,0,34,0,496,0,11056, ... counts increasing complete binary trees with e.g.f. sec^2(x/sqrt 2). - Wenjin Woan, Oct 03 2007
a(n) = number of increasing full binary trees on vertex set [2n-1] with the left-largest property: the largest descendant of each non-leaf vertex occurs in its left subtree (Poupard). The first Mathematica recurrence below counts these trees by number 2k-1 of vertices in the left subtree of the root: the root is necessarily labeled 1 and n necessarily occurs in the left subtree and so there are Binomial[2n-3,2k-2] ways to choose the remaining labels for the left subtree. - David Callan, Nov 29 2007
Number of bilabeled unordered increasing trees with 2n labels. - Markus Kuba, Nov 18 2014
Conjecture: taking the sequence modulo an integer k gives an eventually purely periodic sequence with period dividing phi(k). For example, the sequence taken modulo 10 begins [1, 1, 4, 4, 6, 6, 4, 4, 6, 6, 4, 4, 6, 6, ...] with an apparent period [4, 4, 6, 6] of length 4 = phi(10) beginning at a(3). - Peter Bala, May 08 2023
Let c(1), c(2), c(3), ... be a geometric progression and s = (2*c(1)/c(2))^(1/2). Then c(1)*s*tan(x/s) = Sum_{n>0} a(n) * c(n) * x^(2*n-1) / (2*n-1)!. - Michael Somos, Jan 15 2025

Examples

			G.f. = x + x^2 + 4*x^3 + 34*x^4 + 496*x^5 + 11056*x^6 + 349504*x^7 + ...
		

References

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

Crossrefs

Row sums of A008301.
Left edge of triangle A210108.

Programs

  • Magma
    A002105:= func< n | (-1)^(n+1)*2^n*(4^n - 1)*Bernoulli(2*n)/n >;
    [A002105(n): n in [1..30]]; // G. C. Greubel, Sep 20 2024
  • Maple
    S := proc(n, k) option remember; if k=0 then `if`(n=0, 1, 0) else S(n, k-1) + S(n-1, n-k) fi end: A002105 := n -> S(2*n-1, 2*n-1)/2^(n-1): seq(A002105(i),i=1..16); # Peter Luschny, Jul 08 2012
    # The above function written as a formula: a(n) = A008281(2*n-1, 2*n-1)/2^(n-1).
    # Alternatively, based on the triangular numbers A000217:
    T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1) else
    A000217(n - k + 1) * T(n, k - 1) + T(n - 1, k) fi fi end:
    a := n -> T(n, n): seq(a(n), n = 0..18);  # Peter Luschny, Sep 30 2023
  • Mathematica
    u[1] = 1; u[n_]/;n>=2 := u[n] = Sum[Binomial[2n-3,2k-2]u[k]u[n-k],{k,n-1}]; Table[u[n],{n,8}] (* Poupard and also Develin and Sullivant, give a different recurrence that involves a symmetric sum: v[1] = 1; v[n_]/;n>=2 := v[n] = 1/2 Sum[Binomial[2n-2,2k-1]v[k]v[n-k],{k,n-1}] *) (*David Callan, Nov 29 2007 *)
    a[n_] := (-1)^n 2^(n+1) PolyLog[1-2n, -1]; Array[a, 10] (* Vladimir Reshetnikov, Jan 23 2011 *)
    Table[(-1)^(n+1)*2^n*(2^(2n)-1)*BernoulliB[2n]/n,{n,1,20}] (* Vaclav Kotesovec, Nov 03 2014 *)
    eulerCF[f_, len_] := Module[{g}, g[len-1]=1; g[k_]:=g[k]=1-f[k]/(f[k]-1/g[k+1]); CoefficientList[g[0] + O[x]^len, x]]; A002105List[len_] := eulerCF[(1/2) x (#+1) (#+2)&, len]; A002105List[19] (* Peter Luschny, Aug 08 2015 after Sergei N. Gladkovskii *)
    Table[PolyGamma[2n-1, 1/2] 2^(2-n)/Pi^(2n), {n, 1, 10}] (* Vladimir Reshetnikov, Oct 18 2015 *)
    Table[EulerE[2n-1, 0] (-2)^n, {n, 1, 10}] (* Vladimir Reshetnikov, Oct 21 2015 *)
  • PARI
    {a(n) = if( n<1, 0, ((-2)^n - (-8)^n) * bernfrac(2*n) / n)}; /* Michael Somos, Jun 22 2002 */
    
  • PARI
    {a(n) = if( n<0, 0, (2*n)! * polcoeff( -2 * log( cos(x / quadgen(8) + O(x^(2*n + 1)))), 2*n))}; /* Michael Somos, Jul 17 2003 */
    
  • PARI
    {a(n) = if( n<0, 0, -(-2)^(n+1) * sum(i=1, 2*n, 2^-i * sum(j=1, i, (-1)^j * binomial( i-1, j-1) * j^(2*n - 1))))}; /* Michael Somos, Sep 07 2013 */
    
  • PARI
    {a(n)=local(CF=1+x*O(x^n));if(n<1,return(0), for(k=1,n,CF=1/(1-(n-k+1)*(n-k+2)/2*x*CF));return(Vec(CF)[n]))}  /* Paul D. Hanna */
    
  • PARI
    {a(n)=local(X=x+x*O(x^n),Egf);Egf=sum(m=0,n,prod(k=1,m,tanh(k*X)));n!*polcoeff(Egf,n)} /* Paul D. Hanna, May 11 2010 */
    
  • Python
    from sympy import bernoulli
    def A002105(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 A002105_list(n) :
        D = [0]*(n+2); D[1] = 1
        R = []; z = 1/2; b = True
        for i in(0..2*n-1) :
            h = i//2 + 1
            if b :
                for k in range(h-1, 0, -1) : D[k] += D[k+1]
                z *= 2
            else :
                for k in range(1, h+1, 1) :  D[k] += D[k-1]
            b = not b
            if b : R.append(D[h]*z/h)
        return R
    A002105_list(16) # Peter Luschny, Jun 29 2012
    
  • SageMath
    def A002105(n): return (-1)^(n+1)*2^n*(4^n -1)*bernoulli(2*n)/n
    [A002105(n) for n in range(1,31)] # G. C. Greubel, Sep 20 2024
    

Formula

E.g.f.: 2*log(sec(x / sqrt(2))) = Sum_{n>0} a(n) * x^(2*n) / (2*n)!. - Michael Somos, Jun 22 2002
A000182(n) = 2^(n-1) * a(n). - Michael Somos, Jun 22 2002
a(n) = 2^(n-1)/n * A110501(n). - Don Knuth, Jan 16 2007
a(n+1) = Sum_{k = 0..n} A094665(n, k). - Philippe Deléham, Jun 11 2004
O.g.f.: A(x) = x/(1-x/(1-3*x/(1-6*x/(1-10*x/(1-15*x/(... -n*(n+1)/2*x/(1 - ...))))))) (continued fraction). - Paul D. Hanna, Oct 07 2005
sqrt(2) tan( x/sqrt(2)) = Sum_(n>=0) (x^(2n+1)/(2n+1)!) a_n. - Dominique Foata and Guo-Niu Han, Oct 24 2008
Basic hypergeometric generating function: Sum_{n>=0} Product {k = 1..n} (1-exp(-2*k*t))/Product {k = 1..n} (1+exp(-2*k*t)) = 1 + t + 4*t^2/2! + 34*t^3/3! + 496*t^4/4! + ... [Andrews et al., Theorem 4]. For other sequences with generating functions of a similar type see A000364, A000464, A002439, A079144 and A158690. - Peter Bala, Mar 24 2009
E.g.f.: Sum_{n>=0} Product_{k=1..n} tanh(k*x) = Sum_{n>=0} a(n)*x^n/n!. - Paul D. Hanna, May 11 2010
a(n) = (-1)^(n+1)*sum(j!*stirling2(2*n+1,j)*2^(n+1-j)*(-1)^(j),j,1,2*n+1), n>=0. - Vladimir Kruchinin, Aug 23 2010
From Gary W. Adamson, Jul 14 2011: (Start)
a(n) = upper left term in M^n, a(n+1) = sum of top row terms in M^n; where M = the infinite square production matrix:
1, 3, 0, 0, 0, 0, 0, ...
1, 3, 6, 0, 0, 0, 0, ...
1, 3, 6, 10, 0, 0, 0, ...
1, 3, 6, 10, 15, 0, 0, ... (End)
E.g.f. A(x) satisfies differential equation A''(x)=exp(A(x)). - Vladimir Kruchinin, Nov 18 2011
E.g.f.: For E(x)=sqrt(2)* tan( x/sqrt(2))=x/G(0); G(k)= 4*k + 1 - x^2/(8*k + 6 - x^2/G(k+1)); (from continued fraction Lambert's, 2-step). - Sergei N. Gladkovskii, Jan 14 2012
a(n) = (-1)^n*2^(n+1)*Li_{1-2*n}(-1). (See also the Mathematica prog. by Vladimir Reshetnikov.) - Peter Luschny, Jun 28 2012
G.f.: 1/G(0) where G(k) = 1 - x*( 4*k^2 + 4*k + 1 ) - x^2*(k+1)^2*( 4*k^2 + 8*k + 3)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 14 2013
G.f.: 1/Q(0), where Q(k) = 1 - (k+1)*(k+2)/2*x/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: (1/G(0))/sqrt(x) - 1/sqrt(x), where G(k) = 1 - sqrt(x)*(2*k+1)/(1 + sqrt(x)*(2*k+1)/(1 + sqrt(x)*(k+1)/(1 - sqrt(x)*(k+1)/G(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Jul 07 2013
log(2) - 1/1 + 1/2 - 1/3 + ... + (-1)^n / n = (-1)^n / 2 * (1/n - 1 / (2*n^2) + 1 / (2*n^2)^2 - 4 / (2*n^2)^3 + ... + (-1)^k * a(k) / (2*n^2)^k + ...) asymptotic expansion. - Michael Somos, Sep 07 2013
G.f.: T(0), where T(k) = 1-x*(k+1)*(k+2)/(x*(k+1)*(k+2)-2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 24 2013
a(n) ~ 2^(3*n+2) * n^(2*n-1/2) / (exp(2*n) * Pi^(2*n-1/2)). - Vaclav Kotesovec, Nov 03 2014
From Peter Bala, Sep 10 2015: (Start)
The e.g.f. A(x) = sqrt(2)*tan(x/sqrt(2)) satisfies A''(x) = A(x)*A'(x), hence the recurrence a(0) = 0, a(1) = 1, else a(n) = Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the aerated sequence [0,1,0,1,0,4,0,34,0,496,...].
Note that the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence [1,1,1,2,5,16,61,272,...] = A000111. (End)
a(n) = polygamma(2*n-1, 1/2)*2^(2-n)/Pi^(2*n). - Vladimir Reshetnikov, Oct 18 2015
E.g.f.: sqrt(2)*tan(x/sqrt(2)) = Sum_{n>0} a(n) * x^(2*n-1) / (2*n-1)!. - Michael Somos, Mar 05 2017
From Peter Bala, May 05 2017: (Start)
Let B(x) = A(x)/x = 1 + x + 4*x^2 + 34*x^3 + ... denote the shifted o.g.f. Then B(x) = 1/(1 + 2*x - 3*x/(1 - x/(1 + 2*x - 10*x/(1 - 6*x/(1 + 2*x - 21*x/(1 - 15*x/(1 + 2*x - 36*x/(1 - 28*x/(1 + 2*x - ...))))))))), where the coefficient sequence [3, 1, 10, 6, 21, 15, 36, 28, ...] in the partial numerators of the continued fraction is obtained by swapping adjacent triangular numbers. Cf. A079144.
It follows (by means of an equivalence transformation) that the second binomial transform of B(x), with g.f. equal to 1/(1 - 2*x)*B(x/(1 - 2*x)), has the S-fraction representation 1/(1 - 3*x/(1 - x/(1 - 10*x/(1 - 6*x/(1 - 21*x/(1 - 15*x/(1 - 36*x/(1 - 28*x/(1 - ...))))))))). Compare with the S-fraction representation of the g.f. A(x) given above by Hanna, dated Oct 07 2005. (End)
The computation can be based on the triangular numbers, a(n) = T(n, n) where T(n, k) = A000217(n - k + 1) * T(n, k - 1) + T(n - 1, k) for 0 < k < n, and T(n, 0) = 1, T(n, n) = T(n, k-1) if k > 0. This is equivalent to Paul D. Hanna's continued fraction 2005. - Peter Luschny, Sep 30 2023

Extensions

Additional comments from Michael Somos, Jun 25 2002

A022493 Fishburn numbers: number of linearized chord diagrams of degree n; also number of nonisomorphic interval orders on n unlabeled points.

Original entry on oeis.org

1, 1, 2, 5, 15, 53, 217, 1014, 5335, 31240, 201608, 1422074, 10886503, 89903100, 796713190, 7541889195, 75955177642, 810925547354, 9148832109645, 108759758865725, 1358836180945243, 17801039909762186, 243992799075850037, 3492329741309417600, 52105418376516869150, 809029109658971635142
Offset: 0

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Author

Alexander Stoimenow (stoimeno(AT)math.toronto.edu)

Keywords

Comments

Claesson and Linusson calls these the Fishburn numbers, after Peter Fishburn. [Peter Clingerman Fishburn (1936-2021) was an American mathematician and a pioneer in the field of decision-making processes. - Amiram Eldar, Jun 20 2021]
Also, number of unlabeled (2+2)-free posets.
Also, number of upper triangular matrices with nonnegative integer entries and without zero rows or columns such that sum of all entries is equal to n. - Vladeta Jovovic, Mar 10 2008
Row sums of A193387.
Also number of ascent sequences of length n. - N. J. A. Sloane, Dec 10 2011
An ascent sequence is a sequence [d(1), d(2), ..., d(n)] where d(1)=0, d(k)>=0, and d(k) <= 1 + asc([d(1), d(2), ..., d(k-1)]) where asc(.) counts the ascents of its argument. - Joerg Arndt, Oct 17 2012
Replacing the function asc(.) above by a function chg(.) that counts changes in the prefix gives A000522 (arrangements). - Joerg Arndt, May 10 2013
Number of restricted growth strings (RGS) [d(0), d(1), d(2), ..., d(n-1)] such that d(0)=0, 0 <= d(k) <= k, and d(j) != d(k)+1 for j < k. - Joerg Arndt, May 10 2013
Number of permutations p(1),p(2),...,p(n) having no subsequence p(i),p(j),p(k) such that i+1 = j < k and p(k)+1 = p(i) < p(j); this corresponds to the avoidance of bivincular pattern (231,{1},{1}) in the terminology of Parviainen. Also, number of permutations p(1),...,p(n) with no subsequence p(i),p(j),p(k) with i+1 = j < k, p(i)+1 = p(k) < p(j). This is the bivincular pattern (132,{1},{1}). Proved by Bousquet-Mélou et al. and by Parviainen, respectively. - Vít Jelínek, Sep 05 2014

Examples

			From _Emanuele Munarini_, Oct 16 2012: (Start)
There are a(4)=15 ascent sequences (dots for zeros):
  01:  [ . . . . ]
  02:  [ . . . 1 ]
  03:  [ . . 1 . ]
  04:  [ . . 1 1 ]
  05:  [ . . 1 2 ]
  06:  [ . 1 . . ]
  07:  [ . 1 . 1 ]
  08:  [ . 1 . 2 ]
  09:  [ . 1 1 . ]
  10:  [ . 1 1 1 ]
  11:  [ . 1 1 2 ]
  12:  [ . 1 2 . ]
  13:  [ . 1 2 1 ]
  14:  [ . 1 2 2 ]
  15:  [ . 1 2 3 ]
(End)
From _Joerg Arndt_, May 10 2013: (Start)
There are a(4)=15 RGS of length 4 such that d(0)=0, 0 <= d(k) <= k, and d(j) != d(k)+1 for j < k (dots for zeros):
  01:  [ . . . . ]
  02:  [ . . . 1 ]
  03:  [ . . . 2 ]
  04:  [ . . . 3 ]
  05:  [ . . 1 1 ]
  06:  [ . . 1 2 ]
  07:  [ . . 1 3 ]
  08:  [ . . 2 . ]
  09:  [ . . 2 2 ]
  10:  [ . . 2 3 ]
  11:  [ . 1 1 1 ]
  12:  [ . 1 1 2 ]
  13:  [ . 1 1 3 ]
  14:  [ . 1 2 2 ]
  15:  [ . 1 2 3 ]
(End)
G.f. = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + 53*x^5 + 217*x^6 + 1014*x^7 + 5335*x^8 + ...
		

References

  • P. C. Fishburn, Interval Graphs and Interval Orders, Wiley, New York, 1985.

Crossrefs

Cf. A079144 for the labeled analog, A059685, A035378.
Cf. A138265.
Cf. A137251 (ascent sequences with k ascents), A218577 (ascent sequences with maximal element k), A175579 (ascent sequences with k zeros).
Cf. A218579 (ascent sequences with last zero at position k-1), A218580 (ascent sequences with first occurrence of the maximal value at position k-1), A218581 (ascent sequences with last occurrence of the maximal value at position k-1).
Row sums of A294219, main diagonal of A294220.

Programs

  • Maple
    b:= proc(n, i, t) option remember; `if`(n<1, 1,
          add(b(n-1, j, t+`if`(j>i,1,0)), j=0..t+1))
        end:
    a:= n-> b(n-1, 0, 0):
    seq(a(n), n=0..30);  # Alois P. Heinz, Nov 09 2012
  • Mathematica
    max = 22; f[x_] := Sum[ Product[ 1-(1-x)^j, {j, 1, n}], {n, 0, max}]; CoefficientList[ Series[ f[x], {x, 0, max}], x] (* Jean-François Alcover, Dec 27 2011, after g.f. *)
    b[n_, i_, t_] := b[n, i, t] = If[n<1, 1, Sum[b[n-1, j, t + If[j>i, 1, 0]], {j, 0, t+1}] ]; a[n_] := b[n-1, 0, 0]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Apr 09 2015, after Alois P. Heinz *)
  • Maxima
    F(x,n) := remainder(sum(product(1-(1-x)^i,i,1,k),k,0,n),x^(n+1));
    makelist(coeff(F(x,n),x^n),n,0,20); /* Emanuele Munarini, Oct 16 2012 */
    
  • PARI
    {a(n) = polcoeff( sum(i=0, n, prod(j=1, i, 1 - (1-x)^j, 1 + x * O(x^n))), n)}; /* Michael Somos, Jul 21 2006 */
    
  • Sage
    # Program adapted from Alois P. Heinz's Maple code.
    # b(n,i,t) gives the number of length-n postfixes of ascent sequences
    # with a prefix having t ascents and last element i.
    @CachedFunction
    def b(n, i, t):
        if n <= 1: return 1
        return sum(b(n - 1, j, t + (j > i)) for j in range(t + 2))
    def a(n): return b(n, 0, 0)
    [a(n) for n in range(66)]
    # Joerg Arndt, May 11 2013

Formula

Zagier gives the g.f. Sum_{n>=0} Product_{i=1..n} (1-(1-x)^i).
Another formula for the g.f.: Sum_{n>=0} 1/(1-x)^(n+1) Product_{i=1..n} (1-1/(1-x)^i)^2. Proved by Andrews-Jelínek and Bringmann-Li-Rhoades. - Vít Jelínek, Sep 05 2014
Coefficients in expansion of Sum_{k>=0} Product_{j=1..k} (1-x^j) about x=1 give (-1)^n*a(n). - Bill Gosper, Aug 08 2001
G.f.: 1 + x*Q(0), where Q(k) = 1 + (1-(1-x)^(2*k+2))/(1 - (1-(1-x)^(2*k+3))/(1 -(1-x)^(2*k+3) + 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 13 2013
G.f. (conjecture): Q(0), where Q(k) = 1 + (1 - (1-x)^(2*k+1))/(1 - (1-(1-x)^(2*k+2))/(1 -(1-x)^(2*k+2) + 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 13 2013
G.f.: 1 + z(1)/( 1+0 - z(2)/( 1+z(2) - z(3)/( 1+z(3) - z(4)/( 1+z(4) - z(5)/(...))))) where z(k) = 1 - (1-x)^k; this is Euler's continued fraction for 1 + z(1) + z(1)*z(2) + z(1)*z(2)*z(3) + ... . - Joerg Arndt, Mar 11 2014
Asymptotics (proved by Zagier, 2001; see also Brightwell and Keller, 2011): a(n) ~ (12*sqrt(3)*exp(Pi^2/12)/Pi^(5/2)) * n!*sqrt(n)*(6/Pi^2)^n. - Vaclav Kotesovec, May 03 2014 [edited by Vít Jelínek, Sep 05 2014]
For any prime p that is not a quadratic residue mod 23, there is at least one value of j such that for all k and m, we have a(p^k*m - j) mod p^k = 0. E.g., for p=5 one may take j=1 and k=2, and conclude that a(25-1), a(50-1), a(75-1), a(100-1), ... are multiples of 25. See Andrews-Sellers, Garvan, and Straub. - Vít Jelínek, Sep 05 2014
From Peter Bala, Dec 17 2021: (Start)
Conjectural g.f.s:
A(x) = Sum_{n >= 1} n*(1 - x)^n*Product_{k = 1..n-1} (1 - (1 - x)^k).
x*A(x) = 1 - Sum_{n >= 1} n*(1 - x)^(2*n+1)*Product_{k = 1..n-1} (1 - (1 - x)^k). (End)

Extensions

Edited by N. J. A. Sloane, Nov 05 2011

A002439 Glaisher's T numbers.

Original entry on oeis.org

1, 23, 1681, 257543, 67637281, 27138236663, 15442193173681, 11828536957233383, 11735529528739490881, 14639678925928297567703, 22427641105413135505628881, 41393949926819051111431239623, 90592214447886493688036507587681, 231969423543894989257690172433129143
Offset: 0

Views

Author

Keywords

Comments

Kashaev's invariant for the (3,2)-torus knot. See Hikami 2003. For other Kashaev invariants see A208679, A208680, and A208681. - Peter Bala, Mar 01 2012
From Peter Bala, Dec 18 2021: (Start)
Glaisher's T numbers occur in the evaluation of the L-function L(X_12,s) := Sum_{k >= 1} X_12(k)/k^s for positive even values of s, where X_12(n) = A110161(n) is a nonprincipal Dirichlet character mod 12: the result is L(X_12,2*n+2) = a(n)/(6*sqrt(3)*36^n*(2*n+1)!) * Pi^(2*n+2).
We make the following conjectures:
1) Taking the sequence modulo an integer k gives an eventually periodic sequence with period dividing phi(k). For example, the sequence taken modulo 50 begins [1, 23, 31, 43, 31, 13, 31, 33, 31, 3, 31, 23, 31, 43, 31, 13, 31, 33, 31, 3, 31, 23, ...] and appears to have a pre-period of length 1 and a period of length 10 = (1/2)*phi(50).
2) Let i >= 0 and define a_i(n) = a(n+i). Then for each 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 the expansion of exp( Sum_{n >= 1} a_i(n)*x^n/n ) has integer coefficients.
3)(i) a(m*n) == a(m)^n (mod 2^k) for k = 2*v_2(m) + 7, where v_p(i) denotes the p-adic valuation of i.
(ii) a(m*n) == a(m)^n (mod 3^k) for k = 2*v_3(m) + 2.
4)(i) a(2*m*n) == a(n)^(2*m) (mod 2^k) for k = v_2(m) + 7
(ii) a((2*m+1)*n) == a(n)^(2*m+1) (mod 2^k) for k = v_2(m) + 7.
5)(i) a(3*m*n) == a(n)^(3*m) (mod 3^k) for k = v_3(m) + 2
(ii) a((3*m+1)*n) == a(n)^(3*m+1) (mod 3^k) for k = v_3(m) + 2
(iii) a((3*m+2)*n) == a(n)^(3*m+2) (mod 3^2).
6) For prime p >= 5, a((p-1)/2*n*m) == a((p-1)/2*n)^m (mod p^k) for k = v_p(m-1) + 1. (End)

Examples

			G.f. = 1 + 23*x + 1681*x^2 +257543*x^3 + 67637281*x^4 + 27138236663*x^5 + ...
		

References

  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 76.
  • J. W. L. Glaisher, Messenger of Math., 28 (1898), 36-79, see esp. p. 76.
  • J. W. L. Glaisher, On the Bernoullian function, Q. J. Pure Appl. Math., 29 (1898), 1-168.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Bisections: A156175, A156176.
Twice this sequence gives A000191. A208679, A208680, A208681.

Programs

  • Magma
    m:=32; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( Sin(2*x)/(2*Cos(3*x)) )); [Factorial(2*n-1)*b[2*n-1]: n in [1..Floor((m-2)/2)]]; // G. C. Greubel, Jul 04 2019
    
  • Maple
    A002439 := proc(n) option remember; if n = 0 then 1; else (-4)^n-add((-9)^k*binomial(2*n+1, 2*k)*procname(n-k), k=1..n+1) ; end if; end proc:
  • Mathematica
    a[n_] := a[n] = (-4)^n - Sum[(-9)^k*Binomial[2n + 1, 2k]*a[n-k], {k, 1, n}]; a[0] = 1; Table[a[n], {n, 0, 11}] (* Jean-François Alcover, Dec 05 2011, after Maple *)
    With[{nn=30},Take[CoefficientList[Series[Sin[2x]/(2Cos[3x]),{x,0,nn}], x]Range[0,nn-1]!,{2,-1,2}]] (* Harvey P. Dale, Feb 05 2012 *)
    a[n_] := -(-4)^n 3^(1 + 2 n) EulerE[1 + 2 n, 1/6]  (* Bill Gosper, Oct 12 2015 *)
  • PARI
    {a(n) = my(m=n+1); if( m<2, m>0, (-4)^(m-1) - sum(k=1, m, (-9)^k * binomial(2*m-1, 2*k) * a(n-k)))}; /* Michael Somos, Dec 11 1999 */
    
  • Sage
    m = 32; T = taylor(sin(2*x)/(2*cos(3*x)), x, 0, m); [factorial(2*n+1)*T.coefficient(x, 2*n+1) for n in (0..(m-2)/2)] # G. C. Greubel, Jul 04 2019

Formula

Q_{2n+1}(sqrt(3))/sqrt(3), where the polynomials Q_n() are defined in A104035. - N. J. A. Sloane, Nov 06 2009
E.g.f.: sin(2*x)/(2*cos(3*x)) = Sum a(n)*x^(2*n+1)/(2*n+1)!.
With offset 1 instead of 0: a(1)=1, a(n)=(-4)^(n-1) - Sum_{k=1..n} (-9)^k*C(2*n-1, 2*k)*a(n-k).
a(n) = -(-4)^n*3^(2n+1)*E_{2n+1}(1/6), where E is an Euler polynomial. - Bill Gosper, Aug 08 2001, corrected Oct 12 2015.
From Peter Bala, Mar 24 2009: (Start)
Basic hypergeometric generating function: exp(-t)*Sum {n = 0..inf} Product {k = 1..n} (1-exp(-24*k*t)) = 1 + 23*t + 1681*t^2/2! + .... For other sequences with generating functions of a similar type see A000364, A000464, A002105, A079144, A158690.
a(n) = (1/2)*(-1)^(n+1)*L(-2*n-1), where L(s) is a Dirichlet L-function for a Dirichlet character with modulus 12: L(s) = 1 - 1/5^s - 1/7^s + 1/11^s + - - + .... See the Andrew's link. (End)
From Peter Bala, Jan 21 2011: (Start)
Let I = sqrt(-1) and w = exp(2*Pi*I/6). Then
a(n) = I/sqrt(3) *sum {k = 0..2*n+2} w^(n-k) *sum {j = 1..2*n+2} (-1)^(k-j) *binomial(2*n+2,k-j) *(2*j-1)^(2*n+1).
This formula can be used to obtain congruences for a(n). For example, for odd prime p we find a(p-1) = 1 (mod p) and a((p-1)/2) = (-1)^((p-1)/2) (mod p).
Cf. A002437 and A182825. (End)
a(n) = (-1)^n/(4*n+4)*12^(2*n+1)*sum {k = 1..12} X(k)*B(2*n+2,k/12), where B(n,x) is a Bernoulli polynomial and X(n) is a periodic function modulo 12 given by X(n) = 0 except for X(12*n+1) = X(12*n+11) = 1 and X(12*n+5) = X(12*n+7) = -1. - Peter Bala, Mar 01 2012
a(n) ~ n^(2*n+3/2) * 2^(4*n+3) * 3^(2*n+3/2) / (exp(2*n) * Pi^(2*n+3/2)). - Vaclav Kotesovec, Mar 01 2014
From Peter Bala, May 11 2017: (Start)
Let X = 24*x. G.f. A(x) = 1/(1 + x - X/(1 - 2*X/(1 + x - 5*X/(1 - 7*X/(1 + x - 12*X/(1 - ...)))))) = 1 + 23*x + 1681*x^2 + ..., where the sequence [1, 2, 5, 7, 12, ...] of unsigned coefficients in the partial numerators of the continued fraction are generalized pentagonal numbers A001318.
A(x) = 1/(1 + 25*x - 2*X/(1 - X/(1 + 25*x - 7*X/(1 - 5*X/(1 + 25*x - 15*X/(1 - 12*X/(1 + 25*x - 26*X/(1 - 22*X/(1 + 25*x - ...))))))))), where the sequence [2, 1, 7, 5, 15, 12, 26, 22, ...] of unsigned coefficients in the partial numerators is obtained by swapping pairs of adjacent generalized pentagonal numbers.
G.f. as a J-fraction: A(x) = 1/(1 - 23*x - 2*X^2/(1 - 167*x - 5*7*X^2/(1 - 455*x - 12*15*X^2/(1 - 887*x - ...)))).
Let B(x) = 1/(1 - x)*A(x/(1 - x)), that is, B(x) is the binomial transform of A(x). Then B(x/24) is the o.g.f. for A079144. (End)
a(n) == 23^n ( mod (2^7)*(3^2) ). - Peter Bala, Dec 25 2021

Extensions

More terms from Michael Somos
Offset changed from 1 to 0 by N. J. A. Sloane, Dec 11 1999

A000464 Expansion of e.g.f. sin(x)/cos(2*x).

Original entry on oeis.org

1, 11, 361, 24611, 2873041, 512343611, 129570724921, 44110959165011, 19450718635716001, 10784052561125704811, 7342627959965776406281, 6023130568334172003579011, 5858598896811701995459355761, 6667317162352419006959182803611, 8776621742176931117228228227924441
Offset: 0

Views

Author

Keywords

Comments

From Peter Bala, Dec 22 2021: (Start)
Conjectures:
1) 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 [11, 4, 20, 10, 17, 1, 11, 4, 20, 10, 17, 1, 11, 4, 20, 10, 17, 1, ...] with an apparent period of length 6, which divides phi(21) = 12.
2) For i >= 0, define a_i(n) = a(n+i). Then for each 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 the expansion of exp(Sum_{n >= 1} a_i(n)*x^n/n) has integer coefficients.
3) a(m*n) == a(m)^n (mod 2^k) for k = 2*v_2(m) + 4, where v_p(i) denotes the p-adic valuation of i.
4)(i) a(2*m*n) == a(n)^(2*m) (mod 2^k) for k = v_2(m) + 4
(ii) a((2*m+1)*n) == a(n)^(2*m+1) (mod 2^k) for k = v_2(m) + 4. (End)

References

  • H. Cohen, Number Theory - Volume II: Analytic and Modern Tools, Graduate Texts in Mathematics. Springer-Verlag.
  • J. W. L. Glaisher, "On the coefficients in the expansions of cos x/ cos 2x and sin x/ cos 2x", Quart. J. Pure and Applied Math., 45 (1914), 187-222.
  • I. J. Schwatt, Intro. to Operations with Series, Chelsea, p. 278.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row 2 of A235606.
Cf. A064073. Bisection of A000822, A001586.

Programs

  • Maple
    a := n -> (-1)^n*2^(6*n+4)*(Zeta(0, -2*n-1, 5/8)-Zeta(0, -2*n-1, 7/8)):
    seq(a(n), n=0..12); # Peter Luschny, Oct 15 2015
  • Mathematica
    With[{nn=30},Take[CoefficientList[Series[Sin[x]/Cos[2x],{x,0,nn}],x] Range[0,nn-1]!,{2,-1,2}]] (* Harvey P. Dale, Mar 23 2012 *)
    nmax = 15; km0 = 10; d[n_, km_] := Round[(2^(4n-1/2) (2n-1)! Sum[ JacobiSymbol[2, 2k+1]/(2k+1)^(2n), {k, 0, km}])/Pi^(2n)]; dd[km_] := dd[km] = Table[d[n, km], {n, 1, nmax}]; dd[km0]; dd[km = 2*km0]; While[dd[km] != dd[km/2, km = 2*km]]; A000464 = dd[km] (* Jean-François Alcover, Feb 08 2016 *)
  • PARI
    a(n)=if(n<0, 0, n+=n+1; n!*polcoeff(sin(x+x*O(x^n))/cos(2*x+x*O(x^n)),n)) /* Michael Somos, Feb 09 2006 */

Formula

E.g.f.: Sum_{k>=0} a(k)x^(2k+1)/(2k+1)! = sin(x)/cos(2x).
a(n) = (-1)^n*L(X,-2n+1) where L(X,z) is the Dirichlet L-function L(X,z) = Sum_{k>=0} X(k)/k^z and where X(k) is the Dirichlet character Legendre(k,2) which begins 1,0,-1,0,-1,0,1,0,1,0,-1,0,-1,0,1,0,1,0,-1,0.... - Benoit Cloitre, Mar 22 2009 [This Dirichlet character is A091337. - Jianing Song, Oct 22 2023]
From Peter Bala, Mar 24 2009: (Start)
Basic hypergeometric generating function:
2*exp(-t)*Sum_{n = 0..inf} (Product_{k = 1..n} (1-exp(-16*k*t))/Product_{k = 1..n+1} (1+exp(-(16*k-8)*t))) = 1 + 11*t + 361*t^2/2! + 24611*t^3/3! + .... For other sequences with generating functions of a similar type see A000364, A002105, A002439, A079144 and A158690.
a(n) = (-1)^(n+1)*L(-2*n-1), where L(s) is the Dirichlet L-function L(s) = 1 - 1/3^s - 1/5^s + 1/7^s + - - + ... [Andrews et al., Theorem 5]. (End)
From Peter Bala, Jun 18 2009: (Start)
a(n) = (-1)^n*B_(2*n+2)(X)/(2*n+2), where B_n(X) denotes the X-Bernoulli number with X a Dirichlet character modulus 8 given by X(8*n+1) = X(8*n+7) = 1, X(8*n+3) = X(8*n+5) = -1 and X(2*n) = 0. See A161722 for the values of B_n(X).
For the theory and properties of the generalized Bernoulli numbers B_n(X) and the associated generalized Bernoulli polynomials B_n(X,x) see [Cohen, Section 9.4].
The present sequence also occurs in the evaluation of the finite sum of powers Sum_{i = 0..m-1} {(8*i+1)^n - (8*i+3)^n - (8*i+5)^n + (8*i+7)^n}, n = 1,2,... - see A151751 for details. (End)
G.f. 1/G(0) where G(k) = 1 + x - x*(4*k+3)*(4*k+4)/(1 - (4*k+4)*(4*k+5)*x/G(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Aug 11 2012
G.f.: 1/E(0) where E(k) = 1 - 11*x - 32*x*k*(k+1) - 16*x^2*(k+1)^2*(4*k+3)*(4*k+5)/E(k+1) (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 17 2012
a(n) ~ (2*n+1)! * 2^(4*n+7/2) / Pi^(2*n+2). - Vaclav Kotesovec, May 03 2014
a(n) = (-1)^n*2^(6*n+4)*(Zeta(-2*n-1,5/8)-Zeta(-2*n-1,7/8)). - Peter Luschny, Oct 15 2015
From Peter Bala, May 11 2017: (Start)
G.f. A(x) = 1 + 11*x + 361*x^2 + ... = 1/(1 + x - 12*x/(1 - 20*x/(1 + x - 56*x/(1 - 72*x/(1 + x - ... - 4*n*(4*n - 1)*x/(1 - 4*n*(4*n + 1)*x/((1 + x) - ...))))))).
A(x) = 1/(1 + 9*x - 20*x/(1 - 12*x/(1 + 9*x - 72*x/(1 - 56*x/(1 + 9*x - ... - 4*n*(4*n + 1)*x/(1 - 4*n*(4*n - 1)*x/(1 + 9*x - ...))))))).
It follows that the first binomial transform of A(x) and the ninth binomial transform of A(x) have continued fractions of Stieltjes-type (S-fractions). (End)
a(n) = (-1)^(n+1)*4^(2*n+1)*E(2*n+1,1/4), where E(n,x) is the n-th Euler polynomial. Cf. A002439. - Peter Bala, Aug 13 2017
From Peter Bala, Dec 04 2021: (Start)
F(x) = exp(x)*(exp(2*x) - 1)/(exp(4*x) + 1) = x - 11*x^3/3! + 361x^5/5! - 24611*x^7/7! + ... is the e.g.f. for the sequence [1, 0, -11, 0, 361, 0, -24611, 0, ...], a signed and aerated version of this sequence.
The binomial transform exp(x)*F(x) = x + 2*x^2/2! - 8*x^3/3! - 40*x^4/4! + + - - is an e.g.f. for a signed version of A000828 (omitting the initial term). (End)
From Peter Bala, Dec 22 2021: (Start)
a(1) = 1, a(n) = (-1)^(n-1) - Sum_{k = 1..n} (-4)^k*C(2*n-1,2*k)*a(n-k).
a(n) == 1 (mod 10); a(5*n+1) == 0 mod(11);
a(n) == - 23^(n+1) (mod 108); a(n) == (7^2)*59^n (mod 144);
a(n) == 11^n (mod 240); a(n) == (11^2)*131^n (mod 360). (End)

Extensions

Better description, new reference, Aug 15 1995

A138265 Number of upper triangular zero-one matrices with n ones and no zero rows or columns.

Original entry on oeis.org

1, 1, 1, 2, 5, 16, 61, 271, 1372, 7795, 49093, 339386, 2554596, 20794982, 182010945, 1704439030, 17003262470, 180011279335, 2015683264820, 23801055350435, 295563725628564, 3850618520827590, 52514066450469255, 748191494586458700, 11115833059268126770
Offset: 0

Views

Author

Vladeta Jovovic, Mar 10 2008, Mar 11 2008

Keywords

Comments

Row sums of A193357.
This is also the number of rigid unlabeled interval orders with n points (see Brightwell-Keller, Theorem 2; or Dukes-Kitaev-Remmel-Steingrímsson, Theorem 8). - N. J. A. Sloane, Dec 04 2011 [Corrected by Vít Jelínek, Sep 04 2014.]
Number of length-n ascent sequences without flat steps (i.e., no two adjacent digits are equal). An ascent sequence is a sequence [d(1), d(2), ..., d(n)] where d(k)>=0 and d(k) <= 1 + asc([d(1), d(2), ..., d(k-1)]) and asc(.) gives the number of ascents of its argument. [Joerg Arndt, Nov 05 2012]

Examples

			From _Joerg Arndt_, Nov 05 2012: (Start)
The a(4) = 5 such matrices with 4 ones are (dots for zeros):
  1 . . .      1 1 .      1 . 1      1 1 .      1 . .
  . 1 . .      . . 1      . 1 .      . 1 .      . 1 1
  . . 1 .      . . 1      . . 1      . . 1      . . 1
  . . . 1
The a(5)=16 ascent sequences without flat steps are (dots for zeros):
  [ 1]   [ . 1 . 1 . ]
  [ 2]   [ . 1 . 1 2 ]
  [ 3]   [ . 1 . 1 3 ]
  [ 4]   [ . 1 . 2 . ]
  [ 5]   [ . 1 . 2 1 ]
  [ 6]   [ . 1 . 2 3 ]
  [ 7]   [ . 1 2 . 1 ]
  [ 8]   [ . 1 2 . 2 ]
  [ 9]   [ . 1 2 . 3 ]
  [10]   [ . 1 2 1 . ]
  [11]   [ . 1 2 1 2 ]
  [12]   [ . 1 2 1 3 ]
  [13]   [ . 1 2 3 . ]
  [14]   [ . 1 2 3 1 ]
  [15]   [ . 1 2 3 2 ]
  [16]   [ . 1 2 3 4 ]
(End)
		

Crossrefs

Column k=0 of A242153.
Column k=1 of A264909.
Row sums of A137252.

Programs

  • Maple
    g:=sum(product(1-1/(1+x)^i,i=1..n),n=0..35): gser:=series(g,x=0,30): seq(coeff(gser,x,n),n=0..22);  # Emeric Deutsch, Mar 23 2008
    # second Maple program:
    b:= proc(n, i, t) option remember; `if`(n<1, 1, add(
         `if`(i=j, 0, b(n-1, j, t+`if`(j>i, 1, 0))), j=0..t+1))
        end:
    a:= n-> b(n-1, 0$2):
    seq(a(n), n=0..30);  # Alois P. Heinz, Nov 09 2012, Jan 14 2015
  • Mathematica
    max = 25; g = Sum[Product[1 - 1/(1 - x)^i, {i, 1, n}], {n, 0, max}]; gser = Series[g, {x, 0, max}]; a[n_] := SeriesCoefficient[gser, {x, 0, n}]; Table[a[n] // Abs, {n, 0, max-1}] (* Jean-François Alcover, Jan 24 2014, after Emeric Deutsch *)
  • Sage
    # Adaptation of the Maple program by Alois P. Heinz:
    @CachedFunction
    def b(n, i, t):
        if n<1: return 1
        return sum(b(n-1, j, t+(j>i)) for j in range(t+2))
    def a(n):
        if n<1: return 1
        return sum((-1)^(n-k)*binomial(n-1, k-1)*b(k-1, 0, 0) for k in range(n+1))
    [a(n) for n in range(33)]
    # Joerg Arndt, Feb 26 2014

Formula

G.f.: Sum_{n>=0} (Product_{i=1..n} 1-1/(1+x)^i).
G.f.: Sum_{n>=0} (1+x)^(n+1)*Product_{i=1..n} (1-(1+x)^i)^2. Proved by Bringmann-Li-Rhoades, and by Andrews-Jelínek. - Vít Jelínek, Sep 04 2014
a(n) = (1/n!)*Sum_{k=0..n} Stirling1(n,k)*A079144(k). a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n-1,k-1)*A022493(k).
G.f.: B(x/(1+x)) where B(x) is the g.f. of A022493; g.f.: Q(0,u) where u=x/(1+x), Q(k,u) = 1 + (1 - (1-x)^(2*k+1))/(1 - (1-(1-x)^(2*k+2))/(1 -(1-x)^(2*k+2) + 1/Q(k+1,u) )); (continued fraction). - Sergei N. Gladkovskii, Oct 03 2013
Asymptotics (Brightwell and Keller, 2011): a(n) ~ 12*sqrt(3)/(exp(Pi^2/12)*Pi^(5/2)) * n!*sqrt(n)*(6/Pi^2)^n. - Vaclav Kotesovec, May 03 2014
From Vít Jelínek, Sep 04 2014: (Start)
For each m, a(5m+4) mod 5 = 0. Conjectured by Andrews-Sellers, and proved by Garvan (see Remark 1.4(ii) in Garvan's paper).
For each m, a(5m+1) mod 5 = a(5m+2) mod 5 = 3*a(5m+3) mod 5. Proved by Garvan (see (1.17) in Garvan's paper).
The limit a(n)/A022493(n) is equal to exp(-Pi^2/6). This corresponds to the asymptotic probability that a random unlabeled interval order is rigid (See Brightwell-Keller; or Jelínek, Fact 5.2). (End)
Conjectural g.f.: 1 + Sum_{n >= 0} n/(1+x)^(n+1) * (Product_{i = 1..n} 1 - 1/(1+x)^i). Cf. A194530. - Peter Bala, Aug 21 2023

Extensions

More terms from Emeric Deutsch, Mar 23 2008

A158690 Expansion of the basic hypergeometric series 1 + (1 - exp(-t)) + (1 - exp(-t))*(1 - exp(-3*t)) + (1 - exp(-t))*(1 - exp(-3*t))*(1 - exp(-5*t)) + ... as a series in t.

Original entry on oeis.org

1, 1, 5, 55, 1073, 32671, 1431665, 85363615, 6646603073, 654896692351, 79656194515025, 11722538113191775, 2052949879753739873, 421931472111868912831, 100568330857984368195185
Offset: 0

Views

Author

Peter Bala, Mar 24 2009

Keywords

Comments

We appear to get the same sequence by expanding 1 - (1 - exp(t)) + (1 - exp(t))*(1 - exp(2*t)) - (1 - exp(t))*(1 - exp(2*t))*(1 - exp(3*t)) + ... as a series in t. Compare with A079144. For other sequences with generating functions of a similar type see A000364, A000464, A002105 and A002439.
From Peter Bala, Mar 13 2017: (Start)
It appears that the g.f. has two other forms: either F(exp(-t)) where F(q) = Sum_{n >= 0} q^(n+1)*Product_{k = 1..n} 1 - q^(2*k) = q + q^2 + q^3 - q^7 - q^8 - q^10 - q^11 - ... is a g.f. for A003475 or 1/2*G(exp(t)) where G(q) = 1 + Sum_{n >= 0} (-1)^n*q^(n+1)*Product_{k = 1..n} 1 - q^k = 1 + q - q^2 + 2*q^3 - 2*q^4 + q^5 + q^7 - 2*q^8 + ... is a g.f. for A003406. See Zagier, Example 1. (End)
From Peter Bala, Dec 18 2021: (Start)
Conjectures:
1) Taking the sequence modulo an integer k gives an eventually periodic sequence with period dividing phi(k). For example, the sequence taken modulo 16 begins [1, 1, 5, 7, 1, 15, 1, 15, 1, 15, 1, 15, ...] with an apparent pre-period of length 4 and a period of length 2.
2) Let i >= 0 and define a_i(n) = a(n+i). Then for each 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 the expansion of exp( Sum_{n >= 1} a_i(n)*x^n/n ) has integer coefficients. For example, the expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 21*x^3 + 291*x^4 + 6861*x^5 + 246171*x^6 + 12458901*x^7 + 843915891*x^8 + 73640674461*x^9 + 8041227405771*x^10 + ... appears to have integer coefficients. (End)

Examples

			G.f. A(x) = 1 + x + 5*x^2 + 55*x^3 + 1073*x^4 + 32671*x^5 + 1431665*x^6 + ...
		

Crossrefs

Programs

  • Mathematica
    max = 14; se = Series[1 + Sum[ Product[1 - E^(-(2*k - 1)*t), {k, 1, n}], {n, 1, max}], {t, 0, max}]; CoefficientList[se, t]*Range[0, max]! (* Jean-François Alcover, Mar 06 2013 *)
  • PARI
    {a(n)=n!*polcoeff(sum(m=0, n, prod(k=1, m, 1-exp(-(2*k-1)*x+x*O(x^n)))), n)} \\ Paul D. Hanna, Aug 01 2012
    
  • PARI
    {a(n)=n!*polcoeff(sum(m=0, n, prod(k=1, m, exp(k*x+x*O(x^n))-1)), n)} \\ Paul D. Hanna, Aug 01 2012

Formula

Basic hypergeometric generating function: 1 + Sum_{n >= 0} Product_{k = 1..n} (1 - exp(2*k-1)*t) = 1 + t + 5*t^2/2! + 55*t^3/3! + ....
a(n) ~ 6*sqrt(2) * 12^n * (n!)^2 / Pi^(2*n+2). - Vaclav Kotesovec, May 04 2014
Conjectural g.f.: G(exp(t)) as a formal power series in t, where G(q) := Sum_{n >= 0} q^(2*n+1) * Product_{k = 1..2*n} (1 - q^k). - Peter Bala, May 16 2017
E.g.f.: Sum_{n>=0} exp(n*(n+1)/2*x) / Product_{k=0..n} (1 + exp(k*x)). - Paul D. Hanna, Oct 14 2020

A156179 Numerator of Euler(n,1/3).

Original entry on oeis.org

1, -1, -2, 13, 22, -121, -602, 18581, 30742, -305071, -2523002, 61203943, 303692662, -4353296221, -50402079002, 6669149100757, 11030684333782, -206772189255571, -3077986048956602, 128970681211645873, 1066578948824962102, -24697503335329725121, -449342758735568563802
Offset: 0

Views

Author

N. J. A. Sloane, Nov 07 2009

Keywords

Examples

			1, -1/6, -2/9, 13/108, 22/81, -121/486, -602/729, 18581/17496, ...
		

Crossrefs

Cf. A156180 and A079144.
Terms of even indices give A210657.

Programs

  • Maple
    [seq(euler(n,1/3),n=0..50)];
  • Mathematica
    Numerator[EulerE[Range[0,30],1/3]] (* Harvey P. Dale, Apr 29 2012 *)

A208730 Sequence related to Kashaev's invariant for the (5,2)-torus knot.

Original entry on oeis.org

1, 2, 10, 104, 1870, 51632, 2027470, 107354144, 7370645950, 636754087472, 67591284235630, 8647294709864384, 1312197219579059230, 233025643830843282512
Offset: 0

Views

Author

Peter Bala, Mar 01 2012

Keywords

Comments

This is sequence b_n(5) in Table 2 of Hikami 2003.

Crossrefs

Formula

Define F(q) := sum {m,n >= 0} (q^(-m*n)*product {i = 1.. m+n} (1-q^i)).
E.g.f.: F(exp(-t)) = 1 + 2*t + 10*t^2! + 104*t^3/3! + .... For the expansion of F(1-q) see A208733. F(q) also appears in a conjectural e.g.f. for A208679.
a(n) = (9/40)^n*sum {k = 0..n} binomial(n,k)*A208679(k+1)/9^k.
Conjectural S-fraction for the o.g.f.: 1/(1-2*x/(1-3*x/(1-9*x/(1-11*x/(1-...-1/2*n*(5*n-1)*x/(1-1/2*n*(5*n+1)*x/(1- ....

A208731 Sequence related to Kashaev's invariant for the (7,2)-torus knot.

Original entry on oeis.org

1, 3, 21, 303, 7581, 291903, 16004541, 11842303, 113649275901, 13729560793503, 2038432934681661, 364819653173167503, 77453921498088781821, 19245815560568454066303, 5532998260944453966936381
Offset: 0

Views

Author

Peter Bala, Mar 01 2012

Keywords

Comments

This is sequence b_n(7) in Table 2 of Hikami 2003.

Crossrefs

Formula

a(n) = (25/56)^n*sum {k = 0..n} binomial(n,k)*A208680(k+1)/25^k.
Conjectural S-fraction for the o.g.f.: 1/(1-3*x/(1-4*x/(1-13*x/(1-15*x/(1-...-1/2*n*(7*n-1)*x/(1-1/2*n*(7*n+1)*x/(1- ....
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