cp's OEIS Frontend

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

Showing 1-6 of 6 results.

A255884 Expansion of exp( Sum_{n >= 1} A002438(n)*x^n/n ).

Original entry on oeis.org

1, 5, 115, 7955, 1179715, 304888655, 121350927565, 68751844662605, 52528700295424915, 52031089992310711055, 64835758857480094584265, 99249388572274155967996505, 183075972804988649078529524365, 400493686169423616676960341062705, 1025151296160300228944197705742007715
Offset: 0

Views

Author

Peter Bala, Mar 09 2015

Keywords

Comments

A002438(n+1) =(-1)^n*6^(2*n)*E(2*n,1/6), where E(n,x) denotes the n-th Euler polynomial. In general it appears that when k is a nonzero integer, the expansion of exp( Sum_{n >= 1} k^(2*n)*E(2*n,1/k)*(-x)^n/n ) has (positive) integer coefficients. See A255881 (k = 2), A255882(k = 3) and A255883 (k = 4).

Crossrefs

Programs

  • Maple
    #A255884
    k := 6:
    exp(add(k^(2*n)*euler(2*n, 1/k)*(-x)^n/n, n = 1 .. 14)): seq(coeftayl(%, x = 0, n), n = 0 .. 14);
  • Mathematica
    A000243[n_]:= (1 + 9^(n - 1))*Abs[EulerE[2*(n - 1)]]/2; a:= With[{nmax = 75}, CoefficientList[Series[Exp[Sum[A000243[k + 1]*x^(k)/(k), {k, 1, 85}]], {x, 0, nmax}], x]]; Table[a[[n]], {n, 1, 50}] (* G. C. Greubel, Aug 26 2018 *)

Formula

O.g.f.: exp( 5*x + 205*x^2/2 + 22265*x^3/3 + 4544185 *x^4/4 + ... ) = 1 + 5*x + 115*x^2 + 7955*x^3 + 1179715*x^4 + ....
a(0) = 1 and for n >= 1, n*a(n) = Sum_{k = 1..n} (-1)^k*6^(2*k)*E(2*k,1/6)*a(n-k).
a(n) ~ 2^(4*n + 2) * 3^(2*n) * n^(2*n - 1/2) / (exp(2*n) * Pi^(2*n + 1/2)). - Vaclav Kotesovec, Jun 08 2019

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

Original entry on oeis.org

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

Views

Author

Keywords

Comments

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

Examples

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

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 810; gives a version with signs: E_{2n} = (-1)^n*a(n) (this is A028296).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
  • J. M. Borwein and D. M. Bailey, Mathematics by Experiment, Peters, Boston, 2004; p. 49
  • J. M. Borwein, D. H. Bailey, and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 141.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 420.
  • G. Chrystal, Algebra, Vol. II, p. 342.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 110.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 69.
  • L. Euler, Inst. Calc. Diff., Section 224.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 444.
  • L. Seidel, Über eine einfache Entstehungsweise der Bernoulli'schen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapters 5 and 33, pages 41, 314.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 269.

Crossrefs

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

Programs

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

Formula

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

Extensions

Typo in name corrected by Anders Claesson, Dec 01 2015

A083884 a(n) = (3^(2*n) + 1) / 2.

Original entry on oeis.org

1, 5, 41, 365, 3281, 29525, 265721, 2391485, 21523361, 193710245, 1743392201, 15690529805, 141214768241, 1270932914165, 11438396227481, 102945566047325, 926510094425921, 8338590849833285, 75047317648499561, 675425858836496045, 6078832729528464401
Offset: 0

Views

Author

Paul Barry, May 09 2003

Keywords

Comments

Number of compositions of even natural numbers into n parts <= 8. - Adi Dani, May 28 2011
a(n) for n >= 1 gives the number of line segments in the n-th iteration of the Peano curve given by plotting (A163528, A163529) or by (Siromoney 1982) when parallel line segments that are connected end-to-end are counted as a single line segment. - Jason V. Morgan, Oct 08 2021

Examples

			From _Adi Dani_, May 28 2011: (Start)
a(2)=41: there are 41 compositions of even natural numbers into 2 parts <=8:
(0,0);
(0,2),(2,0),(1,1);
(0,4),(4,0),(1,3),(3,1),(2,2);
(0,6),(6,0),(1,5),(5,1),(2,4),(4,2),(3,3);
(0,8),(8,0),(1,7),(7,1),(2,6),(6,2),(3,5),(5,3),(4,4);
(2,8),(8,2),(3,7),(7,3),(4,6),(6,4),(5,5);
(4,8),(8,4),(5,7),(7,5),(6,6);
(6,8),(8,6),(7,7);
(8,8).  (End)
		

References

  • Siromoney, R., & Subramanian, K.G. (1982). Space-filling curves and infinite graphs. Graph-Grammars and Their Application to Computer Science.

Crossrefs

Programs

Formula

a(0) = 1, a(n) = 9*a(n-1) - 4.
a(n) = Sum_{k=0..n} binomial(2*n, 2*k)*4^k.
a(n) = A002438(n) / A000364(n); A000364(n) : Euler numbers.
G.f.: (1-5*x)/((1-x)*(1-9*x)).
a(n) = (3^n + 1^n + (-1)^n + (-3)^n)/4.
E.g.f.: exp(3*x) + exp(x) + exp(-x) + exp(-3*x).
Each term expresses a Pythagorean relationship, along with (a(n)-1) and a power of 3, n>0, such that sqrt((a(n))^2 - (a(n)-1)^2) = 3^n. E.g., 365^2 - 364^2 - 3^3 = 27 (the Pythagorean triangle (365, 364, 27)). - Gary W. Adamson, Jun 25 2006
a(n) = 10*a(n-1) - 9*a(n-2). - Wesley Ivan Hurt, Apr 21 2021

Extensions

Additional comments from Philippe Deléham, Jul 10 2005

A086646 Triangle, read by rows, of numbers T(n,k), 0 <= k <= n, given by T(n,k) = A000364(n-k)*binomial(2*n, 2*k).

Original entry on oeis.org

1, 1, 1, 5, 6, 1, 61, 75, 15, 1, 1385, 1708, 350, 28, 1, 50521, 62325, 12810, 1050, 45, 1, 2702765, 3334386, 685575, 56364, 2475, 66, 1, 199360981, 245951615, 50571521, 4159155, 183183, 5005, 91, 1, 19391512145, 23923317720, 4919032300, 404572168, 17824950, 488488, 9100, 120, 1
Offset: 0

Views

Author

Philippe Deléham, Jul 26 2003

Keywords

Comments

The elements of the matrix inverse are apparently given by T^(-1)(n,k) = (-1)^(n+k)*A086645(n,k). - R. J. Mathar, Mar 14 2013
Let E(y) = Sum_{n >= 0} y^n/(2*n)! = cosh(sqrt(y)). Then this triangle is the generalized Riordan array (1/E(-y), y) with respect to the sequence (2*n)! as defined in Wang and Wang. - Peter Bala, Aug 06 2013
Let P_n be the poset of even size subsets of [2n] ordered by inclusion. Then Sum_{k=0..n}(-1)^(n-k)*T(n,k)*x^k is the characteristic polynomial of P_n. - Geoffrey Critzer, Feb 24 2021

Examples

			Triangle begins:
      1;
      1,     1;
      5,     6,     1;
     61,    75,    15,    1;
   1385,  1708,   350,   28,  1;
  50521, 62325, 12810, 1050, 45, 1;
  ...
From _Peter Bala_, Aug 06 2013: (Start)
Polynomial  |        Real zeros to 5 decimal places
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
R(5,-x)     | 1, 9.18062, 13.91597
R(10,-x)    | 1, 9.00000, 25.03855,  37.95073
R(15,-x)    | 1, 9.00000, 25.00000,  49.00895, 71.83657
R(20,-x)    | 1, 9.00000, 25.00000,  49.00000, 81.00205, 114.87399
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
(End)
		

Crossrefs

Cf. A000281.
Cf. A000795 (row sums).
Cf. A055133, A086645 (unsigned matrix inverse), A103364, A104033.
T(2n,n) give |A214445(n)|.

Programs

  • Maple
    A086646 := proc(n,k)
        if k < 0 or k > n then
            0 ;
        else
            A000364(n-k)*binomial(2*n,2*k) ;
        end if;
    end proc: # R. J. Mathar, Mar 14 2013
  • Mathematica
    R[0, _] = 1;
    R[n_, x_] := R[n, x] = x^n - Sum[(-1)^(n-k) Binomial[2n, 2k] R[k, x], {k, 0, n-1}];
    Table[CoefficientList[R[n, x], x], {n, 0, 8}] // Flatten (* Jean-François Alcover, Dec 19 2019 *)
    T[0, 0] := 1; T[n_, 0] := -Sum[(-1)^k T[n, k], {k, 1, n}]; T[n_, k_]/;0Oliver Seipel, Jan 11 2025 *)

Formula

cosh(u*t)/cos(t) = Sum_{n>=0} S_2n(u)*(t^(2*n))*(1/(2*n)!). S_2n(u) = Sum_{k>=0} T(n,k)*u^(2*k). Sum_{k>=0} (-1)^k*T(n,k) = 0. Sum_{k>=0} T(n,k) = 2^n*A005647(n); A005647: Salie numbers.
Triangle T(n,k) read by rows; given by [1, 4, 9, 16, 25, 36, 49, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938.
Sum_{k=0..n} (-1)^k*T(n,k)*4^(n-k) = A000281(n). - Philippe Deléham, Jan 26 2004
Sum_{k=0..n} T(n,k)*(-4)^k*9^(n-k) = A002438(n+1). - Philippe Deléham, Aug 26 2005
Sum_{k=0..n} (-1)^k*9^(n-k)*T(n,k) = A000436(n). - Philippe Deléham, Oct 27 2006
From Peter Bala, Aug 06 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/(2*n)! = cosh(sqrt(y)). Generating function: E(x*y)/E(-y) = 1 + (1 + x)*y/2! + (5 + 6*x + x^2)*y^2/4! + (61 + 75*x + 15*x^2 + x^3)*y^3/6! + .... The n-th power of this array has a generating function E(x*y)/E(-y)^n. In particular, the matrix inverse is a signed version of A086645 with a generating function E(-y)*E(x*y).
Recurrence equation for the row polynomials: R(n,x) = x^n - Sum_{k = 0..n-1} (-1)^(n-k)*binomial(2*n,2*k)*R(k,x) with initial value R(0,x) = 1.
It appears that for arbitrary complex x we have lim_{n -> infinity} R(n,-x^2)/R(n,0) = cos(x*Pi/2). A stronger result than pointwise convergence may hold: the convergence may be uniform on compact subsets of the complex plane. This would explain the observation that the real zeros of the polynomials R(n,-x) seem to converge to the odd squares 1, 9, 25, ... as n increases. Some numerical examples are given below. Cf. A055133, A091042 and A103364.
R(n,-1) = 0; R(n,-9) = (-1)^n*2*4^n; R(n,-25) = (-1)^n*2*(16^n - 4^n);
R(n,-49) = (-1)^n*2*(36^n - 16^n + 4^n). (End)

A007286 E.g.f.: (sin x + cos 2x) / cos 3x.

Original entry on oeis.org

1, 1, 5, 26, 205, 1936, 22265, 297296, 4544185, 78098176, 1491632525, 31336418816, 718181418565, 17831101321216, 476768795646785, 13658417358350336, 417370516232719345, 13551022195053101056
Offset: 0

Views

Author

Keywords

Comments

Arises in the enumeration of alternating 3-signed permutations.

References

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

Crossrefs

Cf. A006873, A007289, A225109, A002438 (bisection?).

Programs

  • Mathematica
    mx = 17; Range[0, mx]! CoefficientList[ Series[ (Sin[x] + Cos[2x])/Cos[3 x], {x, 0, mx}], x] (* Robert G. Wilson v, Apr 28 2013 *)
  • PARI
    x='x+O('x^66); Vec(serlaplace((sin(x)+cos(2*x))/cos(3*x))) \\ Joerg Arndt, Apr 28 2013
    
  • Sage
    from mpmath import mp, polylog, re
    mp.dps = 32; mp.pretty = True
    def aperm3(n): return 2*((1-I)/(1+I))^n*(1+add(binomial(n,j)*polylog(-j,I)*3^j for j in (0..n)))
    def A007286(n) : return re(aperm3(n))
    [int(A007286(n)) for n in (0..17)] # Peter Luschny, Apr 28 2013

Formula

a(n) = Re(2*((1-I)/(1+I))^n*(1+sum_{j=0..n}(binomial(n,j)*Li_{-j}(I)*3^j))). - Peter Luschny, Apr 28 2013
a(n) ~ n! * 2^(n+1)*3^n/Pi^(n+1). - Vaclav Kotesovec, Jun 15 2013

A225147 a(n) = Im((1-I)^(1-n)*A_{n, 3}(I)) where A_{n, k}(x) are the generalized Eulerian polynomials.

Original entry on oeis.org

-1, 2, 5, -46, -205, 3362, 22265, -515086, -4544185, 135274562, 1491632525, -54276473326, -718181418565, 30884386347362, 476768795646785, -23657073914466766, -417370516232719345, 23471059057478981762, 465849831125196593045, -29279357851856595135406
Offset: 0

Views

Author

Peter Luschny, Apr 30 2013

Keywords

Crossrefs

Cf. A000810 (real part (up to sign)), A212435 (k=2), A122045 (k=1), A002439.
Bisections are A002438 and A000191.

Programs

  • Maple
    B := proc(n, u, k) option remember;
    if n = 1 then if (u < 0) or (u >= 1) then 0 else 1 fi
    else k*u*B(n-1, u, k) + k*(n-u)*B(n-1, u-1, k) fi end:
    EulerianPolynomial := proc(n, k, x) local m; if x = 0 then RETURN(1) fi;
    add(B(n+1, m+1/k, k)*u^m, m = 0..n); subs(u=x, %) end:
    seq(Im((1-I)^(1-n)*EulerianPolynomial(n, 3, I)), n=0..19);
  • Mathematica
    CoefficientList[Series[-E^(-2*x)*Sech[3*x],{x,0,20}],x] * Range[0,20]! (* Vaclav Kotesovec, Sep 29 2014 after Sergei N. Gladkovskii *)
    Table[-6^n EulerE[n,1/6], {n,0,19}] (* Peter Luschny, Nov 16 2016 after Peter Bala *)
  • Sage
    from mpmath import mp, polylog, im
    mp.dps = 32; mp.pretty = True
    def A225147(n): return im(-2*I*(1+add(binomial(n,j)*polylog(-j,I)*3^j for j in (0..n))))
    [int(A225147(n)) for n in (0..19)]

Formula

a(n) = Im(-2*i*(1+Sum_{j=0..n}(binomial(n,j)*Li{-j}(i)*3^j))).
For a recurrence see the Maple program.
G.f.: conjecture -T(0)/(1+2*x), where T(k) = 1 - 9*x^2*(k+1)^2/(9*x^2*(k+1)^2 + (1+2*x)^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 12 2013
a(n) = -(-3)^n*skp(n, 2/3), where skp(n,x) are the Swiss-Knife polynomials A153641. - Peter Luschny, Apr 19 2014
G.f.: A225147 = -1/T(0), where T(k) = 1 + 2*x + (k+1)^2*(3*x)^2/ T(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 29 2014
E.g.f.: -exp(-2*x)*sech(3*x). - Sergei N. Gladkovskii, Sep 29 2014
a(n) ~ n! * (sqrt(3)*sin(Pi*n/2) - cos(Pi*n/2)) * 2^(n+1) * 3^n / Pi^(n+1). - Vaclav Kotesovec, Sep 29 2014
From Peter Bala, Nov 13 2016: (Start)
a(n) = - 6^n*E(n,1/6), where E(n,x) denotes the Euler polynomial of order n.
a(2*n) = (-1)^(n+1)*A002438(n); a(2*n+1) = (1/2)*(-1)^n*A002439(n). (End)
Showing 1-6 of 6 results.