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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A034971 a(n) = floor(E_(n+1)/E_(n)) where E_n is n-th Euler number (see A028296 and A000364).

Original entry on oeis.org

5, 12, 22, 36, 53, 73, 97, 124, 154, 187, 223, 263, 306, 352, 402, 454, 510, 569, 632, 697, 766, 838, 914, 992, 1074, 1159, 1248, 1339, 1434, 1532, 1634, 1738, 1846, 1957, 2071, 2189, 2310, 2434, 2561, 2691, 2825, 2962, 3102, 3246, 3393, 3542, 3696, 3852
Offset: 1

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Author

Keywords

Comments

First nine terms are identical to those of A025740.

References

  • J. Peters J. and J. Stein, Mathematische Tafeln. Revised Russian Edition in 1968, Moscow, Table 9b.

Crossrefs

Programs

  • Mathematica
    Wolfram: ae[ x_ ] := Abs[ EulerE[ 2*x ] ] and fo[ x_ ] := Floor[ N[ ae[ x ]/ae[ x-1 ], 80 ] ]
    Table[Floor[Abs[EulerE[2n+2]]/Abs[EulerE[2n]]],{n,60}] (* Harvey P. Dale, May 01 2011 *)

Formula

a(n) = floor(E(n+1)/E(n)) where E(n) is the n-th coefficient in the expansion of sec(x).

A239322 Interleave (-1)^n*(A000182(n+1) - A000364(n)), -A028296(n+1).

Original entry on oeis.org

0, 1, 1, -5, -11, 61, 211, -1385, -6551, 50521, 303271, -2702765, -19665491, 199360981, 1704396331, -19391512145, -190473830831, 2404879675441, 26684005437391, -370371188237525
Offset: 0

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Author

Paul Curtz, Mar 28 2014

Keywords

Comments

Difference table of a(n):
0, 1, 1, -5, -11, 61, 211, -1385,...
1, 0, -6, -6, 72, 150, -1596,...
-1, -6, 0, 78, 78, -1746,...
-5, -6, 78, 0, -1824,...
11, 72, 78, -1824,...
61, -150, -1746,...
-211, -1596,...
-1385,...
etc.
a(n) is an autosequence (its inverse binomial transform is the signed sequence) of the first kind (its main diagonal is A000004=0's and the first two upper diagonal are the same). Like A000045(n).
Note that e(n) = A000111(n+1) - A000111(n) = 0, 0, 1, 3, 11, 45, 211,... is not in the OEIS. a(2n) = (-1)*(n+1)*e(2n).
Antidiagonals upon A000004:
1,
1,
-6, -5,
-6, -11,
78, 72, 61,
78, 150, 211,
Row sum: 1, 1, -11, -17, 211, 439,... .
b(n) = a(n) mod 9 = 0 followed by period 6: repeat 1, 1, 4, 7, 7, 4 is also an autosequence of the first kind.

Examples

			a(0)=1-1=0, a(1)=-(-1)=1, a(2)=2-1=1, a(3)=-5, a(4)=-(16-5)=-11.
		

Crossrefs

Cf. Zig (A000364) and Zag (A000182) give Euler A000111(n).

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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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

A122045 Euler (or secant) numbers E(n).

Original entry on oeis.org

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

Views

Author

Roger L. Bagula, Sep 13 2006

Keywords

Comments

The convention in the OEIS is that the alternate zeros are normally omitted in such sequences. See A000364 for the official version of this sequence.
Odd primes p such that p | E(p-1) are primes p == 1 (mod 4), A002144. Conjecture: odd composites m such that m | E(m-1) are Carmichael numbers m such that p == 1 (mod 4) for every prime p|m, A265237. - Thomas Ordowski, Feb 06 2020

Examples

			G.f. = 1 - x^2 + 5*x^4 - 61*x^6 + 1385*x^8 - 50521*x^10 + 2702765*x^12 + ...
		

References

  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 5, page 41.

Crossrefs

Programs

  • Magma
    m:=35; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( 1/Cosh(x) )); [Factorial(n-1)*b[n]: n in [1..m-1]]; // G. C. Greubel, Feb 13 2020
    
  • Maple
    seq(euler(n) , n=0..31); # Zerinvary Lajos, Mar 15 2009
    P := proc(n,x) option remember; if n = 0 then 1 else
       (n*x+(1/2)*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x); expand(%) fi end:
    A122045 := n -> (-1)^n*subs(x=-1, P(n,x)):
    seq(A122045(n), n=0..30);  # Peter Luschny, Mar 07 2014
    ptan := proc(n) option remember; if irem(n, 2) = 1 then 0 else
        -add(`if`(k=0, 1, binomial(n, k)*ptan(n - k)), k = 0..n-1,2) fi end:
    A122045 := n -> ifelse(n = 0, 1, ptan(n)):
    seq(A122045(n), n = 0..30); # Peter Luschny, Jun 06 2022
  • Mathematica
    Table[EulerE[n], {n, 0, 30}]
    Range[0, 30]! CoefficientList[ Series[ Sech[x], {x, 0, 30}], x] (* Robert G. Wilson v, Aug 08 2018 *)
    a[0] := 1; a[n_?OddQ] := 0; a[n_?EvenQ] := a[n] = -Sum[a[k] Binomial[n, k], {k, 0, n - 1, 2}]; Map[a, Range[0, 30]] (* Oliver Seipel, May 20 2024 *)
  • Maxima
    a[n]:=if n<2 then 1-n else sum(-a[n-2*k]*binomial(n,2*k),k,1,floor(n/2));
    makelist(a[n],n,0,50); /* Tani Akinari, Sep 15 2023 */
  • PARI
    x='x+O('x^66); Vec(serlaplace(1/cosh(x))) \\ Joerg Arndt, Mar 10 2014
    
  • PARI
    a(n) = 2^n*2^(n+1)*(subst(bernpol(n+1,x), x, 3/4) - subst(bernpol(n+1,x), x, 1/4))/(n+1); \\ Michel Marcus, May 20 2017
    
  • Python
    from sympy import bernoulli as B
    def a(n): return int(2**n*2**(n + 1)*(B(n + 1, 3/4) - B(n + 1, 1/4))/(n + 1))
    print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 24 2017, after PARI code by Michel Marcus
    
  • Python
    from functools import cache
    from math import comb as binomial
    @cache
    def ptan(n):   # see also A331978 and A350972.
        return (0 if n % 2 == 1 else
        -sum(binomial(n,k) * ptan(n-k) if k > 0 else 1 for k in range(0, n-1, 2)))
    def A122045(n): return 1 if n == 0 else ptan(n)
    print([A122045(n) for n in range(31)])  # Peter Luschny, Jun 06 2022
    
  • Sage
    [euler_number(i) for i in range(31)] # Zerinvary Lajos, Mar 15 2009
    

Formula

E.g.f.: sech(x). - Michael Somos, Mar 11 2014
a(n) = (Sum_{k>=0} (-1)^k*(2*k+1)^n)*2. - Gottfried Helms, Mar 09 2012
From Sergei N. Gladkovskii, Oct 14 2012 - Oct 13 2013: (Start)
Continued fractions:
G.f.: 1/U(0) where U(k) = 1 - x + x*(k+1)/(1 - x*(k+1)/U(k+1)).
G.f.: 1/U(0) where U(k) = 1 - x^2 + x^2*(2*k+1)*(2*k+2)/(1 + x^2*(2*k+1)*(2*k+2)/ U(k+1)).
E.g.f.: (1-x)/U(0) where U(k) = 1 - x/(1 - x/(x - (2*k+1)*(2*k+2)/U(k+1))).
E.g.f.: 1 - x^2/U(0) where U(k) = (2*k+1)*(2*k+2) + x^2 - x^2*(2*k+1)*(2*k+2)/U(k+1).
E.g.f.: 1/U(0) where U(k) = 1 + x^2/(2*(2*k+1)*(4*k+1) - 2*x^2*(2*k+1)*(4*k+1)/(x^2 + 4*(4*k+3)*(k+1)/U(k+1))).
E.g.f.: (2 + x^4/(U(0)*(x^2-2) - 2))/(2-x^2) where U(k) = 4*k + 4 + 1/(1 + x^2/(2 - x^2 + (2*k+3)*(2*k+4)/U(k+1))).
G.f.: 1/G(0) where G(k) = 1 + x^2*(2*k+1)^2/(1 + x^2*(2*k+2)^2/G(k+1)) (due to T. J. Stieltjes).
G.f.: 1/S(0) where S(k) = 1 + x^2*(k+1)^2/S(k+1) (due to T. J. Stieltjes).
G.f.: 1 - x/(1+x) + x/(1+x)/Q(0) where Q(k) = 1 - x + x*(k+2)/(1 - x*(k+1)/Q(k+1)).
G.f.: -(1/x)/Q(0) where Q(k) = -1/x + (k+1)^2/Q(k+1) (due to T. J. Stieltjes).
G.f.: (1/(1-x))/Q(0) + 1/(1-x) where Q(k) = 1 - 1/x + (k+1)*(k+2)/Q(k+1).
G.f.: (x/(x-1))/Q(0) + 1/(1-x) where Q(k) = 1 - x + x^2*(k+1)*(k+2)/Q(k+1).
G.f.: 1 - x/(1+x) + (x/(1+x))/Q(0) where Q(k) = 1 + x + (k+1)*(k+2)*x^2/Q(k+1).
E.g.f.: 1 - T(0)*x^2/(2+x^2) where T(k) = 1 - x^2*(2*k+1)*(2*k+2)/(x^2*(2*k+1)*(2*k+2) - ((2*k+1)*(2*k+2) + x^2)*((2*k+3)*(2*k+4) + x^2)/T(k+1)).
G.f.: T(0) where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 + 1/T(k+1)). (End)
a(n) = 2^(2*n+1)*(zeta(-n,1/4) - zeta(-n,3/4)), where zeta(a, z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
a(n) = 2^n*(2^(n+1)/(n+1))*(B(n+1, 3/4) - B(n+1, 1/4)) where B(n,x) is the n-th Bernoulli polynomial. See Liu link. - Michel Marcus, May 20 2017 [This is the same as: a(n) = -4^(n+1)*B(n+1, 1/4)*((n+1) mod 2)/(n+1). Peter Luschny, Oct 30 2020]
a(n) = 2*Im(PolyLog(-n, I)). - Peter Luschny, Sep 29 2020
a(4n) == 5 (mod 60) and a(4n+2) == -1 (mod 60). See Hirschhorn. - Michel Marcus, Jan 11 2022
For n > 1, a(n) = -Sum_{k=1..floor(n/2)} a(n-2*k)*binomial(n,2*k). - Tani Akinari, Sep 15 2023

Extensions

Edited by N. J. A. Sloane, Sep 17 2006

A094088 E.g.f. 1/(2-cosh(x)) (even coefficients).

Original entry on oeis.org

1, 1, 7, 121, 3907, 202741, 15430207, 1619195761, 224061282907, 39531606447181, 8661323866026007, 2307185279184885001, 734307168916191403507, 275199311597682485597221, 119956934012963778952439407
Offset: 0

Views

Author

Ralf Stephan, Apr 30 2004

Keywords

Comments

With alternating signs, e.g.f.: 1/(2-cos(x)).
7 divides a(3n+2). Ira Gessel remarks: For any odd prime p, the coefficients of 1/(2-cosh(x)) as e.g.f. are periodic with period dividing p-1.
Consider the sequence defined by a(0) = 1; thereafter a(n) = c*Sum_{k = 1..n} binomial(2n,2k)*a(n-k). For c = -3, -2, -1, 1, 2, 3, 4 this is A210676, A210657, A028296, A094088, A210672, A210674, A249939.
a(n) is the number of ordered set partitions of {1,2,...,2n} into even size blocks. - Geoffrey Critzer, Dec 03 2012
Except for a(0), row sums of A241171. - Peter Bala, Aug 20 2014
Exp( Sum_{n >= 1} a(n)*x^n/n) is the o.g.f. for A255928. - Peter Bala, Mar 13 2015
Also the 2-packed words of degree n; cf. A011782, A000670, A094088, A243664, A243665, A243666 for k-packed words for 0<=k<=5. - Peter Luschny, Jul 06 2015

Crossrefs

Programs

  • Maple
    f:=proc(n,k) option remember;  local i;
    if n=0 then 1
    else k*add(binomial(2*n,2*i)*f(n-i,k),i=1..floor(n)); fi; end;
    g:=k->[seq(f(n,k),n=0..40)];g(1); # N. J. A. Sloane, Mar 28 2012
  • Mathematica
    nn=30;Select[Range[0,nn]!CoefficientList[Series[1/(2-Cosh[x]),{x,0,nn}],x],#>0&]  (* Geoffrey Critzer, Dec 03 2012 *)
    a[0]=1; a[n_] := Sum[1/2*(1+(-1)^(2*n))*Sum[((-1)^(k-j)*Binomial[k, j]*Sum[(j-2*i )^(2*n)*Binomial[j, i], {i, 0, j}])/2^j, {j, 1, k}], {k, 1, n}]; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Apr 03 2015, after Vladimir Kruchinin *)
  • Maxima
    a(n):=b(2*n+2);
    b(n):=sum(((sum(((sum((j-2*i)^n*binomial(j,i),i,0,j))*(-1)^(k-j)*binomial(k,j))/2^(j),j,1,k))*((-1)^n+1))/2,k,1,n/2); /* Vladimir Kruchinin, Apr 23 2011 */
    
  • Maxima
    a(n):=sum(sum((i-k)^(2*n)*binomial(2*k,i)*(-1)^(i),i,0,k-1)/(2^(k-1)),k,1,2*n); /* Vladimir Kruchinin, Oct 05 2012 */
  • PARI
    a(n) = if (n == 0, 1, sum(k=1, n, binomial(2*n, 2*n-2*k)*a(n-k)));
    
  • Sage
    def A094088(n) :
        @CachedFunction
        def intern(n) :
            if n == 0 : return 1
            if n % 2 != 0 : return 0
            return add(intern(k)*binomial(n,k) for k in range(n)[::2])
        return intern(2*n)
    [A094088(n) for n in (0..14)]  # Peter Luschny, Jul 14 2012
    

Formula

1/(2-cosh(x)) = Sum_{n>=0} a(n)x^(2n)/(2n)! = 1 + x^2/2 + 7x^4/24 + 121x^6/720 + ...
Recurrence: a(0)=1, a(n) = Sum_{k=1..n} C(2n, 2n-2k)*a(n-k).
a(0)=1 and, for n>=1, a(n)=b(2*n) where b(n) = sum(k=1..n/2,((sum(j=1..k, ((sum(i=0..j,(j-2*i)^n*binomial(j,i)))*(-1)^(k-j)*binomial(k,j))/2^(j)))*((-1)^n+1))/2). - Vladimir Kruchinin, Apr 23 2011
E.g.f.: 1/(2-cosh(x))=8*(1-x^2)/(8 - 12*x^2 + 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))) ; (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Sep 30 2012
a(n) = Sum_{k=1..2*n} ( Sum_{i=0..k-1} (i-k)^(2*n) * binomial(2*k,i) * (-1)^i )/2^(k-1), n>0, a(0)=1. - Vladimir Kruchinin, Oct 05 2012
a(n) ~ 2*(2*n)! /(sqrt(3) * (log(2+sqrt(3)))^(2*n+1)). - Vaclav Kotesovec, Oct 19 2013

Extensions

Corrected definition, Joerg Arndt, Apr 26 2011

A065547 Triangle of Salie numbers.

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 3, -3, 1, 0, -17, 17, -6, 1, 0, 155, -155, 55, -10, 1, 0, -2073, 2073, -736, 135, -15, 1, 0, 38227, -38227, 13573, -2492, 280, -21, 1, 0, -929569, 929569, -330058, 60605, -6818, 518, -28, 1, 0, 28820619, -28820619, 10233219, -1879038, 211419, -16086, 882, -36, 1, 0, -1109652905
Offset: 0

Views

Author

Wouter Meeussen, Dec 02 2001

Keywords

Comments

Coefficients of polynomials H(n,x) related to Euler polynomials through H(n,x(x-1)) = E(2n,x).

Examples

			Triangle begins:
 1;
 0,   1;
 0,  -1,    1;
 0,   3,   -3,  1;
 0, -17,   17, -6,   1;
 0, 155, -155, 55, -10, 1;
 ...
		

Crossrefs

Sum_{k>=0} (-1)^(n+k)*2^(n-k)*T(n, k) = A005647(n). Sum_{k>=0} (-1)^(n+k)*2^(2n-k)*T(n, k) = A000795(n). Sum_{k>=0} (-1)^(n+k)*T(n, k) = A006846(n), where A006846 = Hammersley's polynomial p_n(1). - Philippe Deléham, Feb 26 2004.
Column sequences (without leading zeros) give, for k=1..10: A065547 (twice), A095652-9.
See A085707 for unsigned and transposed version.
See A098435 for negative values of n, k.

Programs

  • Mathematica
    h[n_, x_] := Sum[c[k]*x^k, {k, 0, n}]; eq[n_] := SolveAlways[h[n, x*(x - 1)] == EulerE[2*n, x], x]; row[n_] := Table[c[k], {k, 0, n}] /. eq[n] // First; Table[row[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Oct 02 2013 *)
  • PARI
    { S2(n, k) = (1/k!)*sum(i=0,k,(-1)^(k-i)*binomial(k,i)*i^n) }{ Eu(n) = sum(m=0,n,(-1)^m*m!*S2(n+1,m+1)*(-1)^floor(m/4)*2^-floor(m/2)*((m+1)%4!=0)) } T(n,k)=if(nRalf Stephan

Formula

E.g.f.: Sum_{n, k=0..oo} T(n, k) t^k x^(2n)/(2n)! = cosh(sqrt(1+4t) x/2) / cosh(x/2).
T(k, n) = Sum_{i=0..n-k} A028296(i)/4^(n-k)*C(2n, 2i)*C(n-i, n-k-i), or 0 if n
Polynomial recurrences: x^n = Sum_{0<=2i<=n} C(n, 2i)*H(n-i, x); (1/4+x)^n = Sum_{m=0..n} C(2n, 2m)*(1/4)^(n-m)*H(m, x).
Dumont/Zeng give a continued fraction and other formulas.
Triangle T(n, k) read by rows; given by [0, -1, -2, -4, -6, -9, -12, -16, ...] DELTA A000035, where DELTA is Deléham's operator defined in A084938.
Sum_{k=0..n} (-4)^(n-k)*T(n,k) = A000364(n) (Euler numbers). - Philippe Deléham, Oct 25 2006

Extensions

Edited by Ralf Stephan, Sep 08 2004

A210657 a(0)=1; thereafter a(n) = -2*Sum_{k=1..n} binomial(2n,2k)*a(n-k).

Original entry on oeis.org

1, -2, 22, -602, 30742, -2523002, 303692662, -50402079002, 11030684333782, -3077986048956602, 1066578948824962102, -449342758735568563802, 226182806795367665865622, -134065091768709178087428602, 92423044260377387363207812342, -73323347841467639992211297199002
Offset: 0

Author

N. J. A. Sloane, Mar 28 2012

Keywords

Comments

The version without signs has an interpretation as a sum over marked Schröder paths. See the Josuat-Verges and Kim reference.
Consider the sequence defined by a(0)=1; thereafter a(n) = c*Sum_{k=1..n} binomial(2n,2k)*a(n-k). For c = -3, -2, -1, 1, 2, 3, 4 this is A210676, A210657, A028296, A094088, A210672, A210674, A249939.
Apparently a(n) = 2*(-1)^n*A002114(n). - R. J. Mathar, Mar 01 2015

Programs

  • Maple
    A210657:=proc(n) option remember;
       if n=0 then 1
       else -2*add(binomial(2*n,2*k)*procname(n-k),k=1..floor(n)); fi;
    end;
    [seq(f(n),n=0..20)];
    # Second program:
    a := (n) -> 2*36^n*(Zeta(0,-n*2,1/6)-Zeta(0,-n*2,2/3)):
    seq(a(n), n=0..15); # Peter Luschny, Mar 11 2015
  • Mathematica
    nmax=20; Table[(CoefficientList[Series[1/(2*Cosh[x]-1), {x, 0, 2*nmax}], x] * Range[0, 2*nmax]!)[[2*n+1]], {n,0,nmax}] (* Vaclav Kotesovec, Mar 14 2015 *)
    Table[9^n EulerE[2 n, 1/3], {n, 0, 20}] (* Vladimir Reshetnikov, Jun 05 2016 *)
  • PARI
    a(n)=polcoeff(sum(m=0, n, (2*m)!*(-x)^m/prod(k=1, m, 1-k^2*x +x*O(x^n)) ), n)
    for(n=0,20,print1(a(n),", ")) \\ Paul D. Hanna, Sep 17 2012

Formula

O.g.f.: Sum_{n>=0} (2*n)! * (-x)^n / Product_{k=1..n} (1 - k^2*x). - Paul D. Hanna, Sep 17 2012
E.g.f.: 1/(2*cosh(x) - 1) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!. - Paul D. Hanna, Oct 30 2014
E.g.f.: cos(z/2)/cos(3z/2) = Sum_{n>=0} abs(a(n))*x^(2*n)/(2*n)!. - Olivier Gérard, Feb 12 2014
From Peter Bala, Mar 09 2015: (Start)
a(n) = 3^(2*n)*E(2*n,1/3), where E(n,x) is the n-th Euler polynomial.
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 - x*(3*k + 1)^2).
O.g.f. as a continued fraction: 1/(1 + (3^2 - 1^2)*x/(4 + 12^2*x/(4 + (18^2 - 2^2)*x/(4 + 24^2*x/(4 + (30^2 - 2^2)*x/(4 + 36^2*x/(4 + ... ))))))) = 1 - 2*x + 22*x^2 - 602*x^3 + 30742*x^4 - .... See Josuat-Vergès and Kim, p. 23.
The expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) appears to have integer coefficients. See A255882. (End)
a(n) = 2*36^n*(zeta(-n*2,1/6)-zeta(-n*2,2/3)), where zeta(a,z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
a(n) ~ 2 * (-1)^n * (2*n)! * 3^(2*n+1/2) / Pi^(2*n+1). - Vaclav Kotesovec, Mar 14 2015
a(n) = Sum_{k=0..n} A241171(n, k)*(-2)^k. - Peter Luschny, Sep 03 2022

A249939 E.g.f.: 1/(5 - 4*cosh(x)).

Original entry on oeis.org

1, 4, 100, 6244, 727780, 136330084, 37455423460, 14188457293924, 7087539575975140, 4514046217675793764, 3570250394992512270820, 3433125893070920512725604, 3944372161432193963534198500, 5336301013125557989981503385444, 8396749419933421378024498580446180
Offset: 0

Author

Paul D. Hanna, Nov 19 2014

Keywords

Comments

a(n) = 4*A242858(2*n) for n>0.
a(n) = A249940(n)/3.
a(n) == 4 (mod 96) for n>0.

Examples

			E.g.f.: E(x) = 1 + 4*x^2/2! + 100*x^4/4! + 6244*x^6/6! + 727780*x^8/8! +...
where E(x) = 1/(5 - 4*cosh(x)) = -exp(x) / (2 - 5*exp(x) + 2*exp(2*x)).
ALTERNATE GENERATING FUNCTION.
E.g.f.: A(x) = 1 + 4*x + 100*x^2/2! + 6244*x^3/3! + 727780*x^4/4! +...
where 3*A(x) = 1 + 2*exp(x)/2 + 2*exp(4*x)/2^2 + 2*exp(9*x)/2^3 + 2*exp(16*x)/2^4 + 2*exp(25*x)/2^5 + 2*exp(36*x)/2^6 + 2*exp(49*x)/2^7 +...
		

Programs

  • PARI
    /* E.g.f.: 1/(5 - 4*cosh(x)) */
    {a(n) = local(X=x+O(x^(2*n+1))); (2*n)!*polcoeff( 1/(5 - 4*cosh(X)), 2*n)}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* Formula for a(n): */
    {Stirling2(n, k)=n!*polcoeff(((exp(x+x*O(x^n))-1)^k)/k!, n)}
    {a(n) = if(n==0, 1, sum(k=1, (2*n+1)\3, 2*(3*k-1)! * Stirling2(2*n+1, 3*k)))}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* Formula for a(n): */
    {Stirling2(n, k)=n!*polcoeff(((exp(x+x*O(x^n))-1)^k)/k!, n)}
    {a(n) = if(n==0, 1, (4/3)*sum(k=0, 2*n, k! * Stirling2(2*n, k) ))}
    for(n=0, 20, print1(a(n), ", "))

Formula

E.g.f.: 1/3 + (2/3)*Sum_{n>=1} exp(n^2*x) / 2^n = Sum_{n>=0} a(n)*x^n/n!.
a(n) = (4/3) * Sum_{k=0..2*n} k! * Stirling2(2*n, k) for n>0 with a(0)=1.
a(n) = Sum_{k=1..[(2*n+1)/3]} 2 * (3*k-1)! * Stirling2(2*n+1, 3*k) for n>0 with a(0)=3, after Vladimir Kruchinin in A242858.

A210672 a(0)=1; thereafter a(n) = 2*Sum_{k=1..n} binomial(2n,2k)*a(n-k).

Original entry on oeis.org

1, 2, 26, 842, 50906, 4946282, 704888186, 138502957322, 35887046307866, 11855682722913962, 4863821092813045946, 2425978759725443056202, 1445750991051368583278426, 1014551931766896667943384042, 828063237870027116855857421306, 777768202388460616924079724057482
Offset: 0

Author

N. J. A. Sloane, Mar 28 2012

Keywords

Comments

Consider the sequence defined by a(0) = 1; thereafter a(n) = c*Sum_{k = 1..n} binomial(2n,2k)*a(n-k). For c = -3, -2, -1, 1, 2, 3, 4 this is A210676, A210657, A028296, A094088, A210672, A210674, A249939.
Exp( Sum_{n >= 1} a(n)*x^n/n) is the o.g.f. for A255929. - Peter Bala, Mar 13 2015
The Stirling-Bernoulli transform of Fibonacci(n+1) = 1, 1, 2, 3, 5, 8, 13, ... is 1, 0, 2, 0, 26, 0, 842, 0, 50906, 0, ... - Philippe Deléham, May 25 2015

Programs

  • Maple
    f:=proc(n,k) option remember;  local i;
    if n=0 then 1
    else k*add(binomial(2*n,2*i)*f(n-i,k),i=1..floor(n)); fi; end;
    g:=k->[seq(f(n,k),n=0..40)];
    g(2);
  • Mathematica
    nmax=20; Table[(CoefficientList[Series[1/(3-2*Cosh[x]), {x, 0, 2*nmax}], x] * Range[0, 2*nmax]!)[[2*n+1]], {n,0,nmax}] (* Vaclav Kotesovec, Mar 14 2015 *)

Formula

a(n) ~ 2*sqrt(Pi/5) * n^(2*n+1/2) / (exp(2*n) * (log((1+sqrt(5))/2))^(2*n+1)). - Vaclav Kotesovec, Mar 13 2015
E.g.f.: 1/(3-2*cosh(x)) (even coefficients). - Vaclav Kotesovec, Mar 14 2015
a(n) = Sum_{k = 0..2*n} A163626(2*n,k)*A000045(n+1). - Philippe Deléham, May 25 2015
a(n) = Sum_{k=0..n} A241171(n, k)*2^k. - Peter Luschny, Sep 03 2022

A210676 a(0)=1; thereafter a(n) = -3*Sum_{k=1..n} binomial(2n,2k)*a(n-k).

Original entry on oeis.org

1, -3, 51, -2163, 171231, -21785223, 4065116811, -1045879150683, 354837765112791, -153492920593758543, 82453488412268175171, -53850296379425229208803, 42020794900180632536559951, -38611325264740403135096141463, 41264215393801752999038147563131, -50749285521783354479522581233836523
Offset: 0

Author

N. J. A. Sloane, Mar 28 2012

Keywords

Comments

Consider the sequence defined by a(0) = 1; thereafter a(n) = c*Sum_{k=1..n} binomial(2n,2k)*a(n-k). For c = -3, -2, -1, 1, 2, 3, 4 this is A210676, A210657, A028296, A094088, A210672, A210674, A249939.
Exp( Sum_{n >= 1} a(n)*x^n/n) is the o.g.f. for A255926. - Peter Bala, Mar 13 2015
In general, for c<>0 is e.g.f. = 1/(c+1-c*cosh(x)) (even coefficients). For c > 0 is a(n) ~ 2 * (2*n)! / (sqrt(2*c+1) * (arccosh((c+1)/c))^(2*n+1)). For c < 0 is a(n) ~ (-1)^n * (2*n)! / (sqrt(-2*c-1) * 2^(2*n) * arccos(sqrt((2*c + 1) / (2*c)))^(2*n+1)). - Vaclav Kotesovec, Mar 14 2015

Programs

  • Maple
    f:=proc(n,k) option remember;  local i;
    if n=0 then 1
    else k*add(binomial(2*n,2*i)*f(n-i,k),i=1..floor(n)); fi; end;
    g:=k->[seq(f(n,k),n=0..40)];
    g(-3);
  • Mathematica
    nmax=20; Table[(CoefficientList[Series[1/(3*Cosh[x]-2), {x, 0, 2*nmax}], x] * Range[0, 2*nmax]!)[[2*n+1]], {n,0,nmax}] (* Vaclav Kotesovec, Mar 14 2015 *)

Formula

E.g.f.: 1/(3*cosh(x)-2) (even coefficients). - Vaclav Kotesovec, Mar 14 2015
a(n) ~ (-1)^n * (2*n)! / (sqrt(5) * 2^(2*n) * (arccos(sqrt(5/6)))^(2*n+1)). - Vaclav Kotesovec, Mar 14 2015
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