A158690 -id:A158690 - cp's OEIS frontend

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

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

Offset: 0

N. J. A. Sloane

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

			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)

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Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
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N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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K. Dilcher and C. Vignat, Euler and the Strong Law of Small Numbers, Amer. Math. Mnthly, 123 (May 2016), 486-490.
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Index entries for "core" sequences.
Index entries for sequences related to boustrophedon transform.

Cf. A000111, A000182, A011248, A019669, A034947, A060075, A013525, A000816, A002436, A000464, A002105, A002439, A079144, A158690.
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).

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

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 *)

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 */

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 */

{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

{a(n) = if( n<0, 0, (2*n)! * polcoeff( 1 / cos(x + O(x^(2*n + 1))), 2*n))}; /* Michael Somos, Jun 18 2002 */

{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 */

{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

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

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

a(n)=abs(eulerfrac(2*n)) \\ Charles R Greathouse IV, Mar 23 2022

\\ 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

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

# 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

# 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

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

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

N. J. A. Sloane

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

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

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).

T. D. Noe, Table of n, a(n) for n = 1..100
G. E. Andrews, J. Jimenez-Urroz, and K. Ono, q-series identities and values of certain L-functions, Duke Math J. Volume 108, Number 3 (2001), 395-419.
R. C. Archibald, Review of Terrill-Terrill paper, Math. Comp., 1 (1945), pp. 385-386.
Peter Bala, A continued fraction expansion related to A002105
Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.
M. P. Develin and S. P. Sullivant, Markov Bases of Binary Graph Models, Annals of Combinatorics 7 (2003) 441-466.
Dominique Foata and Guo-Niu Han, The doubloon polynomial triangle, Ram. J. 23 (2010), 107-126
Dominique Foata and Guo-Niu Han, Doubloons and new q-tangent numbers, Quart. J. Math. 62 (2) (2011) 417-432
Dominique Foata and Guo-Niu Han, Tree Calculus for Bivariable Difference Equations, 2012. - From _N. J. A. Sloane_, Feb 02 2013
R. L. Graham and Nan Zang, Enumerating split-pair arrangements, J. Combin. Theory Ser. A 115 (2008), no. 2, 293-303.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Markus Kuba and Alois Panholzer, Combinatorial families of multilabelled increasing trees and hook-length formulas, arXiv:1411.4587 [math.CO], 2014.
B. A. Kupershmidt, On skew-symmetric and general deformations of Lax pseudodifferential operators, J. Math. Sci. 151 (4) (2008) 3139, eq. (1.13).
Zhicong Lin, Shi-Mei Ma, David G. L. Wang, and Liuquan Wang, Positivity and divisibility of alternating descent polynomials, arXiv:2011.02685 [math.CO], 2020.
E. Norton, Symplectic Reflection Algebras in Positive Characteristic as Ore Extensions, arXiv preprint arXiv:1302.5411 [math.RA], 2013.
Ronald Orozco López, Solution of the Differential Equation y^(k)= e^(a*y), Special Values of Bell Polynomials and (k,a)-Autonomous Coefficients, Universidad de los Andes (Colombia 2021).
C. Poupard, Deux proprietes des arbres binaires ordonnes stricts, European J. Combin., 10 (1989), 369-374.
M. Safaryan, Problem 11725, Amer. Math. Monthly, 120 (2013), 661.
H. M. Terrill and E. M. Terrill, Tables of numbers related to the tangent coefficients, J. Franklin Inst., 239 (1945), 66-67.
H. M. Terrill and E. M. Terrill, Tables of numbers related to the tangent coefficients, J. Franklin Inst., 239 (1945), 64-67. [Annotated scanned copy]
A. Vieru, Agoh's conjecture: its proof, its generalizations, its analogues, arXiv preprint arXiv:1107.2938 [math.NT], 2011.
Index entries for sequences related to Bernoulli numbers.

Row sums of A008301.
Left edge of triangle A210108.
Cf. A000111, A000217, A000364, A000464, A002439, A008281, A079144, A158690.

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

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

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 *)

{a(n) = if( n<1, 0, ((-2)^n - (-8)^n) * bernfrac(2*n) / n)}; /* Michael Somos, Jun 22 2002 */

{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 */

{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 */

{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 */

{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 */

from sympy import bernoulli
def A002105(n): return abs(((2-(2<<(m:=n<<1)))*bernoulli(m)<Chai Wah Wu, Apr 14 2023

# 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

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

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

Additional comments from Michael Somos, Jun 25 2002

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

N. J. A. Sloane

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)

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

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).

T. D. Noe, Table of n, a(n) for n=0..100
G. E. Andrews, J. Jimenez-Urroz and K. Ono, q-series identities and values of certain L-functions, Duke Math J., Volume 108, No.3 (2001), 395-419.
Peter Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)
J. Bryson, K. Ono, S. Pitman and R. C. Rhoades, Unimodal Sequences and Quantum and Mock Modular Forms, P. 2.
Frank Garvan, Congruences and relations for r-Fishburn numbers, arXiv:1406.5611 [math.NT], 2014.
J. W. L. Glaisher, On a set of coefficients analogous to the Eulerian numbers, Proc. London Math. Soc., 31 (1899), 216-235.
K. Hikami, Volume Conjecture and Asymptotic Expansion of q-Series, Experimental Mathematics Vol. 12, Issue 3 (2003).
Michael E. Hoffman, Derivative Polynomials, Euler Polynomials, and Associated Integer Sequences, The Electronic Journal of Combinatorics, Volume 6.1 (1999): Research paper R21, 13 p.
Hsien-Kuei Hwang and Emma Yu Jin, Asymptotics and statistics on Fishburn matrices and their generalizations, arXiv:1911.06690 [math.CO], 2019.
J. C. P. Miller, Letter to N. J. A. Sloane, Mar 26 1971
Hjalmar Rosengren, Elliptic pfaffians and solvable lattice models, arXiv preprint arXiv:1605.02915 [math-ph], 2016.
A. Vieru, Agoh's conjecture: its proof, its generalizations, its analogues, arXiv preprint arXiv:1107.2938 [math.NT], 2011.
Don Zagier, Vassiliev invariants and a strange identity related to the Dedekind eta-function, Topology, vol.40, pp.945-960 (2001).
Index entries for sequences related to Glaisher's numbers

Cf. A000364, A000464, A002105, A079144, A158690.
Bisections: A156175, A156176.
Twice this sequence gives A000191. A208679, A208680, A208681.

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

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:

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 *)

{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 */

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

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

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

N. J. A. Sloane

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)

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).

Matthew House, Table of n, a(n) for n = 0..215 (terms 0..50 from T. D. Noe)
G. E. Andrews, J. Jimenez-Urroz, and K. Ono, q-series identities and values of certain L-functions, Duke Math Jour., Volume 108, No.3 (2001), 395-419.
Peter Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)
D. Dumont, Further triangles of Seidel-Arnold type and continued fractions related to Euler and Springer numbers, Adv. Appl. Math., 16 (1995), 275-296.
Simon Plouffe, Table of n, a(n) for n=0..1023
D. Shanks, Generalized Euler and class numbers. Math. Comp. 21 (1967) 689-694.
D. Shanks, Corrigenda to: "Generalized Euler and class numbers", Math. Comp. 22 (1968), 699
D. Shanks, Generalized Euler and class numbers, Math. Comp. 21 (1967), 689-694; 22 (1968), 699. [Annotated scanned copy]
Alan D. Sokal, The Euler and Springer numbers as moment sequences, arXiv:1804.04498 [math.CO], 2018.
A. Vieru, Agoh's conjecture: its proof, its generalizations, its analogues, arXiv preprint arXiv:1107.2938 [math.NT], 2011.

Row 2 of A235606.
Cf. A064073. Bisection of A000822, A001586.
Cf. A000364, A000828, A002105, A002439, A079144, A158690.

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

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 *)

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 */

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)

Better description, new reference, Aug 15 1995

A179525 G.f.: A(x) = Sum_{n>=0} Product_{k=1..n} ((1+x)^k - 1).

Original entry on oeis.org

1, 1, 2, 7, 33, 197, 1419, 11966, 115575, 1257718, 15223822, 202860828, 2950665011, 46516215168, 790009447590, 14379745626739, 279256447482090, 5763290215111558, 125960271446527241, 2906289188751628643, 70594767279197608011, 1800695322331687800336, 48122711251655255426539, 1344617808976210991187090, 39206731897407002624384182, 1190905492485213830900901986

Offset: 0

Paul D. Hanna, Jul 17 2010

nonn

From Vít Jelínek, Feb 12 2012: (Start)
a(n) has the following combinatorial interpretations:
(1) the number of upper-triangular matrices over {0,1} having at least one '1'-entry in each row and having n '1'-entries in total. E.g., for n=2, this corresponds to these two matrices (with zeros represented as dots):
1.  .1
.1  .1
(2) the number of upper-triangular matrices over {0,1} that are symmetric with respect to the northeast diagonal, have at least one '1'-entry in each row and column, have no '1'-entry on the northeast diagonal, and have 2n '1'-entries in total. For n=2, those are the two matrices
11.  1...
..1  .1..
..1  ..1.
     ...1
(3) the number of upper-triangular matrices over {0,1} that are symmetric with respect to the northeast diagonal, have at least one '1'-entry in each row and column, have at least one '1'-entry on the northeast diagonal, and have n '1'-entries on or above the northeast diagonal. For n=2, this corresponds to
11  1..
.1  .1.
    ..1
(End)
This is an example of Peter Bala's identity (cf. A158690):
Sum_{n>=0} Product_{k=1..n} (q^k - 1) = Sum_{n>=0} q^(-n^2) * Product_{k = 1..n} (q^(2*k-1) - 1) at q = 1 + x. See cross-references for other examples.

			G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 33*x^4 + 197*x^5 + 1419*x^6 +...
A(x) = 1 + ((1+x)-1) + ((1+x)-1)*((1+x)^2-1) + ((1+x)-1)*((1+x)^2-1)*((1+x)^3-1) +...
Let q = 1+x, then g.f. also equals:
A(x) = 1 + (q-1)/q + (q-1)*(q^3-1)/q^4 + (q-1)*(q^3-1)*(q^5-1)/q^9 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)/q^16 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)*(q^9-1)/q^25 +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..360
Kathrin Bringmann, Yingkun Li, Robert C. Rhoades, Asymptotics for the number of row-Fishburn matrices, European Journal of Combinatorics, vol.41, pp.183-196, (October-2014); preprint.
Stuart A. Hannah, Sieved Enumeration of Interval Orders and Other Fishburn Structures, arXiv:1502.05340 [math.CO], (18-February-2015).
Hsien-Kuei Hwang, Emma Yu Jin, Asymptotics and statistics on Fishburn matrices and their generalizations, arXiv:1911.06690 [math.CO], 2019.
Vít Jelínek, Counting general and self-dual interval orders, Journal of Combinatorial Theory, Series A, Volume 119, Issue 3, April 2012, pp. 599-614.
Sherry H. F. Yan and Yuexiao Xu, Self-dual interval orders and row-Fishburn matrices, arXiv:1111.4723 [math.CO], 2011.

Cf. variants: A022493, A122400, A158691, A158690, A207386, A207397, A207433.

Cf. A207434 (log).

a[ n_] := SeriesCoefficient[ Sum[ Product[ (1 + x)^j - 1, {j, k}], {k, 0, n}], {x, 0, n}]; (* Michael Somos, Jun 27 2017 *)

{a(n) = polcoeff(sum(i=0, n, prod(j=1, i, (1+x)^j-1, 1+x*O(x^n))), n)};
for(n=0, 30, print1(a(n), ", "))

/* G.f. as a continued fraction: */
{a(n) = local(CF=1+x*O(x)); for(k=0, n, CF=1/((1+x)^(n-k+1)-((1+x)^(n-k+2)-1)*CF)); polcoeff(1/(1-x*CF), n, x)}
for(n=0, 30, print1(a(n), ", "))

{a(n) = local(A=1+x, q=(1+x +x*O(x^n))); A = sum(m=0, n, q^(-m^2)*prod(k=1, m, (q^(2*k-1)-1))); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

/* Sum_{n>=0} n!*Product_{k=0..n-1} [Integral (1+x)^k dx] */
{a(n) = my(A=1); A = sum(m=0,n, m! * prod(k=0,m-1, intformal((1+x)^k) +x*O(x^n)) );polcoeff(A,n)}
for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 16 2016

G.f.: 1/(1 - ((1+x)-1)/((1+x) - ((1+x)^2-1)/((1+x)^2 - ((1+x)^3-1)/((1+x)^3 - ((1+x)^4-1)/((1+x)^4 - ((1+x)^5-1)/((1+x)^5 -...)))))), (continued fraction) [Paul D. Hanna, Jan 29 2012]
G.f.: Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1) where q = 1+x. [Based on Peter Bala's identity in comments]
Conjecturally, a(n) is asymptotically c*n!*(12/Pi^2)^n, where c=6*sqrt(2)*exp(-Pi^2/24)/Pi^2. - Vít Jelínek, Feb 12 2012 [This is correct: see Hwang and Jin, Table 3, p. 26. - Peter Bala, Jan 31 2021]
G.f.: 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). Conjecture. - Sergei N. Gladkovskii, May 13 2013
From Peter Bala, May 16 2017: (Start)
G.f.: A(x) = 1/2*( 1 + Sum_{n >= 0} (1 + x)^(n+1)*Product_{k = 1..n} ((1 + x)^k - 1) ).
Conjectural g.f.: Sum_{n >= 0} 1/(1 + x)^(n+1)*Product_{k = 1..n} (1 - 1/(1 + x)^(2*k)).
Conjectural g.f.: Sum_{n >= 0} (1 + x)^(2*n+1)*Product_{k = 1..2*n} (1 - (1 + x)^k). Cf. A158690, which has e.g.f. A(exp(x) - 1). (End)

A003406 Expansion of Ramanujan's function R(x) = 1 + Sum_{n >= 1} { x^(n*(n+1)/2) / ((1+x)(1+x^2)(1+x^3)...(1+x^n)) }.

Original entry on oeis.org

1, 1, -1, 2, -2, 1, 0, 1, -2, 0, 2, 0, -1, -2, 2, 1, 0, -2, 2, -2, 0, 0, 3, 0, -2, -2, 1, 0, 2, 0, 0, 0, -2, 0, 0, 1, 0, 0, 0, 2, -1, 0, -2, -2, 0, 4, 0, 2, -2, 0, -2, -1, 2, 0, -2, 2, 0, 1, 0, 0, 0, 0, -2, 0, 0, 0, 0, -2, 4, 2, -1, 0, 0, -2, -2, -2, 2, 1, 2, 0, 0, 0, 0, -2, 2, 0, 0, -2, 2, -2, -2, 0, 3, 0, 0, 2, 0, 0, 0, -2, 1, -2, 0, -2, 0

Offset: 0

N. J. A. Sloane

a(n) = A117192(n) - A117193(n) for n>0 (number of partitions into distinct parts with even rank minus those with odd rank); see also A000025. - Reinhard Zumkeller, Mar 03 2006

Ramanujan showed that R(x) = 2*Sum_{n>=0} (S(x) - P(n,x)) - 2*S(x)*D(x), where P(n,x) = Product_{k=1..n} (1+x^k), S(x) = g.f. A000009 = P(oo,x) and D(x) = -1/2 + Sum_{n>=1} x^n/(1-x^n) = -1/2 + g.f. A000005. - Michael Somos

			1 + x - x^2 + 2*x^3 - 2*x^4 + x^5 + x^7 - 2*x^8 + 2*x^10 - x^12 - 2*x^13 + ...
q + q^25 - q^49 + 2*q^73 - 2*q^97 + q^121 + q^169 - 2*q^193 + 2*q^241 - ...

G. E. Andrews, Ramanujan's "lost" notebook V: Euler's partition identity, Adv. in Math. 61 (1986), no. 2, 156-164; Math. Rev. 87i:11137. [ The expansion in (2.8) is incorrect. ]
F. J. Dyson, A walk through Ramanujan's garden, pp. 7-28 of G. E. Andrews et al., editors, Ramanujan Revisited. Academic Press, NY, 1988.
F. J. Dyson, Selected Papers, Am. Math. Soc., 1996, p. 200.
B. Gordon and D. Sinor, Multiplicative properties of eta-products, Number theory, Madras 1987, pp. 173-200, Lecture Notes in Math., 1395, Springer, Berlin, 1989. see page 182. MR1019331 (90k:11050)
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..2000 from T. D. Noe)
G. E. Andrews, Questions and conjectures in partition theory, Amer. Math. Monthly, 93 (1986), 708-711.
G. E. Andrews, Some debts I owe, Séminaire Lotharingien de Combinatoire, Paper B42a, Issue 42, 2000.
G. E. Andrews, F. J. Dyson and D. Hickerson, Partitions and indefinite quadratic forms, Invent. Math. 91 (1988) 391-407.
S.-Y. Kang, Generalizations of Ramanujan's reciprocity theorem and their applications, J. London Math. Soc., 75 (2007), 18-34.
Alexander E. Patkowski, A note on the rank parity function, Discrete Math. 310 (2010), 961-965.
D. Zagier, Quantum modular forms, Example 1 in Quanta of Maths: Conference in honor of Alain Connes, Clay Mathematics Proceedings 11, AMS and Clay Mathematics Institute 2010, 659-675

Cf. A003475, A005895, A005896, A158690.

g:=1+sum(x^(n*(n+1)/2)/product(1+x^j,j=1..n),n=1..20): gser:=series(g,x=0,110): seq(coeff(gser,x,n),n=0..104); # Emeric Deutsch, Mar 30 2006
t1:= add( (-1)^n*q^(n*(3*n+1)/2)*(1-q^(2*n+1))* add( (-1)^j*q^(-j^2),j=-n..n), n=0..20); t2:=series(t1,q,40); # N. J. A. Sloane, Jun 27 2011

max = 105; f[x_] := 1 + Sum[ x^(n*(n+1)/2) / Product[ 1+x^j, {j, 1, n}], {n, 1, max}]; CoefficientList[ Series[ f[x], {x, 0, max}], x] (* Jean-François Alcover, Dec 02 2011 *)
max = 105; s = 1 + Sum[2*q^(n*(n+1)/2)/QPochhammer[-1, q, n+1], {n, 1, Ceiling[Sqrt[2 max]]}] + O[q]^max; CoefficientList[s, q] (* Jean-François Alcover, Nov 25 2015 *)

{a(n) = local(t); if( n<0, 0, t = 1 + O(x^n); polcoeff( sum( k=1, n, t *= if( k>1, x^k - x, x) + O(x^(n-k+2)), 1), n))} /* Michael Somos, Mar 07 2006 */

{a(n) = local(t); if( n<0, 0, t = 1 + O(x^n); polcoeff( sum( k=1, (sqrtint(8*n + 1)-1)\2, t *= x^k / (1 + x^k) + x * O(x^(n - (k^2-k)/2)), 1), n))} /* Michael Somos, Aug 17 2006 */

{a(n) = local(A, p, e, x, y); if( n<0, 0, n = 24*n+1; A = factor(n); prod( k=1, matsize(A)[1], if( p=A[k, 1], e=A[k, 2]; if( p<5, 0, if( p%24>1 && p%24<23, if(e%2, 0, if( p%24==7 || p%24==17, (-1)^(e/2), 1)), x=y=0; if( p%24==1, forstep(i=1, sqrtint(p), 2, if( issquare( (i^2+p)/2, &y), x=i; break)), for( i=1, sqrtint(p\2), if( issquare(2*i^2 + p, &x), y=i; break))); (e+1)*(-1)^( (x + if((x-y)%6, y, -y))/6*e))))))} /* Michael Somos, Aug 17 2006 */

G.f.: 1 - Sum_{n > 0} (-x)^n * (1 - x) * (1 - x^2) * ... * (1  -x^(n-1)).
G.f.: 1 + Sum_{n>=1}(x^(n(n+1)/2)/Product_{j=1..n}(1+x^j)). - Emeric Deutsch, Mar 30 2006
Define c(24*k + 1) = A003406(k), c(24*k - 1) = -2*A003475(k), c(n) = 0 otherwise. Then c(n) is multiplicative with c(2^e) = c(3^e) = 0^e, c(p^e) = (-1)^(e/2) * (1+(-1)^e)/2 if p == 7, 17 (mod 24), c(p^e) = (1+(-1)^e)/2 if p == 5, 11, 13, 19 (mod 24), c(p^e) = (e+1)*(-1)^(y*e) where p == 1, 23 (mod 24) and p = x^2 - 72*y^2 . - Michael Somos, Aug 17 2006
Also R(x) = -2 + Sum_{n>=0} (n+1)*x^(n(n-1)/2)/(Product_{k=1..n} (1+x^k)). - Paul D. Hanna, May 22 2010

A158691 The number of upper-triangular matrices with at least one nonzero entry in each row and whose entries sum to n.

Original entry on oeis.org

1, 1, 3, 12, 61, 380, 2815, 24213, 237348, 2612681, 31915787, 428481472, 6271362282, 99388642292, 1695614865711, 30984649882928, 603790447393402, 12498732438500663, 273902239550757626, 6334968666307580051, 154211723833861061644, 3941258052200287007636

Offset: 0

Peter Bala, Mar 24 2009

There is a connection with the alternating series 1 - (1-x) + (1-x)(1-x^2) - (1-x)(1-x^2)(1-x^3) + .... Namely, if we replace x with 1/(1-x) in the partial sum 1 - (1-x) + (1-x)(1-x^2) - (1-x)(1-x^2)(1-x^3) + ... + (-1)^n(1-x)(1-x^2)...(1-x^n) and then expand about x = 0 we get a series whose first n+1 coefficients agree with the first n+1 terms of the present sequence.
From Vít Jelínek, Sep 04 2014: (Start)
The above remark, first conjectured by Bala, is a consequence of the identities satisfied by the generating function for a(n). More precisely, the generating function F(x)=Sum_{n>=0} a(n)x^n can be expressed by any of these three formulas:
F(x) = Sum_{n>=0} Product_{i=1..n} (1-(1-x)^(2*i-1))
F(x) = Sum_{n>=0} Product_{i=1..n} (1/(1-x)^i - 1)
F(x) = Sum_{n>=0} (1-x)^(n+1) Product_{i=1..n} (1-(1-x)^(2*i))
The first two formulas were conjectured to be equal by Bala. This was confirmed by Andrews and Jelínek, who also derived the third formula.
Bringmann, Li and Rhoades have independently derived the three formulas above, and additionaly, they proved that
F(x) = (1/2)*Sum_{n>=0} Product_{i=1..n} (1/(1+(1-x)^i)).
(End)
From Vít Jelínek, Feb 12 2012: (Start)
a(n) has the following combinatorial interpretations:
(1) the number of upper-triangular nonnegative integer matrices with at least one positive entry in each row, whose entries sum to n. E.g., for n=2 this corresponds to these three matrices (with zeros represented as dots):
   (2),  (1.)  and  (.1)
         (.1)       (.1)
(2) the number of upper-triangular nonnegative matrices that are symmetric along the northeast diagonal, have no positive entry on the northeast diagonal, have at least one positive entry in every row and column, and their entries sum to 2n. These are the three such matrices with n=2:
   (2.),  (11.) and (1...)
   (.2)   (..1)     (.1..)
          (..1)     (..1.)
                    (...1)
(3) the number of upper-triangular nonnegative matrices that are symmetric along the northeast diagonal, have at least one positive entry on the northeast diagonal, have at least one positive entry in every row and column, and their entries on or above the northeast diagonal sum to n. These are the three such matrices with n=2:
   (2), (11) and (1..)
        (.1)     (.1.)
                 (..1)
(End)

			G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 61*x^4 + 380*x^5 + 2815*x^6 +...

Seiichi Manyama, Table of n, a(n) for n = 0..200
George E. Andrews, Vít Jelínek, On q-Series Identities Related to Interval Orders, arXiv:1309.6669 [math.CO], 2013.
George E. Andrews, Vít Jelínek, On q-Series Identities Related to Interval Orders, European Journal of Combinatorics, Volume 39, July 2014, 178-187.
Kathrin Bringmann, Yingkun Li, Robert C. Rhoades, Asymptotics for the number of row-Fishburn matrices, European Journal of Combinatorics, Volume 41, October 2014, Pages 183-196; preprint.
Hsien-Kuei Hwang, Emma Yu Jin, Asymptotics and statistics on Fishburn matrices and their generalizations, arXiv:1911.06690 [math.CO], 2019.
Florian Ingels, Romain Azaïs, Enumeration of Unordered Forests, arXiv:2003.08144 [cs.DM], 2020.
Vít Jelínek, Counting self-dual interval orders, arXiv:1106.2261 [math.CO], 2011.
Vít Jelínek, Counting general and self-dual interval orders, Journal of Combinatorial Theory, Series A, Volume 119, Issue 3, April 2012, pp. 599-614.
Sherry H. F. Yan and Yuexiao Xu, Self-dual interval orders and row-Fishburn matrices, arXiv:1111.4723 [math.CO], 2011.
Sherry H. F. Yan and Yuexiao Xu, Self-dual interval orders and row-Fishburn matrices, Electronic Journal of Combinatorics, 19(2):P5 (2012).

Cf. A022493, A138265, A158690, A179525.

g:=sum(product(1-(1-x)^(2*i-1), i= 1..n), n = 0..20): gser:=series(g, x = 0,20): seq(coeff(gser, x, n), n = 0..19); # by definition
g:=sum((-1)^n*product(1-1/(1-x)^i, i= 1..n), n = 0..20): gser:=series(g, x = 0,20): seq(coeff(gser, x, n), n = 0..19);

a[ n_] := SeriesCoefficient[ Sum[ Product[ 1 - (1 - x)^(2 i - 1), {i, k}], {k, 0, n}], {x, 0, n}]; (* Michael Somos, Jun 27 2017 *)

{a(n)=polcoeff(sum(m=0, n, prod(k=1, m, 1/(1-x)^k-1, 1+x*O(x^n))), n)} /* Paul D. Hanna, Jan 29 2012 */

/* G.f. as a Continued Fraction: */
{a(n)=local(CF=1+x*O(x)); for(k=0, n, CF=1/(1 - (1-x)^(n-k+1)*(1-(1-x)^(n-k+2))*CF)); polcoeff(1/(1-x*CF), n, x)} /* Paul D. Hanna, Jan 29 2012 */

Sum_{n >= 0} Product_{i= 1..n} (1-(1-x)^(2*i-1)) = 1 + x + 3*x^2 + 12*x^3 + 61*x^4 + .... Compare with A022493, A138265 and A158690.
G.f.: Sum_{n>=0} Product_{k=1..n} [1/(1-x)^k - 1].
G.f.: 1/(1 - (1-(1-x))/(1 - (1-x)*(1-(1-x)^2)/(1 - (1-x)^2*(1-(1-x)^3)/(1 - (1-x)^3*(1-(1-x)^4)/(1 - (1-x)^4*(1-(1-x)^5)/(1 -...)))))), a continued fraction. - Paul D. Hanna, Jan 29 2012
By results of Bringmann, Li and Rhoades, a(n) is asymptotically c*n!*(12/Pi^2)^n, with c=6*sqrt(2)*exp(Pi^2/24)/Pi^2, and the ratio a(n)/A179525(n) tends to exp(Pi^2/12). - Vít Jelínek, Sep 04 2014
From Peter Bala, May 16 2017: (Start)
G.f. 1/2*( 1 + Sum_{n >= 0} 1/(1 - x)^((n+1)*(n+2)/2) * Product_{i = 1..n} (1 - (1 - x)^i) ).
Conjectural g.f.: Sum_{n >= 0} 1/(1 - x)^((n+1)*(2*n+1)) * Product_{i = 1..2*n} ((1 - x)^i - 1). (End)

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

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

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A002439 Glaisher's T numbers.

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

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