A002114 -id:A002114 - cp's OEIS frontend

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

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

Offset: 0

N. J. A. Sloane, Mar 28 2012

sign

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

Seiichi Manyama, Table of n, a(n) for n = 0..200
Matthieu Josuat-Vergès and Jang Soo Kim, Touchard-Riordan formulas, T-fractions, and Jacobi's triple product identity, arXiv:1101.5608 [math.CO], 2011.
Zhi-Hong Sun, On the further properties of U_n, arXiv:1203.5977 [math.NT], 2012.

Cf. A002114, A028296, A094088, A210657, A210672, A210674, A210676, A249939, A255882, A241171.

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

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 *)

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

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

A000436 Generalized Euler numbers c(3,n).

Original entry on oeis.org

1, 8, 352, 38528, 7869952, 2583554048, 1243925143552, 825787662368768, 722906928498737152, 806875574817679474688, 1118389087843083461066752, 1884680130335630169428983808, 3794717805092151129643367268352

Offset: 0

N. J. A. Sloane

nonn

			G.f. = 1 + 8*x + 352*x^2 + 38528*x^3 + 7869952*x^4 + 2583554048*x^5 + ...

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
Michael E. Hoffman, Derivative polynomials, Euler Polynomials, and associated integer sequences, El. J. Combinat. 6 (see Th. 3.3).
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]

Row 3 of A235605.
Bisections: A156177 and A156178.
Cf. A000191, A007289, overview in A349264.
Cf. A210657, A255882.

A000436 := proc(nmax) local a,n,an; a := [1] : n := 1 : while nops(a)< nmax do an := 1-sum(binomial(2*n,2*i)*3^(2*n-2*i)*(-1)^i*op(i+1,a),i=0..n-1) : a := [op(a),an*(-1)^n] ; n := n+1 ; od ; RETURN(a) ; end:
A000436(10) ; # R. J. Mathar, Nov 19 2006
a := n -> 2*(-144)^n*(Zeta(0,-2*n,1/6)-Zeta(0,-2*n,2/3)):
seq(a(n), n=0..12); # Peter Luschny, Mar 11 2015

a[0] = 1; a[n_] := a[n] = (-1)^n*(1 - Sum[(-1)^i*Binomial[2n, 2i]*3^(2n - 2i)*a[i], {i, 0, n-1}]); Table[a[n], {n, 0, 12}] (* Jean-François Alcover, Jan 31 2012, after R. J. Mathar *)
With[{nn=30},Take[CoefficientList[Series[Cos[x]/Cos[3x],{x,0,nn}], x] Range[ 0,nn]!,{1,-1,2}]] (* Harvey P. Dale, May 22 2012 *)

x='x+O('x^66); v=Vec(serlaplace( cos(x) / cos(3*x) ) ); vector(#v\2,n,v[2*n-1]) \\ Joerg Arndt, Apr 27 2013

from mpmath import mp, lerchphi
mp.dps = 32; mp.pretty = True
def A000436(n): return abs(3^(2*n)*2^(2*n+1)*lerchphi(-1,-2*n,1/3))
[A000436(n) for n in (0..12)]  # Peter Luschny, Apr 27 2013

E.g.f.: cos(x) / cos(3*x) (even powers only).
For n>0, a(n) = A002114(n)*2^(2n+1) = (1/3)*A002112(n)*2^(2n+1). - Philippe Deléham, Jan 17 2004
a(n) = Sum_{k=0..n} (-1)^k*9^(n-k)*A086646(n,k). - Philippe Deléham, Oct 27 2006
(-1)^n a(n) = 1 - Sum_{i=0..n-1} (-1)^i*binomial(2n,2i)*3^(2n-2i)*a(i). - R. J. Mathar, Nov 19 2006
a(n) = P_{2n}(sqrt(3))/sqrt(3) (where the polynomials P_n() are defined in A155100). - N. J. A. Sloane, Nov 05 2009
E.g.f.: E(x) = cos(x)/cos(3*x) = 1 + 4*x^2/(G(0)-2*x^2); G(k) = (2*k+1)*(k+1) - 2*x^2 + 2*x^2*(2*k+1)*(k+1)/G(k+1); (continued fraction, Euler's kind, 1-step). - Sergei N. Gladkovskii, Jan 02 2012
G.f.: 1 / (1 - 2*4*x / (1 - 6*6*x / (1 - 8*10*x / (1 - 12*12*x / (1 - 14*16*x / (1 - 18*18*x / ...)))))). - Michael Somos, May 12 2012
a(n) = | 3^(2*n)*2^(2*n+1)*lerchphi(-1,-2*n,1/3) |. - Peter Luschny, Apr 27 2013
a(n) = (-1)^n*6^(2*n)*E(2*n,1/3), where E(n,x) denotes the n-th Euler polynomial. Calculation suggests that the expansion exp( Sum_{n >= 1} a(n)*x^n/n ) = exp( 8*x + 352*x^2/2 + 38528*x^3/3 + ... ) = 1 + 8*x + 208*x^2 + 14336*x^3 + ... has integer coefficients. Cf. A255882. - Peter Bala, Mar 10 2015
a(n) = 2*(-144)^n*(zeta(-2*n,1/6)-zeta(-2*n,2/3)), where zeta(a,z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
From Vaclav Kotesovec, May 05 2020: (Start)
For n>0, a(n) = (2*n)! * (zeta(2*n+1, 1/6) - zeta(2*n+1, 5/6)) / (sqrt(3)*Pi^(2*n+1)).
For n>0, a(n) = (-1)^(n+1) * 2^(2*n-1) * Bernoulli(2*n) * (zeta(2*n+1, 1/6) - zeta(2*n+1, 5/6)) / (Pi*sqrt(3)*zeta(2*n)). (End)
Conjecture: for each positive integer k, the sequence defined by a(n) (mod k) is eventually periodic with period dividing phi(k). For example, modulo 13 the sequence becomes [1, 8, 1, 9, 12, 10, 0, 8, 1, 9, 12, 10, 0, ...]; after the initial term 1 this appears to be a periodic sequence of period 6, a divisor of phi(13) = 12. - Peter Bala, Dec 11 2021

A079144 Number of labeled interval orders on n elements: (2+2)-free posets.

Original entry on oeis.org

1, 1, 3, 19, 207, 3451, 81663, 2602699, 107477247, 5581680571, 356046745023, 27365431508779, 2494237642655487, 266005087863259291, 32815976815540917183, 4636895313201764853259, 743988605732990946684927

Offset: 0

Detlef Pauly (dettodet(AT)yahoo.de), Dec 27 2002

From Peter Bala, Dec 26 2021: (Start)
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 8 begins [1, 1, 3, 3, 7, 3, 7, 3, 7,  ...] and appears to have a pre-period of length 3 and a period of length 2 = (1/2)*phi(8).
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. (End)

			1 + x + 3*x^2 + 19*x^3 + 207*x^4 + 3451*x^5 + 81663*x^6 + 2602699*x^7 + ...

Alois P. Heinz, Table of n, a(n) for n = 0..260
Peter Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)
Graham Brightwell and Mitchel T. Keller, Asymptotic Enumeration of Labelled Interval Orders, arXiv:1111.6766 [math.CO], 2011.
Anders Claesson, Mark Dukes and Martina Kubitzke, Partition and composition matrices, arXiv:1006.1312 [math.CO], 2010-2011.
Hsien-Kuei Hwang and Emma Yu Jin, Asymptotics and statistics on Fishburn matrices and their generalizations, arXiv:1911.06690 [math.CO], 2019.
Don Zagier, Vassiliev invariants and a strange identity related to the Dedekind eta-function, Topology, vol.40, pp.945-960 (2001); see p. 952.
Yan X Zhang, Four Variations on Graded Posets, arXiv preprint arXiv:1508.00318 [math.CO], 2015.

Cf. A022493 (unlabeled interval orders).
Cf. A002439 (Glaisher's T numbers), A002114 (Glaisher's H numbers).
Cf. A001318, A138265.

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:
seq(1/(24^n)*add(binomial(n,k)*A002439(k), k = 0..n), n = 0..20); # Peter Bala, Dec 26 2021

nmax=20; rk=Rest[CoefficientList[Series[Sum[Product[1-1/(1+x)^j,{j,1,n}],{n,0,nmax}],{x,0,nmax}],x]]; Flatten[{1,Table[Sum[rk[[k]] * k! * StirlingS2[n,k],{k,1,n}],{n,1,nmax}]}] (* Vaclav Kotesovec, May 03 2014 *)

{a(n) = if( n<0, 0, n! * polcoeff( subst( sum( i=0, n, prod( j=1, i, 1 - (1 - x + O(x^(n - i + 2)))^j )), x, 1 - exp( -x + x * O(x^n))), n))} /* Michael Somos, Apr 01 2012 */

a(n) = (1/(24^n))*Sum_{k=0..n} binomial(n, k)*A002439(k). Zagier 2001, p. 954.
G.f.: Sum(Product(1-exp(-k*x),k = 1 .. n),n = 0 .. infinity). a(n) = Sum_{k=0..n} k!*Stirling2(n,k)*A138265(k). - Vladeta Jovovic, Mar 11 2008
From Peter Bala, Mar 19 2009: (Start)
Conjectural form for the o.g.f. as a continued fraction:
1/(1-x/(1-2*x/(1-5*x/(1-7*x/(1-12*x/(1-15*x/(1- ...))))))) = 1 + x + 3*x^2 + 19*x^3 + 207*x^4 + ..., where the sequence [1,2,5,7,12,15,..] is the sequence of generalized pentagonal numbers A001318. Compare with the continued fraction form of the o.g.f. of A002105. (End)
E.g.f.: 1+(exp(x)-1)/(G(0)+1-exp(x)), where G(k)= 2*exp(x*(k+1))-1-exp(x*(k+1))*(exp(x*(k+2))-1)/G(k+1); (continued fraction, Euler's kind, 1-step). - Sergei N. Gladkovskii, Jan 06 2012
Asymptotics (Brightwell and Keller, 2011): a(n) ~ 12*sqrt(3)/Pi^(5/2) * (n!)^2 * sqrt(n) * (6/Pi^2)^n. - Vaclav Kotesovec, May 03 2014
From Peter Bala, May 11 2017: (Start)
For a proof of above conjectural continued fraction representation of the o.g.f. see the Bala link.
G.f.: 1/(1 + x - 2*x/(1 - 1*x/(1 + x - 7*x/(1 - 5*x/(1 + x - 15*x/(1 - 12*x/(1 + x - 26*x/(1 - 22*x/(1 + x - ...))))))))), where the sequence of unsigned partial numerators [2, 1, 7, 5, 15, 12, ...] is obtained from A001318 by swapping adjacent terms.
E.g.f.: F(q) = Sum_{n >= 0} q^(n+1)*Product_{i = 1..n} (1 - q^i)^2 at q = exp(t). Note that F(q) at q = 1/(1 - t) is a g.f. for unlabeled interval orders A022493, and at q = 1 - t gives a g.f. for A138265. (End)
From Peter Bala, Dec 26 2021: (Start)
a(6*n + 5) == 0 (mod 7); a(10*n + 7) == 0 (mod 11);
a(12*n + 11) == 0 (mod 13); a(16*n + 5) == a(16*n + 8) == 0 (mod 17);
a(18*n + 3) == 0 (mod 19); a(22*n + 4) == 0 (mod 23). (End)

A334479 Decimal expansion of Product_{k>=1} (1 + 1/A007528(k)^3).

Original entry on oeis.org

1, 0, 0, 9, 1, 3, 4, 5, 0, 8, 6, 3, 8, 4, 7, 4, 4, 7, 8, 0, 7, 1, 1, 3, 7, 5, 3, 9, 5, 8, 9, 2, 0, 5, 5, 8, 8, 1, 7, 4, 5, 6, 4, 7, 8, 5, 2, 9, 5, 2, 5, 5, 9, 9, 3, 0, 7, 2, 3, 6, 2, 0, 8, 1, 4, 8, 7, 9, 6, 2, 8, 3, 5, 9, 1, 6, 3, 6, 0, 3, 2, 1, 1, 9, 3, 2, 6, 6, 4, 3, 5, 2, 6, 4, 0, 4, 9, 6, 5, 9, 7, 5, 6, 1, 6

Offset: 1

Vaclav Kotesovec, May 02 2020

In general, for s > 0, Product_{k>=1} (1 + 1/A007528(k)^(2*s+1)) / (1 - 1/A007528(k)^(2*s+1)) = (1 - 1/2^(2*s + 1)) * (3^(2*s + 1) - 1) * (2*s)! * zeta(2*s + 1) / (sqrt(3) * A002114(s) * Pi^(2*s + 1)).
For s > 1, Product_{k>=1} (1 + 1/A007528(k)^s) / (1 - 1/A007528(k)^s) = (2^s - 1) * (3^s - 1) * zeta(s) / (zeta(s, 1/6) - zeta(s, 5/6)).
For s > 1, Product_{k>=1} (1 + 1/A002476(k)^s) * (1 + 1/A007528(k)^s) = 6^s * zeta(s) / ((2^s + 1) * (3^s + 1) * zeta(2*s)).

			1.0091345086384744780711375395892055881745647852...

Cf. A007528, A175646, A334424, A334426, A334480, A334482.

A334479 / A334480 = 91*sqrt(3)*zeta(3)/(6*Pi^3).

A334477 * A334479 = 810*zeta(3)/Pi^6.

More digits from Vaclav Kotesovec, Jun 27 2020

A334477 Decimal expansion of Product_{k>=1} (1 + 1/A002476(k)^3).

Original entry on oeis.org

1, 0, 0, 3, 6, 0, 2, 5, 4, 0, 2, 2, 1, 2, 5, 9, 8, 9, 6, 7, 0, 4, 3, 2, 3, 9, 3, 3, 3, 3, 2, 1, 8, 7, 8, 5, 9, 1, 7, 0, 5, 3, 9, 4, 7, 7, 1, 1, 7, 5, 0, 8, 7, 2, 1, 3, 7, 0, 2, 2, 4, 0, 2, 6, 4, 1, 6, 5, 2, 3, 7, 1, 7, 3, 7, 1, 7, 3, 6, 2, 6, 1, 4, 6, 6, 2, 7, 5, 2, 0, 4, 0, 8, 1, 5, 1, 4, 8, 2, 9, 8, 9, 1, 5, 7

Offset: 1

Vaclav Kotesovec, May 02 2020

In general, for s > 0, Product_{k>=1} (1 + 1/A002476(k)^(2*s+1)) / (1 - 1/A002476(k)^(2*s+1)) = sqrt(3) * (2*Pi)^(2*s + 1) * zeta(2*s + 1) * A002114(s) / ((2^(2*s + 1) + 1) * (3^(2*s + 1) + 1) * (2*s)! * zeta(4*s + 2)).
For s > 1, Product_{k>=1} (1 + 1/A002476(k)^s) / (1 - 1/A002476(k)^s) = (zeta(s, 1/6) - zeta(s, 5/6))*zeta(s) / ((2^s + 1)*(3^s + 1)*zeta(2*s)).
For s > 1, Product_{k>=1} (1 + 1/A002476(k)^s) * (1 + 1/A007528(k)^s) = 6^s * zeta(s) / ((2^s + 1) * (3^s + 1) * zeta(2*s)).
For s > 0, Product_{k>=1} ((A007528(k)^(2*s+1) - 1) / (A007528(k)^(2*s+1) + 1)) * ((A002476(k)^(2*s+1) + 1) / (A002476(k)^(2*s+1) - 1)) = 6 * A002114(s)^2 * (4*s + 2)! / ((2^(4*s + 2) - 1) * (3^(4*s + 2) - 1) * Bernoulli(4*s + 2) * (2*s)!^2) = Bernoulli(2*s)^2 * (4*s + 2)! * (zeta(2*s + 1, 1/6) - zeta(2*s + 1, 5/6))^2 / (8*Pi^2 * (2^(4*s + 2) - 1) * (3^(4*s + 2) - 1) * Bernoulli(4*s + 2) * (2*s)!^2 * zeta(2*s)^2).

			1.0036025402212598967043239333321878591705394771...

Cf. A002476, A175646, A334424, A334426, A334478, A334481.

A334477 / A334478 = 15*sqrt(3)*zeta(3)/Pi^3.

A334477 * A334479 = 810*zeta(3)/Pi^6.

More digits from Vaclav Kotesovec, Jun 27 2020

A334478 Decimal expansion of Product_{k>=1} (1 - 1/A002476(k)^3).

Original entry on oeis.org

9, 9, 6, 4, 0, 1, 6, 9, 2, 8, 1, 6, 0, 3, 6, 6, 3, 2, 2, 6, 2, 3, 6, 1, 1, 2, 2, 3, 8, 4, 7, 1, 8, 7, 9, 9, 9, 6, 5, 5, 7, 3, 8, 1, 8, 7, 1, 4, 0, 5, 3, 1, 5, 3, 7, 8, 6, 9, 8, 8, 9, 7, 4, 9, 3, 0, 1, 5, 9, 1, 3, 3, 2, 5, 3, 4, 3, 0, 6, 8, 4, 2, 5, 6, 2, 1, 9, 1, 9, 7, 2, 9, 9, 7, 7, 5, 2, 3, 2, 2, 1, 2, 3, 0, 1, 9

Offset: 0

Vaclav Kotesovec, May 02 2020

In general, for s > 0, Product_{k>=1} (1 + 1/A002476(k)^(2*s+1)) / (1 - 1/A002476(k)^(2*s+1)) = sqrt(3) * (2*Pi)^(2*s + 1) * zeta(2*s + 1) * A002114(s) / ((2^(2*s + 1) + 1) * (3^(2*s + 1) + 1) * (2*s)! * zeta(4*s + 2)).
For s > 1, Product_{k>=1} (1 + 1/A002476(k)^s) / (1 - 1/A002476(k)^s) = (zeta(s, 1/6) - zeta(s, 5/6))*zeta(s) / ((2^s + 1)*(3^s + 1)*zeta(2*s)).
For s > 1, Product_{k>=1} (1 - 1/A002476(k)^s) * (1 - 1/A007528(k)^s) = 6^s / ((2^s - 1)*(3^s - 1)*zeta(s)).

			0.996401692816036632262361122384718799965573818714...

R. J. Mathar, Table of Dirichlet L-series and Prime Zeta Modulo Functions for Small Moduli, arXiv:1008.2547 [math.NT], 2010-2015, p. 26 (case 3 1 3 = 1/A334478).

Cf. A002476, A175646, A334425, A334427, A334477.

A334477 / A334478 = 15*sqrt(3)*zeta(3)/Pi^3.

A334478 * A334480 = 108/(91*zeta(3)).

More digits from Vaclav Kotesovec, Jun 27 2020

A334480 Decimal expansion of Product_{k>=1} (1 - 1/A007528(k)^3).

Original entry on oeis.org

9, 9, 0, 8, 8, 4, 1, 4, 5, 5, 2, 5, 2, 1, 3, 3, 5, 6, 5, 6, 3, 4, 0, 3, 1, 7, 3, 5, 5, 9, 4, 3, 2, 7, 5, 1, 6, 4, 3, 4, 8, 3, 1, 2, 1, 7, 5, 0, 0, 7, 6, 1, 3, 3, 0, 4, 8, 6, 7, 7, 4, 7, 8, 4, 9, 4, 3, 1, 7, 8, 8, 8, 2, 5, 7, 6, 7, 4, 3, 1, 7, 7, 5, 2, 7, 6, 3, 4, 5, 2, 1, 7, 8, 9, 8, 8, 9, 2, 9, 2, 1, 3, 5, 4, 6, 7

Offset: 0

Vaclav Kotesovec, May 02 2020

In general, for s > 0, Product_{k>=1} (1 + 1/A007528(k)^(2*s+1)) / (1 - 1/A007528(k)^(2*s+1)) = (1 - 1/2^(2*s + 1)) * (3^(2*s + 1) - 1) * (2*s)! * zeta(2*s + 1) / (sqrt(3) * A002114(s) * Pi^(2*s + 1)).
For s > 1, Product_{k>=1} (1 + 1/A007528(k)^s) / (1 - 1/A007528(k)^s) = (2^s - 1) * (3^s - 1) * zeta(s) / (zeta(s, 1/6) - zeta(s, 5/6)).
For s > 1, Product_{k>=1} (1 - 1/A002476(k)^s) * (1 - 1/A007528(k)^s) = 6^s / ((2^s - 1)*(3^s - 1)*zeta(s)).

			0.990884145525213356563403173559432751643483121750... = 1/1.0091997177631243951237...

R. J. Mathar, Table of Dirichlet L-series and Prime Zeta Modulo Functions for Small Moduli, arXiv:1008.2547 [math.NT], 2010-2015, p. 26 (case 6 5 3 = 1/A334480).

Cf. A007528, A175646, A334425, A334427, A334479.

A334479 / A334480 = 91*sqrt(3)*zeta(3)/(6*Pi^3).

A334478 * A334480 = 108/(91*zeta(3)).

More digits from Vaclav Kotesovec, Jun 27 2020

A002112 Glaisher's H numbers.

Original entry on oeis.org

3, 33, 903, 46113, 3784503, 455538993, 75603118503, 16546026500673, 4616979073434903, 1599868423237443153, 674014138103352845703, 339274210193051498798433, 201097637653063767131142903, 138634566390566081044811718513

Offset: 1

N. J. A. Sloane

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.
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).

Vincenzo Librandi, Table of n, a(n) for n = 1..100
J. W. L. Glaisher, On a set of coefficients analogous to the Eulerian numbers, Proc. London Math. Soc., 31 (1899), 216-235.
Michael E. Hoffman, Derivative polynomials, Euler polynomials, and associated integer sequences, The Electronic Journal of Combinatorics 6.1 (1999).
Index entries for sequences related to Glaisher's numbers

e[0] = 1; e[n_] := e[n] = (-1)^n*(1 - Sum[(-1)^i*Binomial[2n, 2i]*3^(2n-2i)*e[i], {i, 0, n-1}]); a[n_] := 3*e[n]/2^(2n+1); Table[a[n], {n, 1, 14}] (* Jean-François Alcover, Jan 31 2012, after Philippe Deléham *)

H(n) = 2^(2n+1)*I(n), where e.g.f. for (-1)^n*I(n) is (3/2)/(1+exp(x)+exp(-x)) (see A047788, A047789).
H(n) = 3*A000436(n)/2^(2n+1) = 3*A002114(n). - Philippe Deléham, Jan 17 2004
E.g.f.: E(x) = 3*x^2/(G(0)-x^2); G(k) = 2*(2*k+1)*(k+1) - x^2 + 2*x^2*(2*k+1)*(k+1)/G(k+1); (continued fraction Euler's kind, 1-step ). - Sergei N. Gladkovskii, Jan 03 2012
If E(x) = Sum_{k>=0} a(k+1)*x^(2k+2), then A002112(k) = a(k+1)*(2*k+2)!. - Sergei N. Gladkovskii, Jan 09 2012
From Vaclav Kotesovec, May 05 2020: (Start)
a(n) = sqrt(3) * (2*n)! * (zeta(2*n+1, 1/6) - zeta(2*n+1, 5/6)) / (2*Pi)^(2*n+1).
a(n) = (-1)^(n+1) * sqrt(3) * Bernoulli(2*n) * (zeta(2*n+1, 1/6) - zeta(2*n+1, 5/6)) / (4*Pi*zeta(2*n)). (End)

A060082 Coefficients of even-indexed Euler polynomials (falling powers without zeros).

Original entry on oeis.org

1, 1, -1, 1, -2, 1, 1, -3, 5, -3, 1, -4, 14, -28, 17, 1, -5, 30, -126, 255, -155, 1, -6, 55, -396, 1683, -3410, 2073, 1, -7, 91, -1001, 7293, -31031, 62881, -38227, 1, -8, 140, -2184, 24310, -177320, 754572, -1529080, 929569, 1, -9, 204, -4284, 67626, -753610, 5497596, -23394924, 47408019

Offset: 0

Wolfdieter Lang, Mar 29 2001

E(2n,x) = x^(2n) + Sum_{k=1..n} a(n,k)*x^(2n-2k+1).

			E(0,x) = 1.
E(2,x) = x^2 - x.
E(4,x) = x^4 - 2*x^3 + x.
E(6,x) = x^6 - 3*x^5 + 5*x^3 - 3*x.
E(8,x) = x^8 - 4*x^7 + 14*x^5 - 28*x^3 + 17*x.
E(10,x) = x^10 - 5*x^9 + 30*x^7 - 126*x^5 + 255*x^3 - 155*x.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 809.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Z.-W. Sun, Introduction to Bernoulli and Euler polynomials

E(2n, 1/2)*(-4)^n = A000364(n) (signless Euler numbers without zeros).
-E(2n, -1/2)*(-4)^n/3 = A076552(n), -E(2n, 1/3)*(-9)^n/2 = A002114(n).
Cf. A060083 (rising powers), A060096-7 (Euler polynomials), A004172 (with zeros).
Columns (left edge) include A000330, A053132. Columns (right edge) include A001469.

Table[ CoefficientList[ EulerE[2*n, x], x] // Reverse // DeleteCases[#, 0]&, {n, 0, 9}] // Flatten (* Jean-François Alcover, Jun 21 2013 *)

{B(n,v='x)=sum(i=0,n,binomial(n,i)*bernfrac(i)*v^(n-i))} E(n,v='x)=2/(n+1)*(B(n+1,v)-2^(n+1)*B(n+1,v/2)) \\ Ralf Stephan, Nov 05 2004

E(n, x) = 2/(n+1) * [B(n+1, x) - 2^(n+1)*B(n+1, x/2) ], with B(n, x) the Bernoulli polynomials.

Edited by Ralf Stephan, Nov 05 2004

A186432 Triangle associated with the set S of squares {0,1,4,9,16,...}.

Original entry on oeis.org

1, 1, 1, 1, 12, 1, 1, 30, 30, 1, 1, 56, 140, 56, 1, 1, 90, 420, 420, 90, 1, 1, 132, 990, 1848, 990, 132, 1, 1, 182, 2002, 6006, 6006, 2002, 182, 1, 1, 240, 3640, 16016, 25740, 16016, 3640, 240, 1, 1, 306, 6120, 37128, 87516, 87516, 37128, 6120, 306, 1, 1, 380, 9690, 77520, 251940, 369512, 251940, 77520, 9690, 380, 1

Offset: 0

Peter Bala, Feb 22 2011

Given a subset S of the integers Z, Bhargava [1] has shown how to associate with S a generalized factorial function, denoted n!_S, sharing many properties of the classical factorial function n! (which corresponds to the choice S = Z). In particular, he shows that the generalized binomial coefficients n!_S/(k!_S*(n-k)!_S) are always integral for any choice of S. Here we take S = {0,1,4,9,16,...}, the set of squares.
The associated generalized factorial function n!_S is given by the formula
n!S = Product{k=0..n} (n^2 - k^2), with the convention 0!S = 1. This should be compared with n! = Product{k=0..n} (n - k).
For n >= 1, n!_S = (2*n)!/2 = A002674(n).
Compare this triangle with A086645 and also A186430 - the generalized binomial coefficients for the set S of prime numbers {2,3,5,7,11,...}.

			Triangle begins
n/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....1
.2..|..1....12.....1
.3..|..1....30....30.....1
.4..|..1....56...140....56.....1
.5..|..1....90...420...420....90.....1
.6..|..1...132...990..1848...990...132.....1
.7..|..1...182..2002..6006..6006..2002...182.....1
...

M. Bhargava, The factorial function and generalizations, Amer. Math. Monthly, 107 (2000), 783-799.

Cf. A002114, A086645, A186430, A186433 (inverse).

Table[2 Binomial[2 n, 2 k] - Boole[Or[k == 0, k == n]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, May 23 2017 *)

TABLE ENTRIES
T(n,k) = n!_S/(k!_S*(n-k)!_S),
which simplifies to
T(n,k) = 2*binomial(2*n,2*k) for 1 <= k < n,
with boundary conditions T(n,0) = 1 and T(n,n) = 1 for n >= 0.
RELATIONS WITH OTHER SEQUENCES
Denote this triangle by T. The first column of the inverse T^-1 (see A186433) begins [1, -1, 11, -301, 15371, ...] and, apart from the initial 1, is a signed version of the Glaisher's H' numbers A002114.
The first column of (1/2)*T^2 begins [1/2, 1, 7, 31, 127, ...] and, apart from the initial term, equals A000225(2*n-1), counting the preferential arrangements on (2*n - 1) labeled elements having less than or equal to two ranks.
The first column of (1/3)*T^3 begins [1/3, 1, 13, 181, 1933, ...] and, apart from the initial term, is A101052(2*n-1), which gives the number of preferential arrangements on (2*n-1) labeled elements having less than or equal to three ranks.

cp's OEIS Frontend

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

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A000436 Generalized Euler numbers c(3,n).

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A079144 Number of labeled interval orders on n elements: (2+2)-free posets.

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A334479 Decimal expansion of Product_{k>=1} (1 + 1/A007528(k)^3).

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A334477 Decimal expansion of Product_{k>=1} (1 + 1/A002476(k)^3).

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A334478 Decimal expansion of Product_{k>=1} (1 - 1/A002476(k)^3).

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A334480 Decimal expansion of Product_{k>=1} (1 - 1/A007528(k)^3).

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A002112 Glaisher's H numbers.

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