cp's OEIS Frontend

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

Showing 1-10 of 10 results.

A159600 E.g.f. C(x) satisfies: C(x) = (1 - 2*S(x)^2)^(1/4), where S'(x) = C(x)^3 and C'(x) = -S(x) with C(0)=1.

Original entry on oeis.org

1, -1, 3, -27, 441, -11529, 442827, -23444883, 1636819569, -145703137041, 16106380394643, -2164638920874507, 347592265948756521, -65724760945840254489, 14454276753061349098587
Offset: 0

Views

Author

Paul D. Hanna, May 07 2009

Keywords

Comments

See A104203 for the expansion of the sine lemniscate function sl(x).
E.g.f. C(x) is an even function; zero terms are omitted.
Radius of convergence is |x| <= r:
r = sqrt(2)*(Pi/2)^(3/2)/gamma(3/4)^2 with
C(r) = gamma(3/4)^2/(Pi/2)^(3/2) where:
r = L/sqrt(2) where L=Lemniscate constant;
r = 1.8540746773013719184338503471952600...
C(r) = 0.76275976350181318806232598096361579...

Examples

			E.g.f.: C(x) = 1 - x^2/2! + 3*x^4/4! - 27*x^6/6! + 441*x^8/8! -+ ...
C(x)^2 = 1 - 2*x^2/2! + 12*x^4/4! - 144*x^6/6! + 3024*x^8/8! -+ ...
C(x)^3 = 1 - 3*x^2/2! + 27*x^4/4! - 441*x^6/6! + 11529*x^8/8! -+ ...
C(x)^4 = 1 - 4*x^2/2! + 48*x^4/4! - 1008*x^6/6! + 32256*x^8/8! -+ ...
C(x)^4 + 2*S(x)^2 = 1 where:
S(x) = x - 3*x^3/3! + 27*x^5/5! - 441*x^7/7! + 11529*x^9/9! + ...
S(x)^2 = 2*x^2/2! - 24*x^4/4! + 504*x^6/6! - 16128*x^8/8! +-...
From _Paul D. Hanna_, Jul 29 2011: (Start)
O.g.f.: 1 - x + 3*x^2 - 27*x^3 + 441*x^4 - 11529*x^5 + 442827*x^6 -+ ... + a(n)*x^n + ...
O.g.f.: 1/(1 + x/(1 + 2*x/(1 + 9*x/(1 + 8*x/(1 + 25*x/(1 + 18*x/(1 + 49*x/(1 + 32*x/(1-...))))))))) (continued fraction). (End)

Crossrefs

Cf. A159601 (S(x)); A193541, A193544: All of these have the same |a(n)|. - M. F. Hasler, Aug 31 2012
Cf. A129194.

Programs

  • Mathematica
    a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ JacobiCN[ x, 1/2], {x, 0, m}]]]; (* Michael Somos, Apr 25 2011 *)
a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, With[ {s = InverseSeries[ Integrate[ Series[(1 - x^4 / 4) ^ (-1/2), {x, 0, m + 1}], x]]}, m! SeriesCoefficient[ Sqrt[ (2 - s^2) / (2 + s^2)], {x, 0, m}]]]]; (* Michael Somos, Apr 25 2011 *)
  • PARI
    {a(n)=local(S=x,C);for(i=0,2*n,S=intformal((1-2*S^2+O(x^(2*n+2)))^(3/4))); C=(1-2*S^2)^(1/4) ;(2*n)!*polcoeff(C,2*n)}
  • PARI
    {a(n) = my(A, m); if( n<0, 0, m = 2*n; A = serreverse( intformal( (1 - x^4 / 4 + x * O(x^m)) ^ (-1/2))); m! * polcoeff( sqrt( (2 - A^2) / (2 + A^2)), m))}; /* Michael Somos, Apr 25 2011 */
  • PARI
    {a(n) = local(C=1+x); for(i=1,n, C = exp( intformal( C * intformal(-1/C^3 + x*O(x^n)) ) ) ); n!*polcoeff(C,n)}
for(n=0,20,print1(a(2*n),", "))
  • PARI
    {a(n) = local(C=1+x); for(i=1,n, C = exp( intformal( -1/C * intformal(C^3 + x*O(x^n)) ) ) ); n!*polcoeff(C,n)}
for(n=0,20,print1(a(2*n),", "))

Formula

E.g.f. C(x) satisfies: C(x)^4 + 2*S(x)^2 = 1 where S(x) = Integral [1 - 2*S(x)^2]^(3/4) dx with S(0)=0; Left-shift of the Laplace transform of e.g.f. C(x) equals the Laplace transform of S(x).
E.g.f.: Sum_{k>=0} 2^k * a(k) * x^(2*k) / (2*k)! = cos lemn(x) where cos lemn(x) is the cosine lemniscate function of Gauss. - Michael Somos, Apr 25 2011
O.g.f.: 1/(1 + 1^2*x/(1 + 2^2/2*x/(1 + 3^2*x/(1 + 4^2/2*x/(1 + 5^2*x/(1 + 6^2/2*x/(1 + 7^2*x/(1 + 8^2/2*x/(1+...))))))))) (continued fraction). - Paul D. Hanna, Jul 29 2011
a(n) = (-1)^floor(n/2) * A193544(n) = (-1)^ceiling(n/2) * A193544(n) = -A159601(n). - M. F. Hasler, Aug 31 2012
G.f.: 1/Q(0) where p=1/2, Q(k) = 1 + x*(2*k+1)^2/( 1 + p*x*(2*k+2)^2/Q(k+1) ); (continued fraction due to Stieltjes T.J.). - Sergei N. Gladkovskii, Mar 22 2013
From Paul D. Hanna, Jun 02 2015: (Start)
E.g.f. C(x) satisfies:
(1) C(x) = exp( Integral C(x) * Integral -1/C(x)^3 dx dx ).
(2) C(x) = exp( Integral -1/C(x) * Integral C(x)^3 dx dx ).
(End)
G.f.: 1 / (1 + b(1)*x / (1 + b(2)*x / (1 + b(3)*x / ... ))) where b = A129194. - Michael Somos, Jan 03 2013

A193543 E.g.f.: Pi/(sqrt(2)*L) * (1 + 2*Sum_{n>=1} cosh(2*Pi*n*x/L)/cosh(n*Pi)) where L = Lemniscate constant.

Original entry on oeis.org

1, 1, 9, 153, 4977, 261009, 20039481, 2121958377, 296297348193, 52750142341281, 11662264481073129, 3134732109393169593, 1006734732695870345937, 380718482718134681818929, 167456229155543640166939161, 84761007600911799530893148937, 48919649166315485705652984573633
Offset: 0

Views

Author

Paul D. Hanna, Jul 29 2011

Keywords

Comments

L = Lemniscate constant = 2*(Pi/2)^(3/2)/gamma(3/4)^2 = 2.62205755429...
Compare the definition with that of the dual sequence A193540.

Examples

			E.g.f.: A(x) = 1 + x^2/2! + 9*x^4/4! + 153*x^6/6! + 4977*x^8/8! + 261009*x^10/10! + 20039481*x^12/12! +...+ a(n)*x^(2*n)/(2*n)! +...
where
A(x)*sqrt(2)*L/Pi = 1 + 2*cosh(2*Pi*x/L)/cosh(Pi) + 2*cosh(4*Pi*x/L)/cosh(2*Pi) + 2*cosh(6*Pi*x/L)/cosh(3*Pi) +...
Let B(x) equal the e.g.f. of A193540, where:
B(x)*sqrt(2)*L/Pi = 1 + 2*cos(2*Pi*x/L)/cosh(Pi) + 2*cos(4*Pi*x/L)/cosh(2*Pi) + 2*cos(6*Pi*x/L)/cosh(3*Pi) +...
explicitly,
B(x) = 1 - x^2/2! + 9*x^4/4! - 153*x^6/6! + 4977*x^8/8! - 261009*x^10/10! + 20039481*x^12/12! +...
then A(x)^-2 + B(x)^-2 = 2
as illustrated by:
A(x)^-2 = 1 - 2*x^2/2! + 144*x^6/6! - 96768*x^10/10! + 268240896*x^14/14! +...
B(x)^-2 = 1 + 2*x^2/2! - 144*x^6/6! + 96768*x^10/10! - 268240896*x^14/14! +...
...
O.g.f.: 1 + x + 9*x^2 + 153*x^3 + 4977*x^4 + 261009*x^5 + 20039481*x^6 +...+ a(n)*x^n +...
O.g.f.: 1/(1 - x/(1 - 8*x/(1 - 9*x/(1 - 32*x/(1 - 25*x/(1 - 72*x/(1 - 49*x/(1 - 128*x/(1-...))))))))).

Links

Crossrefs

Cf. A193540, A193541, A193542, A193544, A193545.

Programs

  • Mathematica
    nmax = 20; s = CoefficientList[Series[JacobiDN[Sqrt[2]*x, 1/2], {x, 0, 2*nmax}], x] * Range[ 0, 2*nmax]!; Table[(-1)^n * s[[2*n + 1]], {n, 0, nmax}] (* Vaclav Kotesovec, Nov 29 2020 *)
  • PARI
    {a(n)=local(L=2*(Pi/2)^(3/2)/gamma(3/4)^2);if(n==0,1,sqrt(2)*Pi/L*suminf(k=1,(2*k*Pi/L)^(2*n)/cosh(k*Pi)))} \\ Paul D. Hanna, Aug 29 2012
for(n=0, 20, print1(a(n), ", "))
  • PARI
    {a(n)=local(R,L=2*(Pi/2)^(3/2)/gamma(3/4)^2);
R=Pi/(sqrt(2)*L)*(1 + 2*suminf(m=1,cosh(2*Pi*m*x/L +O(x^(2*n+1)))/cosh(m*Pi)));
round((2*n)!*polcoeff(R,2*n))}
  • PARI
    {a(n)=local(R,L=2*(Pi/2)^(3/2)/gamma(3/4)^2);
R=Pi/(sqrt(2)*L)*(1 + 2*suminf(m=1,1/(1 - (2*m*Pi/L)^2*x+x*O(x^n))/cosh(m*Pi)));
round(polcoeff(R,n))} \\ Paul D. Hanna, Aug 29 2012
  • PARI
    {a(n) = my(C=1); C = cosh( serreverse( intformal( 1/sqrt( cosh(2*x +O(x^(2*n+1))) ) ) ) ); (2*n)!*polcoeff(C,2*n)}
for(n=0,20,print1(a(n),", ")) \\ Paul D. Hanna, Aug 14 2017

Formula

E.g.f.: cosh( Series_Reversion( Integral 1/sqrt( cosh(2*x) ) dx ) ). - Paul D. Hanna, Aug 14 2017
E.g.f.: sqrt(1 + S(x)^2), where S(x) is the e.g.f. of A289695. - Paul D. Hanna, Aug 14 2017
E.g.f.: 1 + Integral S(x) * sqrt(1 + 2*S(x)^2) dx, where S(x) is the e.g.f. of A289695. - Paul D. Hanna, Aug 14 2017
...
Given e.g.f. A(x), define the e.g.f. of A193540:
B(x) = Pi/(sqrt(2)*L) * (1 + 2*Sum_{n>=1} cos(2*Pi*n*x/L) / cosh(n*Pi)),
then A(x)^-2 + B(x)^-2 = 2 by Ramanujan's cos/cosh identity.
...
E.g.f. equals the reciprocal of the e.g.f. of A193544.
...
O.g.f.: 1/(1 - 1^2*x/(1 - 2*2^2*x/(1 - 3^2*x/(1 - 2*4^2*x/(1 - 5^2*x/(1 - 2*6^2*x/(1 - 7^2*x/(1 - 2*8^2*x/(1-...))))))))) (continued fraction).
O.g.f.: Pi/(sqrt(2)*L) * (1 + 2*Sum_{n>=1} 1/(1 - (2*n*Pi/L)^2*x) / cosh(n*Pi)) where L = Lemniscate constant. - Paul D. Hanna, Aug 29 2012
...
a(n) = sqrt(2)*Pi/L * Sum_{k>=1} (2*k*Pi/L)^(2*n) / cosh(k*Pi) for n>0 where L = Lemniscate constant. - Paul D. Hanna, Aug 29 2012
...
G.f.: Q(0), where Q(k) = 1 - x*(2*k+1)^2/(x*(2*k+1)^2 - 1/(1 - 2*x*(2*k+2)^2/(2*x*(2*k+2)^2 - 1/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2013
a(n) ~ 2^(7*n + 4) * Pi^(n+1) * n^(2*n + 1/2) / (exp(2*n) * Gamma(1/4)^(4*n + 2)). - Vaclav Kotesovec, Nov 29 2020

A193541 E.g.f.: sqrt(2)*L / (Pi*(1 + 2*Sum_{n>=1} cos(2*Pi*n*x/L)/cosh(n*Pi) )) where L = Lemniscate constant.

Original entry on oeis.org

1, 1, -3, -27, 441, 11529, -442827, -23444883, 1636819569, 145703137041, -16106380394643, -2164638920874507, 347592265948756521, 65724760945840254489, -14454276753061349098587, -3658147171522531111996803, 1055646229815910768764248289
Offset: 0

Views

Author

Paul D. Hanna, Jul 29 2011

Keywords

Comments

L = Lemniscate constant = 2*(Pi/2)^(3/2)/gamma(3/4)^2 = 2.62205755429...
Compare the definition with that of the dual sequence A193544.

Examples

			E.g.f.: A(x) = 1 + x^2/2! - 3*x^4/4! - 27*x^6/6! + 441*x^8/8! + 11529*x^10/10! - 442827*x^12/12! +...+ a(n)*x^(2*n)/(2*n)! +...
where
A(x) = sqrt(2)*L/(Pi*(1 + 2*cos(2*Pi*x/L)/cosh(Pi) + 2*cos(4*Pi*x/L)/cosh(2*Pi) + 2*cos(6*Pi*x/L)/cosh(3*Pi) +...)).
Let B(x) equal the e.g.f. of A193544, where:
B(x) = sqrt(2)*L/(Pi*(1 + 2*cosh(2*Pi*x/L)/cosh(Pi) + 2*cosh(4*Pi*x/L)/cosh(2*Pi) + 2*cosh(6*Pi*x/L)/cosh(3*Pi) +...))
explicitly,
B(x) = 1 - x^2/2! - 3*x^4/4! + 27*x^6/6! + 441*x^8/8! - 11529*x^10/10! - 442827*x^12/12! +...
then A(x)^2 + B(x)^2 = 2
as illustrated by:
A(x)^2 = 1 + 2*x^2/2! - 144*x^6/6! + 96768*x^10/10! - 268240896*x^14/14! +...
B(x)^2 = 1 - 2*x^2/2! + 144*x^6/6! - 96768*x^10/10! + 268240896*x^14/14! +...
...
O.g.f.: 1 + x - 3*x^2 - 27*x^3 + 441*x^4 + 11529*x^5 - 442827*x^6 +...+ a(n)*x^n +...
O.g.f.: 1/(1 - x/(1 + 4*x/(1 - 9*x/(1 + 16*x/(1 - 25*x/(1 + 36*x/(1 - 49*x/(1 + 64*x/(1-...))))))))).

Links

Crossrefs

Cf. A159600, A193540, A193542, A193543, A193544, A193545.

Programs

  • Mathematica
    a[ n_] := If[ n < 0, 0, With[{m = 2 n}, 2^n m! SeriesCoefficient[ JacobiND[ x, 1/2], {x, 0, m}]]]; (* Michael Somos, Oct 18 2011 *)
a[ n_] := If[ n < 0, 0, With[{m = 2 n}, m! SeriesCoefficient[ JacobiDN[ x, -1], {x, 0, m}]]]; (* Michael Somos, Jun 17 2016 *)
  • PARI
    {a(n)=local(R,L=2*(Pi/2)^(3/2)/gamma(3/4)^2);
R=(sqrt(2)*L/Pi)/(1 + 2*suminf(m=1,cos(2*Pi*m*x/L +O(x^(2*n+1)))/cosh(m*Pi)));
round((2*n)!*polcoeff(R,2*n))}

Formula

Given e.g.f. A(x), define the e.g.f. B(x) of A193544:
B(x) = sqrt(2)*L / (Pi*(1 + 2*Sum_{n>=1} cosh(2*Pi*n*x/L)/cosh(n*Pi) )),
then A(x)^2 + B(x)^2 = 2 by Ramanujan's cos/cosh identity.
E.g.f. equals the reciprocal of the e.g.f. of A193540.
O.g.f.: 1/(1 - 1^2*x/(1 + 2^2*x/(1 - 3^2*x/(1 + 4^2*x/(1 - 5^2*x/(1 + 6^2*x/(1 - 7^2*x/(1 + 8^2*x/(1-...))))))))) (continued fraction).
G.f.: 1/U(0) where U(k)= 1 - x*(2*k+1)^2/(1 + x*(2*k+2)^2/U(k+1)); (continued fraction). - Sergei N. Gladkovskii, Jun 28 2012
G.f.: Q(0), where Q(k) = 1 - x*(2*k+1)^2/(x*(2*k+1)^2 - 1/(1 - x*(2*k+2)^2/(x*(2*k+2)^2 + 1/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2013

A193540 E.g.f.: Pi/(sqrt(2)*L) * (1 + 2*Sum_{n>=1} cos(2*Pi*n*x/L)/cosh(n*Pi)) where L = Lemniscate constant.

Original entry on oeis.org

1, -1, 9, -153, 4977, -261009, 20039481, -2121958377, 296297348193, -52750142341281, 11662264481073129, -3134732109393169593, 1006734732695870345937, -380718482718134681818929, 167456229155543640166939161, -84761007600911799530893148937
Offset: 0

Views

Author

Paul D. Hanna, Jul 29 2011

Keywords

Comments

L = Lemniscate constant = 2*(Pi/2)^(3/2)/gamma(3/4)^2 = 2.62205755429...
Compare the definition with that of the dual sequence A193543.

Examples

			E.g.f.: A(x) = 1 - x^2/2! + 9*x^4/4! - 153*x^6/6! + 4977*x^8/8! - 261009*x^10/10! + 20039481*x^12/12! +...+ a(n)*x^(2*n)/(2*n)! +...
where
A(x)*sqrt(2)*L/Pi = 1 + 2*cos(2*Pi*x/L)/cosh(Pi) + 2*cos(4*Pi*x/L)/cosh(2*Pi) + 2*cos(6*Pi*x/L)/cosh(3*Pi) +...
Let B(x) equal the e.g.f. of A193543, where:
B(x)*sqrt(2)*L/Pi = 1 + 2*cosh(2*Pi*x/L)/cosh(Pi) + 2*cosh(4*Pi*x/L)/cosh(2*Pi) + 2*cosh(6*Pi*x/L)/cosh(3*Pi) +...
explicitly,
B(x) = 1 + x^2/2! + 9*x^4/4! + 153*x^6/6! + 4977*x^8/8! + 261009*x^10/10! + 20039481*x^12/12! +...
then A(x)^-2 + B(x)^-2 = 2
as illustrated by:
A(x)^-2 = 1 + 2*x^2/2! - 144*x^6/6! + 96768*x^10/10! - 268240896*x^14/14! +...
B(x)^-2 = 1 - 2*x^2/2! + 144*x^6/6! - 96768*x^10/10! + 268240896*x^14/14! +...
...
O.g.f.: 1 - x + 9*x^2 - 153*x^3 + 4977*x^4 - 261009*x^5 + 20039481*x^6 +...+ a(n)*x^n +...
O.g.f.: 1/(1 + x/(1 + 8*x/(1 + 9*x/(1 + 32*x/(1 + 25*x/(1 + 72*x/(1 + 49*x/(1 + 128*x/(1+...))))))))).

Links

Crossrefs

Cf. A193541, A193542, A193543, A193544, A193545.

Programs

  • Mathematica
    a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ Tan[ JacobiAmplitude[ x, -1]] / Tan[ JacobiAmplitude[ 2 x, -1] / 2], {x, 0, m}]]]; (* Michael Somos, Oct 18 2011 *)
a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ JacobiND[ x, -1], {x, 0, m}]]]; (* Michael Somos, Oct 18 2011 *)
Table[SeriesCoefficient[InverseSeries[Series[EllipticF[x, 1/2], {x, 0, 32}]], 2 n + 1] (2 n + 1)! 2^n, {n, 0, 15}] (* Benedict W. J. Irwin, Apr 04 2017 *)
Table[SeriesCoefficient[JacobiDN[Sqrt[2] x, 1/2], {x, 0, 2 k}] (2 k)!, {k, 0, 20}] (* Jan Mangaldan, Nov 28 2020 *)
nmax = 20; s = CoefficientList[Series[JacobiDN[Sqrt[2] x, 1/2], {x, 0, 2*nmax}], x] * Range[ 0, 2*nmax]!; Table[s[[2*n + 1]], {n, 0, nmax}] (* Vaclav Kotesovec, Nov 29 2020 *)
  • PARI
    {a(n)=local(R,L=2*(Pi/2)^(3/2)/gamma(3/4)^2);
R=Pi/(sqrt(2)*L)*(1 + 2*suminf(m=1,cos(2*Pi*m*x/L +O(x^(2*n+1)))/cosh(m*Pi)));
round((2*n)!*polcoeff(R,2*n))}

Formula

Given e.g.f. A(x), define the e.g.f. of A193543:
B(x) = sqrt(2)*Pi/(2*L) * (1 + 2*Sum_{n>=1} cosh(2*Pi*n*x/L) / cosh(n*Pi)),
then A(x)^-2 + B(x)^-2 = 2 by Ramanujan's cos/cosh identity.
...
E.g.f. equals the reciprocal of the e.g.f. of A193541.
O.g.f. = 1/(1 + 1^2*x/(1 + 2*2^2*x/(1 + 3^2*x/(1 + 2*4^2*x/(1 + 5^2*x/(1 + 2*6^2*x/(1 + 7^2*x/(1 + 2*8^2*x/(1+...))))))))) (continued fraction).
G.f.: 1/Q(0) where p=2, Q(k) = 1 + x*(2*k+1)^2/( 1 + p*x*(2*k+2)^2/Q(k+1) ); (continued fraction due to T. J. Stieltjes). - Sergei N. Gladkovskii, Mar 22 2013
a(n) ~ (-1)^n * 2^(7*n + 4) * Pi^(n+1) * n^(2*n + 1/2) / (exp(2*n) * Gamma(1/4)^(4*n + 2)). - Vaclav Kotesovec, Nov 29 2020

A193542 E.g.f.: 2*L^2/(Pi^2*(1 + 2*Sum_{n>=1} cos(2*Pi*n*x/L)/cosh(n*Pi) )^2) where L = Lemniscate constant.

Original entry on oeis.org

1, 0, 2, 0, 0, 0, -144, 0, 0, 0, 96768, 0, 0, 0, -268240896, 0, 0, 0, 2111592333312, 0, 0, 0, -37975288540299264, 0, 0, 0, 1353569484565546795008, 0, 0, 0, -86498911610371173437669376, 0, 0, 0, 9198407234012051081051108278272, 0, 0, 0, -1536583522302562247445395779495133184
Offset: 0

Views

Author

Paul D. Hanna, Jul 29 2011

Keywords

Comments

L = Lemniscate constant = 2*(Pi/2)^(3/2)/gamma(3/4)^2 = 2.62205755429...
Compare the definition with that of the dual sequence A193545.

Examples

			E.g.f.: A(x) = 1 + 2*x^2/2! - 144*x^6/6! + 96768*x^10/10! - 268240896*x^14/14! +...+ a(n)*x^n/n! +...
which equals the square of the e.g.f. B(x) of A193541:
B(x) = 1 + x^2/2! - 3*x^4/4! - 27*x^6/6! + 441*x^8/8! + 11529*x^10/10! - 442827*x^12/12! +...

Links

Crossrefs

Cf. A193540, A193541, A193543, A193544, A193545.

Programs

  • Mathematica
    a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ JacobiDN[ x, -1]^2, {x, 0, n}]]; (* Michael Somos, Jun 17 2016 *)
  • PARI
    {a(n)=local(R,L=2*(Pi/2)^(3/2)/gamma(3/4)^2);
R=(sqrt(2)*L/Pi)/(1 + 2*suminf(m=1,cos(2*Pi*m*x/L +x*O(x^n))/cosh(m*Pi)));
round(n!*polcoeff(R^2,n))}

Formula

a(n) = -A193545(n) for n>=1.
E.g.f.: dn(x, -1)^2 where dn() is a Jacobi elliptic function. - Michael Somos, Jun 17 2016

A193545 E.g.f.: 2*L^2/(Pi^2*(1 + 2*Sum_{n>=1} cosh(2*Pi*n*x/L)/cosh(n*Pi) )^2) where L = Lemniscate constant.

Original entry on oeis.org

1, 0, -2, 0, 0, 0, 144, 0, 0, 0, -96768, 0, 0, 0, 268240896, 0, 0, 0, -2111592333312, 0, 0, 0, 37975288540299264, 0, 0, 0, -1353569484565546795008, 0, 0, 0, 86498911610371173437669376, 0, 0, 0, -9198407234012051081051108278272, 0, 0, 0, 1536583522302562247445395779495133184
Offset: 0

Views

Author

Paul D. Hanna, Jul 29 2011

Keywords

Comments

L = Lemniscate constant = 2*(Pi/2)^(3/2)/gamma(3/4)^2 = 2.62205755429...
Compare the definition with that of the dual sequence A193542.

Examples

			E.g.f.: A(x) = 1 - 2*x^2/2! + 144*x^6/6! - 96768*x^10/10! + 268240896*x^14/14! +...+ a(n)*x^n/n! +...
which equals the square of the e.g.f. B(x) of A193544:
B(x) = 1 - x^2/2! - 3*x^4/4! + 27*x^6/6! + 441*x^8/8! - 11529*x^10/10! - 442827*x^12/12! +...

Links

Crossrefs

Cf. A193540, A193541, A193542, A193543, A193544.

Programs

  • PARI
    {a(n)=local(R,L=2*(Pi/2)^(3/2)/gamma(3/4)^2);
R=(sqrt(2)*L/Pi)/(1 + 2*suminf(m=1,cosh(2*Pi*m*x/L +x*O(x^n))/cosh(m*Pi)));
round(n!*polcoeff(R^2,n))}

Formula

a(n) = -A193542(n) for n>=1.

A190904 a(n) = Sum_{k=0..n-1} cos(Pi*k/2)*binomial(n-1,k)*a(n-1-k)*a(k) for n > 0, a(0) = 1.

Original entry on oeis.org

1, 1, 1, 0, -3, -12, -27, 0, 441, 3024, 11529, 0, -442827, -4390848, -23444883, 0, 1636819569, 21224560896, 145703137041, 0, -16106380394643, -257991277243392, -2164638920874507, 0, 347592265948756521
Offset: 0

Views

Author

Peter Luschny, Jul 26 2011

Keywords

Links

Crossrefs

Cf. A060627, A060628, A104203, A159600, A193541, A193544.

Programs

  • Maple
    A190904 := proc(n) option remember; `if`(n=0,1,add(((1-irem(k,2))*(-1)^ iquo(k,2))*binomial(n-1,k)*A190904(n-1-k)*A190904(k),k=0..n-1)) end:
  • Mathematica
    a[0] = 1;
a[n_] := a[n] =
  Sum[Mod[(k+1)^3, 4, -1] Binomial[n-1, k] a[n-k-1] a[k], {k, 0, n-1}];
Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Jun 24 2019 *)

Formula

Let F(n,x) = Sum_{k=0..n-1} cos(Pi*k*x)*binomial(n-1,k)*F(n-1-k,x)* F(k,x), then
F(n, 0) = n! = A000142(n),
F(n, 1/2) = a(n),
F(n, 1) = 2^n*Euler_{n}(1) = A_{n}(-1) = A155585(n).
a(2*n) = A159601(n)*(-1)^floor((n-1)/2).
a(2*n+1) = A104203(2*n+1).
From Peter Bala, Aug 25 2011: (Start)
The sequence entries may be calculated as follows: Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0. The coefficients in the expansion of D^n[f](x) in powers of f(x) can be found in A145271. Then we have
a(2*n) = D^(2*n)[sqrt(1+sin^2(x))](0)
a(2*n+1) = D^(2*n)[sqrt(1-x^4)](0).
The generating function involves the Jacobian elliptic functions. Define E(u,k) := cn(i*u,k)-i*sn(i*u,k) = 1+u+u^2/2!+(1+k^2)*u^3/3!+(1+4*k^2)*u^4/4!+..., where cn(u,k) and sn(u,k) are Jacobian elliptic functions of modulus k (see A060627 and A060628). Then the e.g.f. A(u) for this sequence is
A(u) := E(u,i) = 1+u+u^2/2!-3*u^4/4!-12*u^5/5!-27*u^6/6!+....
Proof: Using well-known properties of the Jacobian elliptic functions (see for example Abramowitz and Stegun, Chapter 16) we find the generating function A(u) satisfies the differential equation
(d/du)A(u) = dn(i*u,i)*A(u) = 1/2*(A(i*u)+A(-i*u))*A(u), which leads to a recurrence for the coefficients of A(u):
a(n+1) = sum{k=0..floor(n/2)} (-1)^k*binomial(n,2*k)*a(2*k)*a(n-2*k) with a(0) = 1. This recurrence is equivalent to the defining recurrence for this sequence given above.
End proof.
The generating function A(u) satisfies 1/A(u) = A(-u).
Compare entries of this sequence with those of A104203, A159600, A193541 and A193544.
(End)

A322218 E.g.f.: C(x,q) = 1 + Integral S(x,q) * C(q*x,q) dx, such that C(x,q)^2 - S(x,q)^2 = 1, where C(x,q) = Sum_{n>=0} sum_{k=0..n*(n-1)/2} T(n,k)*x^n*y^k/n!, as an irregular triangle of coefficients T(n,k) read by rows.

Original entry on oeis.org

1, 1, 1, 4, 1, 20, 16, 24, 1, 56, 336, 288, 384, 128, 192, 1, 120, 2352, 6448, 12736, 5888, 10176, 5760, 3840, 1280, 1920, 1, 220, 10032, 93280, 214016, 472704, 385472, 431616, 294912, 341504, 141056, 164352, 69120, 46080, 15360, 23040, 1, 364, 32032, 740168, 4072640, 11702912, 18676672, 30112640, 23848704, 27599616, 17884032, 20958208, 13595136, 11074560, 5992448, 5945856, 2673664, 2300928, 967680, 645120, 215040, 322560, 1, 560, 84448, 3952832, 53301248, 230161152, 738249344, 1166436352, 1970874368, 2196244480, 2459786240, 1804101632, 2061498368, 1537437696, 1437724672, 989968384, 921092096, 487923712, 499621888, 282034176, 211599360, 117383168, 108036096, 42778624, 36814848, 15482880, 10321920, 3440640, 5160960
Offset: 0

Views

Author

Paul D. Hanna, Dec 16 2018

Keywords

Comments

Compare to Jacobi's elliptic function cn(x,k) = 1 - Integral sn(x,k)*dn(x,k) dx such that cn(x,k)^2 + sn(x,k)^2 = 1 and dn(x,k)^2 + k^2*sn(x,k)^2 = 1.
Right border equals A002866.
Row sums equal the secant numbers (A000364).
Last n terms in row n of this triangle and of triangle A322219 are equal for n>0.

Examples

			E.g.f. C(x,q) = Sum_{n>=0} sum_{k=0..n*(n-1)/2} T(n,k) * x^(2*n)*q^(2*k)/(2*n)! starts
C(x,q) = 1 + x^2/2! + (4*q^2 + 1)*x^4/4! + (24*q^6 + 16*q^4 + 20*q^2 + 1)*x^6/6! + (192*q^12 + 128*q^10 + 384*q^8 + 288*q^6 + 336*q^4 + 56*q^2 + 1)*x^8/8! + (1920*q^20 + 1280*q^18 + 3840*q^16 + 5760*q^14 + 10176*q^12 + 5888*q^10 + 12736*q^8 + 6448*q^6 + 2352*q^4 + 120*q^2 + 1)*x^10/10! + (23040*q^30 + 15360*q^28 + 46080*q^26 + 69120*q^24 + 164352*q^22 + 141056*q^20 + 341504*q^18 + 294912*q^16 + 431616*q^14 + 385472*q^12 + 472704*q^10 + 214016*q^8 + 93280*q^6 + 10032*q^4 + 220*q^2 + 1)*x^12/12! + ...
such that C(x,q) = cosh( Integral C(q*x,q) dx ).
This irregular triangle of coefficients T(n,k) of x^(2*n)*q^(2*k)/(2*n)! in C(x,q) begins:
1;
1;
1, 4;
1, 20, 16, 24;
1, 56, 336, 288, 384, 128, 192;
1, 120, 2352, 6448, 12736, 5888, 10176, 5760, 3840, 1280, 1920;
1, 220, 10032, 93280, 214016, 472704, 385472, 431616, 294912, 341504, 141056, 164352, 69120, 46080, 15360, 23040;
1, 364, 32032, 740168, 4072640, 11702912, 18676672, 30112640, 23848704, 27599616, 17884032, 20958208, 13595136, 11074560, 5992448, 5945856, 2673664, 2300928, 967680, 645120, 215040, 322560;
1, 560, 84448, 3952832, 53301248, 230161152, 738249344, 1166436352, 1970874368, 2196244480, 2459786240, 1804101632, 2061498368, 1537437696, 1437724672, 989968384, 921092096, 487923712, 499621888, 282034176, 211599360, 117383168, 108036096, 42778624, 36814848, 15482880, 10321920, 3440640, 5160960; ...
RELATED SERIES.
S(x,q) = x + (q^2 + 1)*x^3/3! + (4*q^6 + q^4 + 10*q^2 + 1)*x^5/5! + (24*q^12 + 16*q^10 + 20*q^8 + 85*q^6 + 91*q^4 + 35*q^2 + 1)*x^7/7! + (192*q^20 + 128*q^18 + 384*q^16 + 288*q^14 + 1200*q^12 + 632*q^10 + 2737*q^8 + 1324*q^6 + 966*q^4 + 84*q^2 + 1)*x^9/9! + (1920*q^30 + 1280*q^28 + 3840*q^26 + 5760*q^24 + 10176*q^22 + 16448*q^20 + 19776*q^18 + 27568*q^16 + 49872*q^14 + 69816*q^12 + 64329*q^10 + 50941*q^8 + 26818*q^6 + 5082*q^4 + 165*q^2 + 1)*x^11/11! +  ...
where C(x,q)^2 - S(x,q)^2 = 1.

Links

Crossrefs

Cf. A322219 (S(x,q)), A000364 (row sums), A193544.

Programs

  • Mathematica
    rows = 8; m = 2 rows; s[x_, ] = x; c[, ] = 1; Do[s[x, q_] = Integrate[c[x, q] c[q x, q] + O[x]^m // Normal, x]; c[x_, q_] = 1 + Integrate[s[x, q] c[q x, q] + O[x]^m // Normal, x], {m}];
CoefficientList[#, q^2]& /@ (CoefficientList[c[x, q], x] Range[0, m]!) // DeleteCases[#, {}]& // Flatten (* Jean-François Alcover, Dec 17 2018 *)
  • PARI
    {T(n,k) = my(S=x,C=1); for(i=1,2*n,
S = intformal(C*subst(C,x,q*x) +O(x^(2*n+1)));
C = 1 + intformal(S*subst(C,x,q*x)));
(2*n)!*polcoeff( polcoeff(C,2*n,x),2*k,q)}
for(n=0,10, for(k=0,n*(n-1)/2, print1( T(n,k),", "));print(""))

Formula

E.g.f. C(x,q) and related series S(x,q) satisfy:
(1) C(x,q)^2 - S(x,q)^2 = 1.
(2) C(x,q) = 1 + Integral S(x,q) * C(q*x,q) dx.
(3) S(x,q) = Integral C(x,q) * C(q*x,q) dx.
(4a) C(x,q) + S(x,q) = exp( Integral C(q*x,q) dx ).
(4b) C(x,q) = cosh( Integral C(q*x,q) dx ).
(4c) S(x,q) = sinh( Integral C(q*x,q) dx ).
(5) C(q*x,q) = 1 + q * Integral S(q*x,q) * C(q^2*x,q) dx.
(6) S(q*x,q) = q * Integral C(q*x,q) * C(q^2*x,q) dx.
(7a) C(q*x,q) + S(q*x,q) = exp( q * Integral C(q^2*x,q) dx ).
(7b) C(q*x,q) = cosh( q * Integral C(q^2*x,q) dx ).
(7c) S(q*x,q) = sinh( q * Integral C(q^2*x,q) dx ).
PARTICULAR ARGUMENTS.
C(x,q=0) = cosh(x).
C(x,q=1) = 1/cos(x).
C(x,q=i) = cl(i*x), where cl(x) is the cosine lemniscate function (A159600).
FORMULAS FOR TERMS.
T(n, n*(n-1)/2) = 2^(n-1)*n! for n >= 1.
T(n, n*(n-1)/2 - k) = A322219(n, n*(n+1)/2 - k) for k = 0..n-1, n > 0.
Sum_{k=0..n*(n-1)/2} T(n,k) = A000364(n) for n >= 0.
Sum_{k=0..n*(n-1)/2} T(n,k)*(-1)^k = A193544(2*n+1) for n >= 0.

A286306 a(n) = coefficient of x^(2*n)/(2*n)! in exp( integral ( sn(x, 1/2) / cd(x, 1/2) ) dx).

Original entry on oeis.org

1, 1, 3, 27, 441, 11529, 442827, 23444883, 1636819569, 145703137041, 16106380394643, 2164638920874507, 347592265948756521, 65724760945840254489, 14454276753061349098587, 3658147171522531111996803, 1055646229815910768764248289, 344553616791279239828059918881
Offset: 0

Views

Author

Michael Somos, May 05 2017

Keywords

Examples

			G.f. = 1 + x + 3*x^2 + 27*x^3 + 441*x^4 + 11529*x^5 + 442827*x^6 + ...
E.g.f. = 1 + 1*x^2/2! + 3*x^4/4! + 27*x^6/6! + 441*x^8/8! + 11529*x^10/10! + ...

Links

Crossrefs

Cf. A159600, A159601, A190904, A193541, A193544, A242240.

Programs

  • Mathematica
    a[ n_] := If[ n < 0, 0, With[{m = 2 n}, m! SeriesCoefficient[ Exp[ Integrate[ JacobiSN[x, 1/2] / JacobiCD[x, 1/2], x]], {x, 0, m}]]];
a:= With[{nmax = 110}, CoefficientList[Series[Exp[Integrate[JacobiSN[x, 1/2]/JacobiCD[x, 1/2], x]], {x, 0, nmax}], x]*Range[0, nmax]!][[1 ;; ;; 2]]; Table[a[[n]], {n, 1, 50}] (* G. C. Greubel, Jul 29 2018 *)
  • PARI
    {a(n) = my(m); if( n<0, 0, m = 2*n; m! * polcoeff( exp( intformal( serreverse( intformal( (1 + x^4 + x * O(x^m))^(-1/2))))), m))};

Formula

Given e.g.f. A(x) = Sum_{n>=0} a(n) * x^(2*n) / (2*n)!, then 0 = 1 + 2*A'^2 - A*A''.
Given e.g.f. A(x), then A'(x) / A(x) = B(x) where B() is the e.g.f. for A242240.
Given e.g.f. A(x), 1 / A(x) = A(-x).
A159600(n) = (-1)^n * a(n). A159601(n) = -(-1)^n * a(n) if n>0.
A190904(2*n) = A193541(n) = (-1)^floor(n/2) * a(n). A193544(n) = (-1)^floor((n+1)/2) * a(n).

A291207 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of continued fraction 1/(1 + x/(1 - 2^k*x/(1 + 3^k*x/(1 - 4^k*x/(1 + 5^k*x/(1 - ...)))))).

Original entry on oeis.org

1, 1, -1, 1, -1, 0, 1, -1, -1, 1, 1, -1, -3, 5, 0, 1, -1, -7, 27, 17, -2, 1, -1, -15, 167, 441, -121, 0, 1, -1, -31, 1071, 10673, -11529, -721, 5, 1, -1, -63, 6815, 262305, -1337713, -442827, 6845, 0, 1, -1, -127, 42687, 6525377, -161721441, -297209047, 23444883, 58337, -14
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 21 2017

Keywords

Examples

			G.f. of column k: A_k(x) = 1 - x + (1 - 2^k)*x^2 + (2^(k + 1) - 4^k + 6^k - 1)*x^3 + ...
Square array begins:
   1,     1,       1,         1,           1,             1,  ...
  -1,    -1,      -1,        -1,          -1,            -1,  ...
   0,    -1,      -3,        -7,         -15,           -31,  ...
   1,     5,      27,       167,        1071,          6815,  ...
   0,    17,     441,     10673,      262305,       6525377,  ...
  -2,  -121,  -11529,  -1337713,  -161721441,  -19802585281,  ...

Crossrefs

Columns k=0-2 give A105523, A202038, A193544.
Main diagonal gives A292920.
Cf. A290569.

Programs

  • Mathematica
    Table[Function[k, SeriesCoefficient[1/(1 + ContinuedFractionK[-(-1)^i i^k x, 1, {i, 1, n}]), {x, 0, n}]][j - n], {j, 0, 9}, {n, 0, j}] // Flatten

Formula

G.f. of column k: 1/(1 + x/(1 - 2^k*x/(1 + 3^k*x/(1 - 4^k*x/(1 + 5^k*x/(1 - ...)))))), a continued fraction.
Showing 1-10 of 10 results.

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