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.

A193544 E.g.f.: sqrt(2)*(L/Pi) / (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, -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 A193541.

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*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 A193541, 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! - 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+...))))))))).
		

Crossrefs

Programs

  • Mathematica
    L = 2*(Pi/2)^(3/2)/Gamma[3/4]^2; a[0] = 1; a[n_] := 2*Pi/L*NSum[(-1)^k * (2*k*Pi/L)^(2*n)/Cosh[k*Pi], {k, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 50] // Round; Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Sep 29 2017 *)
  • PARI
    {a(n)=local(L=2*(Pi/2)^(3/2)/gamma(3/4)^2); if(n==0, 1, 2*Pi/L*suminf(k=1, (-1)^k*(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=(sqrt(2)*L/Pi)/(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/L)*(1 + 2*suminf(m=1,(-1)^m/(1 - (2*m*Pi/L)^2*x+x*O(x^n))/cosh(m*Pi)));
    round(polcoeff(R,n))} \\ Paul D. Hanna, Aug 29 2012

Formula

Given e.g.f. A(x), define the e.g.f. B(x) of A193541:
B(x) = sqrt(2)*L / (Pi*(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 A193543.
...
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).
O.g.f.: (Pi/L) * (1 + 2*Sum_{n>=1} (-1)^n/(1 - (2*n*Pi/L)^2*x) / cosh(n*Pi)) where L = Lemniscate constant. - Paul D. Hanna, Aug 29 2012
...
a(n) = 2*Pi/L * Sum_{k>=1} (-1)^k*(2*k*Pi/L)^(2*n) / cosh(k*Pi) for n>0 where L = Lemniscate constant. - Paul D. Hanna, Aug 29 2012
G.f.: 1/Q(0), where Q(k)= 1 + x*(2*k+1)^2/(1 - x*(2*k+2)^2/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 27 2013
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

A104203 Expansion of the sine lemniscate function sl(x).

Original entry on oeis.org

1, 0, 0, 0, -12, 0, 0, 0, 3024, 0, 0, 0, -4390848, 0, 0, 0, 21224560896, 0, 0, 0, -257991277243392, 0, 0, 0, 6628234834692624384, 0, 0, 0, -319729080846260095008768, 0, 0, 0, 26571747463798134334265819136, 0, 0, 0, -3564202847752289659513902717468672, 0, 0
Offset: 1

Views

Author

Troy Kessler (tkessler1977(AT)netzero.com), Mar 13 2005

Keywords

Comments

For the series expansion of the cosine lemniscate cl(x) see A159600. The lemniscatic functions sl(x) and cl(x) played a significant role in the development of mathematics in the 18th and 19th centuries. They were the first examples of elliptic functions. In algebraic number theory all abelian extensions of the Gaussian rationals Q(i) are contained in extensions of Q(i) generated by division values of the lemniscatic functions. - Peter Bala, Aug 25 2011

Examples

			G.f. = x - 12*x^5 + 3024*x^9 - 4390848*x^13 + 21224560896*x^17 + ...
Example of the recurrence relation a(n+2) = -2*sum {i+j+k = n} n!/(i!*j!*k!)*a(i)*a(j)*a(k) for n = 13:
There are only 6 compositions of 13-2 = 11 that give a nonzero contribution to the sum, namely 11 = 9+1+1 = 1+9+1 = 1+1+9 and 11 = 5+5+1 = 5+1+5 = 1+5+5
and hence
a(13) = -2*(3*11!/(9!*1!*1*)*a(9)*a(1)*a(1)+3*11!/(5!*5!*1!)*a(5)*a(5)*a(1)) = -4390848.
		

Crossrefs

Cf. A144849, A144853, A159600 (cosine lemniscate).
Taking every fourth term gives A283831.
Cf. A242240.

Programs

  • Mathematica
    Drop[ Range[0, 37]! CoefficientList[ InverseSeries[ Series[ Integrate[1/(1 - x^4)^(1/2), x], {x, 0, 37}]], x], 1] (* Robert G. Wilson v, Mar 16 2005 *)
    a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ JacobiSD[x, 1/2] 2^((n - 1)/2), {x, 0, n}]]; (* Michael Somos, Jan 17 2017 *)
    a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ JacobiSN[x, -1], {x, 0, n}]]; (* Michael Somos, May 26 2021 *)
  • PARI
    x='x+O('x^66);Vec(serlaplace(serreverse( intformal(1/sqrt(1-x^4))))) \\ Joerg Arndt, Mar 24 2017

Formula

From Peter Bala, Aug 25 2011: (Start)
The function sl(x) satisfies the differential equation sl''(x) = -2*sl^3(x) with initial conditions sl(0) = 0, sl'(0) = 1.
Recurrence relation:
a(n+2) = -2*sum {i+j+k = n} n!/(i!*j!*k!)*a(i)*a(j)*a(k).
The inverse of the sine lemniscate function may be defined as the algebraic integral
sl^(-1)(x) := Integral_{s=0..x} 1/sqrt(1-s^4) ds = x + x^5/10 + x^9/24 + 5*x^13/208 + ....
Series reversion produces the expansion
sl(x) = x - 12*x^5/5! + 3024*x^9/9! - 4390848*x^13/13! + ....
The coefficients in this expansion can be calculated using nested derivatives as follows (see [Dominici, Theorem 4.1]): Let f(x) = sqrt(1-x^4). 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) are given in A145271. Then we have a(n) = D^(n-1)[f](0).
a(n) is divisible by 12^n and a(n)/12^n produces (a signed and aerated version of) A144853(n).
(End)
The function sl(x) satisfies the differential equation sl'(x)^2 + sl(x)^4 = 1 with initial conditions sl(0) = 0, sl'(0) = 1. - Michael Somos, Oct 12 2019

Extensions

More terms from Robert G. Wilson v, Mar 16 2005
a(37)- a(39) by Vincenzo Librandi, Mar 24 2017

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-...))))))))).
		

Crossrefs

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

A159601 E.g.f. S(x) satisfies: S(x) = Integral [1 - 2*S(x)^2]^(3/4) dx with S(0)=0.

Original entry on oeis.org

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

Views

Author

Paul D. Hanna, May 08 2009

Keywords

Comments

E.g.f. S(x) is an odd function; zero terms are omitted.
Apart from signs and initial term, same as A159600.
Radius of convergence of S(x) is |x| < r where:
r = (1/2)*Pi^(3/2)/gamma(3/4)^2 ;
r = L/sqrt(2) where L=Lemniscate constant ;
r = 1.8540746773013719184338503471952600...
Although S(x) diverges at |x|=r, the power series expansion:
C(x) = [1 - 2*S(x)^2]^(1/4) converges to
C(r) = gamma(3/4)^2/(Pi/2)^(3/2) = 0.7627597635...

Examples

			E.g.f: 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! +-...
C(x)^4 + 2*S(x)^2 = 1 where:
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! -+...
1/C(x) = C(i*x) = 1 + x^2/2! + 3*x^4/4! + 27*x^6/6! + 441*x^8/8! +...
log(C(x)) = -x^2/2! - 12*x^6/6! - 3024*x^10/10! - 4390848*x^14/14! -...
Coefficients in log(C(x)) are given by A104203 (ignoring signs).
		

Crossrefs

Cf. A159600 (C(x)), A104203 (unsigned e.g.f. = S(x)/C(x)).

Programs

  • PARI
    {a(n)=local(S=x);for(i=0,2*n,S=intformal((1-2*S^2+O(x^(2*n)))^(3/4)));(2*n-1)!*polcoeff(S,2*n-1)}

Formula

E.g.f. S(x) satisfies: C(x)^4 + 2*S(x)^2 = 1 where
S'(x) = C(x)^3 and C'(x) = -S(x) with C(0)=1.
E.g.f. S(x) satisfies: S(x)/C(x) = e.g.f. of unsigned A104203 where C(x)^4 + 2*S(x)^2 = 1.
a(n) = -A159600(n), n>0. - M. F. Hasler, Aug 31 2012
G.f.: (1- 1/Q(0))/x, where Q(k) = 1 + x*(2*k+1)^2/(1 + 2*x*(k+1)^2/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2013

A258659 E.g.f. A(x) satisfies: A(x) = exp( Integral A(x)^(1/2) * Integral 1/A(x)^(3/2) dx dx ).

Original entry on oeis.org

1, 1, 3, 18, 189, 3024, 68607, 2095632, 82908441, 4124203776, 251944606683, 18542621357568, 1618221395188629, 165230649971380224, 19514714407120367127, 2639737292796971845632, 405452689572115086887601, 70178277847381981514366976, 13596354857453497541480646963, 2930800480466007704630652960768
Offset: 0

Views

Author

Paul D. Hanna, Jun 06 2015

Keywords

Comments

More generally, we have the identity for real t:
* if G(x) = exp( Integral G(x)^t * Integral 1/G(x)^(3*t) dx dx ),
then G(x) = exp( Integral 1/G(x)^t * Integral G(x)^(3*t) dx dx ).

Examples

			E.g.f. A(x) = 1 + x^2/2! + 3*x^4/4! + 18*x^6/6! + 189*x^8/8! + 3024*x^10/10! +...
		

Crossrefs

Programs

  • Mathematica
    nmax = 40; A = 1; Do[A = Exp[Integrate[A^(1/2)*Integrate[1/A^(3/2), x], x] + O[x]^nmax], nmax]; CoefficientList[A, x^2]*Range[0, nmax-2, 2]! (* Jean-François Alcover, Nov 27 2017 *)
  • PARI
    {a(n) = local(A=1+x); for(i=1,n, A = exp( intformal( A^(1/2) * intformal(1/A^(3/2) + x*O(x^n)) ) ) ); n!*polcoeff(A,n)}
    for(n=0,20,print1(a(2*n),", "))
    
  • PARI
    {a(n) = local(A=1+x); for(i=1,n, A = exp( intformal( 1/A^(1/2) * intformal(A^(3/2) + x*O(x^n)) ) ) ); n!*polcoeff(A,n)}
    for(n=0,20,print1(a(2*n),", "))
    
  • PARI
    {a(n) = if( n<0, 0, my(m = 2*n); m! * polcoeff( exp( intformal( serreverse( intformal( 1 / sqrt(1 + x^4/4 + x * O(x^m)) ) ) ) ), m))}; /* Michael Somos, Jun 17 2017 */

Formula

E.g.f. A(x) satisfies: A(x) = exp( Integral 1/A(x)^(1/2) * Integral A(x)^(3/2) dx dx ).
a(n) ~ c * d^n * n!^2 * sqrt(n), where d = 32*Pi / Gamma(1/4)^4 = 0.58180245681734198604520486465..., c = 1.31298754327535054303509412... . - Vaclav Kotesovec, Jun 15 2015, updated Mar 16 2024
a(2*n) = A261000(n). - Michael Somos, Jun 17 2017

A258657 E.g.f. A(x) satisfies: A(x) = exp( Integral A(x)^2 * Integral 1/A(x)^6 dx dx ).

Original entry on oeis.org

1, 1, 3, 63, 1449, 79569, 4933467, 538446447, 64308790161, 11702872619361, 2287996769790963, 625623383080392543, 181555646645074399929, 69630517380932205676209, 28096571683140898776853707, 14391876463069615780007165967, 7705942233644896318623038389281, 5082509306223148713110677899062721
Offset: 0

Views

Author

Paul D. Hanna, Jun 06 2015

Keywords

Comments

More generally, we have the identity for real t:
* if G(x) = exp( Integral G(x)^t * Integral 1/G(x)^(3*t) dx dx ),
then G(x) = exp( Integral 1/G(x)^t * Integral G(x)^(3*t) dx dx ).

Examples

			E.g.f. A(x) = 1 + x^2/2! + 3*x^4/4! + 63*x^6/6! + 1449*x^8/8! + 79569*x^10/10! +...
		

Crossrefs

Programs

  • Mathematica
    a[ n_] := If[ n < 0, 0, With[{m = 2 n}, m! SeriesCoefficient[ Sqrt @ JacobiDC[ x, -1], {x, 0, m}]]]; (* Michael Somos, Jun 17 2016 *)
  • PARI
    {a(n) = local(A=1+x); for(i=1,n, A = exp( intformal( A^2 * intformal(1/A^6 + x*O(x^n)) ) ) ); n!*polcoeff(A,n)}
    for(n=0,20,print1(a(2*n),", "))
    
  • PARI
    {a(n) = local(A=1+x); for(i=1,n, A = exp( intformal( 1/A^2 * intformal(A^6 + x*O(x^n)) ) ) ); n!*polcoeff(A,n)}
    for(n=0,20,print1(a(2*n),", "))

Formula

E.g.f. A(x) satisfies: A(x) = exp( Integral 1/A(x)^2 * Integral A(x)^6 dx dx ).
a(n) ~ c * 2^(7*n+1) * Pi^(n+1) * n^(2*n) / (exp(2*n) * Gamma(1/4)^(4*n)), where c = 2^(7/4) / (Gamma(1/4) * Pi^(3/4)) = 0.393150663417355411930715... . - Vaclav Kotesovec, Jun 15 2015, updated Mar 17 2024

A258658 E.g.f. A(x) satisfies: A(x) = exp( Integral A(x)^3 * Integral 1/A(x)^9 dx dx ).

Original entry on oeis.org

1, 1, 3, 123, 3129, 313929, 23062347, 4461062067, 655619300721, 207299066045841, 51139145307699603, 23947711418548452843, 8930007221716001596329, 5800583064218449362313689, 3049682921576823757255485147, 2622347479175212075411836474147, 1849810669560927151969244969258721
Offset: 0

Views

Author

Paul D. Hanna, Jun 06 2015

Keywords

Comments

More generally, we have the identity for real t:
* if G(x) = exp( Integral G(x)^t * Integral 1/G(x)^(3*t) dx dx ),
then G(x) = exp( Integral 1/G(x)^t * Integral G(x)^(3*t) dx dx ).

Examples

			E.g.f. A(x) = 1 + x^2/2! + 3*x^4/4! + 123*x^6/6! + 3129*x^8/8! + 313929*x^10/10! +...
		

Crossrefs

Programs

  • PARI
    {a(n) = local(A=1+x); for(i=1,n, A = exp( intformal( A^3 * intformal(1/A^9 + x*O(x^n)) ) ) ); n!*polcoeff(A,n)}
    for(n=0,20,print1(a(2*n),", "))
    
  • PARI
    {a(n) = local(A=1+x); for(i=1,n, A = exp( intformal( 1/A^3 * intformal(A^9 + x*O(x^n)) ) ) ); n!*polcoeff(A,n)}
    for(n=0,20,print1(a(2*n),", "))

Formula

E.g.f. A(x) satisfies: A(x) = exp( Integral 1/A(x)^3 * Integral A(x)^9 dx dx ).

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

Crossrefs

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.
		

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! + ...
		

Crossrefs

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).
Showing 1-10 of 10 results.