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 1-3 of 3 results.

A092869 Series expansion of the Ramanujan-Goellnitz-Gordon continued fraction.

Original entry on oeis.org

1, -1, 0, 1, -1, 1, 0, -2, 2, -1, 0, 2, -3, 2, 0, -2, 4, -4, 0, 4, -6, 5, 0, -6, 9, -6, 0, 7, -12, 9, 0, -10, 16, -13, 0, 15, -22, 17, 0, -20, 29, -21, 0, 25, -38, 28, 0, -32, 50, -39, 0, 43, -64, 49, 0, -56, 82, -60, 0, 69, -105, 78, 0, -86, 132, -101, 0, 112, -166, 125, 0, -142, 208, -153, 0, 172, -258, 192, 0
Offset: 0

Views

Author

Michael Somos, Mar 07 2004; corrected Jun 09 2004

Keywords

Comments

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
Glaisher (1876) writes "XIII. tan(pi/16) = (e^(-pi/2) - e^(-3 pi/2) - e^(-15 pi/2) + e^(-21 pi/2) + e^(-45 pi/2) - &c.) / (1 - e^(-6 pi/2) - e^(-10 pi/2) + e^(-28 pi/2) + e^(-36 pi/2) - &c.), ..." where the numerator is q * f( -q^2, -q^14) and denominator is f( -q^6, -q^10) where q = e^(-pi/2). - Michael Somos, Jun 22 2012
Berndt writes "[...] v = q^(1/2) f(-q,-q^7) / f(-q^3,-q^5). Then v = q^(1/2) / (1 + q + q^2 / (1 + q^3 + q^4 / (1 + q^5 + q^6 / (1 + x^7 + ...)))). (1.1)". - Michael Somos, Jul 09 2012
Jacobi writes "(7.) (1 - sqrt(k')) / (1 + sqrt(k') + sqrt(2(1+k'))) = (q - q^3 - q^15 + q^21 + q^45 - q^55 - ...) / (1 - q^6 - q^10 + q^28 + q^36 - q^66 - ...)." - Michael Somos, Sep 11 2012

Examples

			G.f. = 1 - x + x^3 - x^4 + x^5 - 2*x^7 + 2*x^8 - x^9 + 2*x^11 - 3*x^12 + ...
G/f. = q - q^3 + q^7 - q^9 + q^11 - 2*q^15 + 2*q^17 - q^19 + 2*q^23 + ...
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part III, Springer-Verlag; see p. 221 Entry 1(ii), eq. (1.1).
  • J. W. L. Glaisher, Identities, Messenger of Mathematics, 5 (1876), 111-112. see Eq. XIII
  • C. G. J. Jacobi, Über die Zur Numerischen Berechnung der Elliptischen Functionen Zweckmaessigsten Formeln, Crelle Bd. 26 (1843), 93-114 = Gesammelte Werke, Bd. 1, 1881, 343-368.

Crossrefs

Programs

  • Magma
    A := Basis( ModularForms( Gamma1(16), 1/2), 159); LS := LaurentSeriesRing( RationalField()); A[3] / A[2]; /* Michael Somos, Aug 31 2018 */
  • Mathematica
    a[ n_] := SeriesCoefficient[ QPochhammer[ x, x^8] QPochhammer[ x^7, x^8] /(QPochhammer[ x^3, x^8] QPochhammer[ x^5, x^8]), {x, 0, n}] (* Michael Somos, Aug 02 2011 *)
    a[ n_] := SeriesCoefficient[ EllipticTheta[ 2, 0, x^2] / (EllipticTheta[ 3, 0, x] + EllipticTheta[ 3, 0, x^2]), {x, 0, n + 1/2}] (* Michael Somos, Aug 02 2011 *)
    a[ n_] := SeriesCoefficient[ Product[(1 - q^k)^KroneckerSymbol[ 8, k], {k, n}], {q, 0, n}] (* Michael Somos, Jul 08 2012 *)
  • PARI
    {a(n) = local(A, u, v); if( n<0, 0, n = 2*n + 1; A = x; forstep( k=3, n, 2, u = A + x * O(x^k); v = subst(u, x, x^2); A -= x^k * polcoeff(u^2 - v + v*u^2 + v^2, k+1) / 2); polcoeff(A, n))}
    
  • PARI
    {a(n) = local(A, m); if( n<0, 0, A = 1 + O(x); m=1; while( m<=n, m*=2; A = x * subst(A, x, x^2); A = sqrt(A * (1 - A) / (1 + A) / x)); polcoeff(A, n))}
    
  • PARI
    {a(n) = local(A, A2); if( n<0, 0, A = eta(x^8 + x * O(x^n))^2 / eta(x^4 + x * O(x^n)); A2 = sum( k=1, sqrtint(n), x^k^2 + x^(2*k^2), 1 + x * O(x^n)); polcoeff(A / A2, n))}
    
  • PARI
    {a(n) = local(A, A2); if( n<0, 0, A = x * O(x^n); A = eta(x + A) * eta(x^4 + A)^2 / eta(x^2 + A)^3; A2 = subst(A, x, x^2); polcoeff( 2 * A^2 * A2^2 / (A^2 + A2), n))}
    
  • PARI
    N=66; x='x+O('x^N); Vec(prod(k=1, N, (1-x^k)^kronecker(2, k))) \\ Seiichi Manyama, Sep 24 2019
    

Formula

Expansion of f(-x, -x^7) / f(-x^3, -x^5) in powers of x where f(, ) is Ramanujan's general theta function. - Michael Somos, Aug 02 2011
Expansion of (phi(x) - phi(x^2)) / (2 * x * psi(x^4)) = 2 * psi(x^4) / (phi(x) + phi(x^2)) in powers of x where phi(), psi() are Ramanujan theta functions. - Michael Somos, Feb 15 2006
Expansion of q^(-1) * (1 - sqrt(k')) / (1 + sqrt(k') + sqrt(2 * (1 + k'))) in powers of q^2 where k' is the complementary elliptic modulus. - Michael Somos, Sep 11 2012
Euler transform of period 8 sequence [-1, 0, 1, 0, 1, 0, -1, 0, ...].
G.f. A(x) satisfies both A(-x) * A(x) = A(x^2) and x * A(x)^2 = B(x * A(x^2)) where B(x) = x * (1 - x) / (1 + x).
Given g.f. A(x), then B(x) = x * A(x^2) satisfies 0 = f(B(x), B(x^2)) where f(u, v) = u^2 - v + v^2 + v*u^2.
Given g.f. A(x), then B(x) = x * A(x^2) satisfies 0 = f(B(x), B(x^3)) where f(u, v) = (1 - u*v) * (u + v)^3 - v * (1 + v^2) * (1 - u^4). - Michael Somos, Feb 15 2006
Given g.f. A(x), then B(x) = x * A(x^2) satisfies 0 = f(B(x), B(x^5)) where f(u, v) = (u - v) * (1 + u*v)^5 - u * (1 - u^4) * (1 + v^2) * (1 - 6*v^2 + v^4). - Michael Somos, Feb 15 2006
G.f.: Product_{k>=0} (1 - x^(8*k + 1)) * (1 - x^(8*k + 7)) / ((1 - x^(8*k + 3)) * (1 - x^(8*k + 5))).
G.f. = continued fraction 1/(1 + x + x^2/(1 + x^3 + x^4/(1 + x^5 + x^6/(1 + x^7 + ...)))). Convolution inverse of A111374.
a(2*n + 1) = -A226559(n). - Michael Somos, Jun 12 2013
a(4*n) = A083365(n). a(4*n + 2) = 0.
G.f. A(x) satisfies x*A(-x^2) = x*B(x^2)/C(x^2) = (F(x) - F(-x))/(F(x) + F(-x)), where B(x) is the g.f. of A069911, C(x) is the g.f. of A069910 and F(x) = Product_{k >= 0} 1 + x^(2*k+1) is the g.f. of A000700. - Peter Bala, Feb 07 2021

A091190 G.f. A(x) satisfies x*A(x)^3 = B(x*A(x^3)) where B(x) = x/(1 - 3*x).

Original entry on oeis.org

1, 1, 2, 5, 13, 35, 97, 273, 778, 2240, 6499, 18976, 55703, 164243, 486130, 1443620, 4299365, 12836825, 38413933, 115184282, 346005073, 1041072108, 3137060983, 9465689545, 28596915843, 86492865522, 261876842801, 793661873276
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2004

Keywords

Comments

More generally, given A(x) satisfies x*A(x)^p = B(x*A(x^p)) where B(x) = x/(1-p*x), then it appears that A(x) is an integer series only when p is prime. This is a special case of sequences with g.f.s that satisfy the more general functional equation x*A(x)^m = B(x*A(x^m)) studied by Michael Somos; some other examples are A085748, A091188 and A091200.

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 13*x^4 + 35*x^5 + 97*x^6 + 273*x^7 + 778*x^8 + 2240*x^9 + 6499*x^10 + 18976*x^11 + 55703*x^12 + ...
where A(x)^3 = A(x^3) / (1 - 3*x*A(x^3)).
RELATED SERIES.
A(x)^3 = 1 + 3*x + 9*x^2 + 28*x^3 + 87*x^4 + 270*x^5 + 839*x^6 + 2607*x^7 + 8100*x^8 + 25169*x^9 + 78207*x^10 + 243009*x^11 + 755095*x^12 + ...
Also, D(x) = x*A(D(x)) is the g.f. of A370441, which begins
D(x) = x + x^2 + 3*x^3 + 12*x^4 + 54*x^5 + 261*x^6 + 1324*x^7 + 6952*x^8 + 37461*x^9 + ... + A370441(n)*x^n + ...
such that D(x)^3 = D( x^3 + 3*D(x)^4 ).
		

Crossrefs

Programs

  • Mathematica
    m = 28; B[x_] = x/(1 - 3 x); A[_] = 1;
    Do[A[x_] = (B[x A[x^3]]/x)^(1/3) + O[x]^m // Normal, {m}];
    CoefficientList[A[x], x] (* Jean-François Alcover, Oct 29 2019 *)
  • PARI
    {a(n) = my(A,p=3,m=1); if(n<0,0, m=1; A=1+O(x); while(m<=n, m*=p; A = x*subst(A,x,x^p); A = (A/(1-p*A)/x)^(1/p)); polcoeff(A,n))}
    for(n=0,30, print1(a(n),", "))

Formula

From Paul D. Hanna, Mar 09 2024: (Start)
G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) A(x)^3 = A(x^3) / (1 - 3*x*A(x^3)).
(2) A(x) = x/Series_Reversion(D(x)) where D(x) = x*A(D(x)) is the g.f. of A370441.
(End)

Extensions

Corrected by T. D. Noe, Oct 25 2006

A091200 G.f. A(x) satisfies xA(x)^5 = B(xA(x^5)) where B(x) = x/(1-5x).

Original entry on oeis.org

1, 1, 3, 11, 44, 185, 802, 3553, 15994, 72886, 335387, 1555487, 7261310, 34083382, 160730900, 761039051, 3616102911, 17235223345, 82372594183, 394648349447, 1894921311499, 9116598414141, 43939539520427, 212124129983285
Offset: 0

Views

Author

Paul D. Hanna, Feb 23 2004

Keywords

Comments

More generally, given A(x) satisfies xA(x)^p = B(xA(x^p)) where B(x) = x/(1-p*x), then it appears that A(x) is an integer series only when p is prime. This is a special case of sequences with g.f.s that satisfy the more general functional equation xA(x)^m = B(xA(x^m)) originated by Michael Somos; some other examples are A085748, A091188 and A091190.

Crossrefs

Programs

  • PARI
    {a(n)=local(A,m); p=5;if(n<0,0,m=1; A=1+O(x); while(m<=n,m*=p; A=x*subst(A,x,x^p); A=(A/(1-p*A)/x)^(1/p));polcoeff(A,n))}
Showing 1-3 of 3 results.