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.

A145267 G.f. satisfies A(x) = Product_{k>0} (1+x^k*A(x)).

Original entry on oeis.org

1, 1, 2, 5, 12, 30, 77, 201, 532, 1427, 3868, 10579, 29161, 80931, 225954, 634197, 1788453, 5064877, 14398536, 41074364, 117541744, 337337862, 970704394, 2800059428, 8095161902, 23452565124, 68076579332, 197965830430
Offset: 0

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Author

Vladeta Jovovic, Oct 05 2008

Keywords

Examples

			From _Paul D. Hanna_, May 20 2011: (Start)
G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 12*x^4 + 30*x^5 + 77*x^6 +...
G.f.: A(x) = (1+x*A(x))*(1+x^2*A(x))*(1+x^3*A(x))*(1+x^4*A(x))*...
G.f.: A(x) = (1+x*A(x)) + x^2*A(x)*(1 + x^3*A(x))*(1+x*A(x))/(1-x) + x^7*A(x)^2*(1 + x^5*A(x))*(1+x*A(x))*(1+x^2*A(x))/((1-x)*(1-x^2)) + x^15*A(x)^3*(1 + x^7*A(x))*(1+x*A(x))*(1+x^2*A(x))*(1+x^3*A(x))/((1-x)*(1-x^2)*(1-x^3)) +... (End)
G.f.: A(x) = 1 + x*A(x)/(1-x) + x^3*A(x)^2/((1-x)*(1-x^2)) + x^6*A(x)^3/((1-x)*(1-x^2)*(1-x^3)) + x^10*A(x)^4/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) +... - _Paul D. Hanna_, Jul 01 2011

Links

Crossrefs

Cf. A145268, A190822.

Programs

  • Mathematica
    nmax = 30; A[] = 0; Do[A[x] = Product[1 + x^k*A[x], {k, 1, nmax}] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] (* Vaclav Kotesovec, Sep 26 2023 *)
(* Calculation of constant d: *) 1/r /. FindRoot[{QPochhammer[-s, r] == s*(1 + s), Log[1 - r] + ((1 + 2*s)*Log[r])/(1 + s) + QPolyGamma[0, Log[-s]/Log[r], r] == 0}, {r, 1/3}, {s, 1}, WorkingPrecision -> 120] (* Vaclav Kotesovec, Sep 26 2023 *)
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=prod(m=1,n,(1+A*x^m+x*O(x^n))));polcoeff(A,n)}  /* Paul D. Hanna, May 20 2011 */
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=sum(m=0,n,x^(m*(3*m+1)/2)*A^m*(1 + x^(2*m+1)*A)*prod(k=1,m,(1+A*x^k)/(1-x^k+x*O(x^n)))));polcoeff(A,n)}  /* Paul D. Hanna, May 20 2011 */
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=sum(m=0,n,x^(m*(m+1)/2)*A^m/prod(k=1,m,1-x^k +x*O(x^n))));polcoeff(A,n)} /* Paul D. Hanna, Jul 01 2011 */

Formula

G.f. satisfies: A(x) = Sum_{n>=0} x^(n*(3n+1)/2)*A(x)^n*(1 + x^(2n+1)*A(x))*Product_{k=1..n} (1 + x^k*A(x))/(1-x^k) due to Sylvester's identity. - Paul D. Hanna, May 20 2011
G.f. satisfies: A(x) = Sum_{n>=0} x^(n*(n+1)/2)*A(x)^n / Product_{k=1..n} (1-x^k). - Paul D. Hanna, Jul 01 2011
a(n) ~ c * d^n / n^(3/2), where d = 3.060735101304296413235... and c = 2.45762465379034328... - Vaclav Kotesovec, Aug 12 2021
Radius of convergence r = 0.32671889820646736561... = 1/d and A(r) = 3.6673575238633912689... satisfy (1) A(r) = 1 / Sum_{n>=1} r^n/(1 + r^n*A(r)) and (2) A(r) = Product_{n>=1} (1 + r^n*A(r)). - Paul D. Hanna, Mar 02 2024

Extensions

More terms from Max Alekseyev, Jan 31 2010

A196151 G.f. satisfies A(x) = Product_{n>=1} (1 + x^n*A(x)^2).

Original entry on oeis.org

1, 1, 3, 11, 43, 179, 778, 3491, 16051, 75235, 358170, 1727124, 8418266, 41408344, 205289265, 1024737905, 5145933602, 25978844478, 131773584768, 671239285119, 3432304205872, 17611565623950, 90652384728648, 467963720803022, 2422110238147351
Offset: 0

Views

Author

Paul D. Hanna, Sep 28 2011

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 11*x^3 + 43*x^4 + 179*x^5 + 778*x^6 + ...
where
(0) A(x) = (1+x*A(x)^2) * (1+x^2*A(x)^2) * (1+x^3*A(x)^2) * (1+x^4*A(x)^2) * ...
(1) A(x) = 1 + x*A(x)^2/(1-x) + x^3*A(x)^4/((1-x)*(1-x^2)) + x^6*A(x)^6/((1-x)*(1-x^2)*(1-x^3)) + x^10*A(x)^8/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) + ...
(2) A(x) = (1+x*A(x)^2) + x^2*A(x)^2*(1 + x^3*A(x)^2)*(1+x*A(x)^2)/(1-x) + x^7*A(x)^4*(1 + x^5*A(x)^2)*(1+x*A(x)^2)*(1+x^2*A(x)^2)/((1-x)*(1-x^2)) + x^15*A(x)^6*(1 + x^7*A(x)^2)*(1+x*A(x)^2)*(1+x^2*A(x)^2)*(1+x^3*A(x)^2)/((1-x)*(1-x^2)*(1-x^3)) + ...

Links

Crossrefs

Cf. A196150, A145267, A145268, A190822.

Programs

  • Mathematica
    nmax = 30; A[] = 0; Do[A[x] = Product[1 + x^k*A[x]^2, {k, 1, nmax}] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] (* Vaclav Kotesovec, Sep 28 2023 *)
(* Calculation of constants {d,c}: *) {1/r, Sqrt[-r*s*(1 + s^2) * Derivative[0, 1][QPochhammer][-s^2, r] / (8*Pi*(s^2 + (1 + s^2)^2 * QPolyGamma[1, Log[-s^2]/Log[r], r]/ Log[r]^2))]} /. FindRoot[{QPochhammer[-s^2, r] == s*(1 + s^2), 1/2 + s^2/(1 + s^2) + (Log[1 - r] + QPolyGamma[0, Log[-s^2]/Log[r], r])/Log[r] == 0}, {r, 1/8}, {s, 2}, WorkingPrecision -> 120] (* Vaclav Kotesovec, Sep 28 2023 *)
  • PARI
    {a(n) = my(A=1+x); for(i=1, n, A=prod(m=1, n, (1+A^2*x^m+x*O(x^n)))); polcoeff(A, n)}
  • PARI
    {a(n) = my(A=1+x); for(i=1, n, A=sum(m=0, n, x^(m*(m+1)/2)*A^(2*m)/prod(k=1, m, 1-x^k +x*O(x^n)))); polcoeff(A, n)}
  • PARI
    {a(n) = my(A=1+x); for(i=1, n, A=sum(m=0, n, x^(m*(3*m+1)/2)*A^(2*m)*(1 + x^(2*m+1)*A^2)*prod(k=1, m, (1+A^2*x^k)/(1-x^k+x*O(x^n))))); polcoeff(A, n)}

Formula

G.f. satisfies:
(1) A(x) = Sum_{n>=0} x^(n*(n+1)/2)*A(x)^(2*n) / Product_{k=1..n} (1-x^k).
(2) A(x) = Sum_{n>=0} x^(n*(3n+1)/2)*A(x)^(2*n)*(1 + x^(2n+1)*A(x)^2)*Product_{k=1..n} (1 + x^k*A(x)^2)/(1-x^k) due to Sylvester's identity.
a(n) ~ c * d^n / n^(3/2), where d = 5.5051727555189932106045782067309509... and c = 0.4987046473347092789085107139372... - Vaclav Kotesovec, Sep 28 2023
Radius of convergence r = 0.181647342310464199522927295317... = 1/d and A(r) = 1.82512871645978495662055342941... satisfy A(r) = 1 / sqrt( Sum_{n>=1} 2*r^n/(1 + r^n*A(r)^2) ) and A(r) = Product_{n>=1} (1 + r^n*A(r)^2). - Paul D. Hanna, Mar 03 2024

A192785 G.f. satisfies: A(x) = Product_{n>=1} (1 + x^n)*(1 + x^(2*n)*A(x)^2).

Original entry on oeis.org

1, 1, 2, 5, 11, 29, 75, 203, 557, 1561, 4427, 12706, 36819, 107576, 316579, 937471, 2791487, 8352973, 25104573, 75749240, 229379444, 696851166, 2123304184, 6487295518, 19870096689, 61001089214, 187673207413, 578532522637, 1786712575547
Offset: 0

Views

Author

Paul D. Hanna, Jul 10 2011

Keywords

Comments

Related q-series identity of Lebesgue:
Product_{n>=1} (1 + q^n)*(1 + z*q^(2*n)) = 1 + Sum_{n>=1} q^(n*(n+1)/2) * Product_{k=1..n} (1 + z*q^k)/(1 - x^k); here q=x, z=A(x)^2.

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 11*x^4 + 29*x^5 + 75*x^6 + ...
The g.f. A = A(x) satisfies the following relations:
A = (1+x)*(1+x^2*A^2)* (1+x^2)*(1+x^4*A^2)* (1+x^3)*(1+x^6*A^2)* ...
A = 1 + x*(1+x*A^2)/(1-x) + x^3*(1+x*A^2)*(1+x^2*A^2)/((1-x)*(1-x^2)) + x^6*(1+x*A^2)*(1+x^2*A^2)*(1+x^3*A^2)/((1-x)*(1-x^2)*(1-x^3)) + ...

Crossrefs

Cf. A190822, A145267.

Programs

  • Mathematica
    nmax = 30; A[] = 0; Do[A[x] = Product[(1 + x^k)*(1 + x^(2*k)*A[x]^2), {k, 1, nmax}] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] (* Vaclav Kotesovec, Mar 04 2024 *)
(* Calculation of constants {d,c}: *) Chop[{1/r, Sqrt[(-r*s*(1 + s^2) * Log[r]^2 * ((s + s^3)*Derivative[0, 1][QPochhammer][-1, r] + r*QPochhammer[-1, r]^2 * Derivative[0, 1][QPochhammer][-s^2, r^2]))/ (2*Pi * QPochhammer[-1, r]*(4*s^2*Log[r]^2 + (1 + s^2)^2 * QPolyGamma[1, Log[-s^2]/(2*Log[r]), r^2]))]} /. FindRoot[{(QPochhammer[-1, r]*QPochhammer[-s^2, r^2])/(2 + 2*s^2) == s, 1 + 3*s^2 + (1 + s^2)*((Log[1 - r^2] + QPolyGamma[0, Log[-s^2]/(2*Log[r]), r^2])/Log[r]) == 0}, {r, 1/3}, {s, 2}, WorkingPrecision -> 70]] (* Vaclav Kotesovec, Mar 04 2024 *)
  • PARI
    {a(n) = my(A=1+x); for(i=1, n, A = prod(m=1, n, (1 + x^m)*(1 + x^(2*m)*A^2 +x*O(x^n)))); polcoeff(A, n)}
  • PARI
    {a(n) = my(A=1+x); for(i=1, n, A = 1 + sum(m=1, sqrtint(2*n), x^(m*(m+1)/2)*prod(k=1, m, (1 + A^2*x^k)/(1 - x^k+x*O(x^n))))); polcoeff(A, n)}

Formula

G.f. satisfies: A(x) = 1 + Sum_{n>=1} x^(n*(n+1)/2)*Product_{k=1..n} (1 + x^k*A(x)^2)/(1 - x^k) due to a Lebesgue identity.
From Vaclav Kotesovec, Mar 04 2024: (Start)
Let A(x) = y, then 2*y*(1 + y^2) = QPochhammer(-1, x) * QPochhammer(-y^2, x^2).
a(n) ~ c * d^n / n^(3/2), where
d = 3.25215123067662173854186425074452291189580485719079882122325713176...,
c = 1.30862302149248708183423553797270804891358016970005788341511105232...
Radius of convergence:
r = 1/d = 0.307488775604062671485504670197489134974315527740973676344144395...
A(r) = s = 2.80682163771231540175973628784430270489737819467327067575665055...
(End)
The values r and A(r) given above also satisfy A(r) = 1 / sqrt( Sum_{n>=1} 2*r^(2*n)/(1 + r^(2*n)*A(r)^2) ). - Paul D. Hanna, Mar 06 2024
Showing 1-3 of 3 results.

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