A190790 G.f. satisfies: A(x) = 1 + Sum_{n>=1} q^(2n-1)/(1 - q^(2n-1)) where q = x*A(x).
1, 1, 2, 6, 18, 58, 198, 696, 2506, 9205, 34344, 129792, 495834, 1911640, 7428444, 29064650, 114404410, 452719183, 1799994588, 7187148262, 28807364008, 115865980972, 467497031164, 1891710323324, 7675031497682, 31215088847239
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 18*x^4 + 58*x^5 + 198*x^6 +... Let q = x*A(x), then the g.f. A(x) satisfies the following series: * A(x) = 1 + q/(1 - q) + q^3/(1 - q^3) + q^5/(1 - q^5) + q^7/(1 - q^7) +... * A(x) = 1 + q/(1 - q) + q^3/(1 - q^2) + q^6/(1 - q^3) + q^10/(1 - q^4) +... * A(x) = 1 + q/(1 - q^2) + q^2/(1 - q^4) + q^3/(1 - q^6) + q^4/(1 - q^8) +... * A(x) = 1 + q + q^2 + 2*q^3 + q^4 + 2*q^5 + 2*q^6 +...+ A001227(n)*q^n +...
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..1000
Programs
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Mathematica
nmax = 30; A[] = 0; Do[A[x] = 1 + Sum[(x*A[x])^(2*k-1)/(1 - (x*A[x])^(2*k-1)), {k, 1, nmax}] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] (* Vaclav Kotesovec, Sep 28 2023 *) (* Calculation of constant d: *) val = r /. FindRoot[{(Log[1 - r^2*s^2] + QPolyGamma[0, -1/2, r^2*s^2]) / Log[r^2*s^2] == s - 1 - 1/(1 - r*s), r*(1/(-1 + r*s)^2 + (2*r* s*(1/(-1 + r^2*s^2) + Derivative[0, 0, 1][QPolyGamma][0, -1/2, r^2*s^2]))/ Log[r^2*s^2]) == 1 + (2*(s - 1 - 1/(1 - r*s)))/(s*Log[r^2*s^2])}, {r, 1/4}, {s, 2}, WorkingPrecision -> 200]; N[ 1/Chop[val], -Floor[Log[10, Abs[Im[val]]]] - 3] (* Vaclav Kotesovec, Sep 28 2023 *)
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PARI
{a(n)=local(A=1+x);for(i=1,n,A=1+sum(m=1,n,(x*A)^(2*m-1)/(1-(x*A)^(2*m-1)+x*O(x^n))));polcoeff(A,n)} -
PARI
{a(n)=local(A=1+x);for(i=1,n,A=1+sum(m=1,n,(x*A)^(m*(m+1)/2)/(1-(x*A)^m+x*O(x^n))));polcoeff(A,n)} -
PARI
{a(n)=local(A=1+x);for(i=1,n,A=1+sum(m=1,n,(x*A)^m/(1-(x*A)^(2*m)+x*O(x^n))));polcoeff(A,n)} -
PARI
{a(n)=local(D=1+sum(m=1,n,sumdiv(m, d, d%2)*x^m)+x*O(x^n));polcoeff(1/x*serreverse(x/D),n)} -
PARI
{a(n)=local(D=1+sum(m=1,n,sumdiv(m, d, d%2)*x^m)+x*O(x^n));polcoeff(D^(n+1)/(n+1),n)}
Formula
G.f. A(x) satisfies:
* A(x) = 1 + Sum_{n>=1} q^(n*(n+1)/2)/(1 - q^n), where q = x*A(x);
* A(x) = 1 + Sum_{n>=1} q^n/(1 - q^(2n)), where q = x*A(x);
Let D(x) = 1 + Sum_{n>=1} A001227(n)*x^n, then
* A(x) = D(x*A(x)) and D(x) = A(x/D(x));
* A(x) = (1/x)*Series_Reversion(x/D(x));
* a(n) = [x^n] D(x)^(n+1)/(n+1), the coefficient of x^n in D(x)^(n+1)/(n+1) for n>=0.
a(n) ~ c * d^n / n^(3/2), where d = 4.3154117906555438598489327064282723007551580672340735654761205324876... and c = 0.545252538971019249263783268322061859441589544238489362753993274... - Vaclav Kotesovec, Sep 28 2023