A317134 G.f.: Sum_{n>=0} binomial(4*(n+1), n)/(n+1) * x^n / (1+x)^(2*(n+1)).

1, 2, 9, 44, 236, 1336, 7862, 47608, 294720, 1856748, 11865684, 76731572, 501176237, 3301501694, 21909634763, 146337236580, 982962605577, 6635968279354, 45001173711683, 306406562117884, 2093909763907401, 14356806252396614, 98735015302171955, 680906548260420320, 4707709357806093085, 32625093782844333722, 226588405850230665429, 1576882804780751603092

Offset: 0

Author

Paul D. Hanna, Jul 22 2018

Keywords

• nonn

Comments

Note that: binomial(4*(n+1), n)/(n+1) = A002293(n+1) for n >= 0, where F(x) = Sum_{n>=0} A002293(n)*x^n satisfies F(x) = 1 + x*F(x)^4.

Compare the g.f. to:

(C1) 1 = Sum_{n>=0} binomial(m*(n+1), n)/(n+1) * x^n / (1+x)^(m*(n+1)) holds for fixed m.

(C2) If S(x,p,q) = Sum_{n>=0} binomial(p*(n+1),n)/(n+1) * x^n/(1+x)^(q*(n+1)), then Series_Reversion ( x*S(x,p,q) ) = x*S(x,q,p) holds for fixed p and q.

Examples

			G.f.: A(x) = 1 + 2*x + 9*x^2 + 44*x^3 + 236*x^4 + 1336*x^5 + 7862*x^6 + 47608*x^7 + 294720*x^8 + 1856748*x^9 + 11865684*x^10 + ...
such that
A(x) = 1/(1+x)^2 + 4*x/(1+x)^4 + 22*x^2/(1+x)^6 + 140*x^3/(1+x)^8 + 969*x^4/(1+x)^10 + 7084*x^5/(1+x)^12 + ... + A002293(n+1)*x^n/(1+x)^(2*(n+1)) + ...
RELATED SERIES.
Series_Reversion( x*A(x) )  =  4*x/((1+x)^2 + sqrt( (1+x)^4 - 4*x ))^2  =  x - 2*x^2 - x^3 + 6*x^4 + 3*x^5 - 20*x^6 - 18*x^7 + 74*x^8 + 111*x^9 - 278*x^10 - 657*x^11 + 980*x^12 + 3739*x^13 + ...
which equals the sum:
Sum_{n>=0} binomial(2*(n+1), n)/(n+1) * x^(n+1)/(1+x)^(4*(n+1)).
The square-root of the g.f. is an integer series:
sqrt(A(x)) = 1 + x + 4*x^2 + 18*x^3 + 92*x^4 + 504*x^5 + 2897*x^6 + 17235*x^7 + 105233*x^8 + 655687*x^9 + 4152461*x^10 + ... + A317135(n)*x^n + ...
which equals the sum:
Sum_{n>=0} binomial(4*n+2, n)/(2*n+1) * x^(n+1)/(1+x)^(2*n+1).
		

Crossrefs

Cf. A316371, A127897, A316987, A317133.

Programs

Mathematica

Rest[CoefficientList[InverseSeries[Series[4*x/((1 + x)^2 + Sqrt[(1 + x)^4 - 4*x])^2, {x, 0, 30}], x], x]](* Vaclav Kotesovec, Jul 22 2018 *)

PARI

{a(n) = my(A = sum(m=0, n, binomial(4*(m+1), m)/(m+1) * x^m / (1+x +x*O(x^n))^(2*(m+1)))); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

PARI

{a(n) = my(A = (1/x) * serreverse( 4*x/((1+x)^2 + sqrt( (1+x)^4 - 4*x  + x*O(x^n)))^2 )); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) satisfies:

(1) A(x) = ( (1 + x*A(x))^2 + sqrt( (1 + x*A(x))^4 - 4*x*A(x) ) )^2 / 4.

(2) A(x) = (1/x) * Series_Reversion( 4*x/((1+x)^2 + sqrt( (1+x)^4 - 4*x ))^2 ).

(3) A(x) = Sum_{n>=0} binomial(4*(n+1), n)/(n+1) * x^n / (1+x)^(2*(n+1)).

a(n) ~ 37^(1/4) * (101 + 16*sqrt(37))^(n+1) / (2*sqrt(Pi) * n^(3/2) * 3^(3*n + 9/2)). - Vaclav Kotesovec, Jul 22 2018