A264227 G.f. A(x) satisfies: A(x)^2 = A( x^2/(1-10*x) ), with A(0) = 0.

1, 5, 40, 350, 3220, 30500, 294625, 2886875, 28598035, 285786575, 2876602225, 29131678625, 296574083425, 3033183585125, 31148390740375, 321040368434375, 3319845741478030, 34433523106882550, 358129419509956150, 3734203057793066750, 39027568927659117700, 408777143934160983500, 4290195975642644398000, 45111124579414224095000

Offset

1,2

Comments

Radius of convergence is r = 1/11, where r = r^2/(1-10*r), with A(r) = 1.

Compare to a g.f. M(x) of Motzkin numbers: M(x)^2 = M(x^2/(1-2*x)) where M(x) = (1-x - sqrt(1-2*x-3*x^2))/(2*x).

Links

Paul D. Hanna, Table of n, a(n) for n = 1..300

Formula

G.f. also satisfies:

(1) A(x) = -A( -x/(1-10*x) ).

(2) A( x/(1+5*x) ) = -A( -x/(1-5*x) ), an odd function.

(3) A( x/(1+5*x) )^2 = A( x^2/(1-25*x^2) ), an even function.

(4) A(x)^4 = A( x^4/((1-10*x)*(1-10*x-10*x^2)) ).

(5) [x^(2*n+1)] (x/A(x))^(2*n) = 0 for n>=0.

(6) [x^(2^n+k)] (x/A(x))^(2^n) = 0 for k=1..2^n-1, n>=1.

Given g.f. A(x), let F(x) denote the g.f. of A264415, then:

(7) A(x) = F(A(x))^2 * x/(1+25*x),

(8) A(x) = F(A(x)^2) * x/(1-5*x),

(9) A( x/(F(x)^2 - 25*x) ) = x,

(10) A( x/(F(x^2) + 5*x) ) = x,

where F(x)^2 = F(x^2) + 30*x.

Sum_{k=0..n} binomial(n,k) * (-5)^(n-k) * a(k+1) = 0 for odd n.

Sum_{k=0..n} binomial(n,k) *(-10)^(n-k) * a(k+1) = (-1)^n * a(n+1) for n>=0.

Example

G.f.: A(x) = x + 5*x^2 + 40*x^3 + 350*x^4 + 3220*x^5 + 30500*x^6 + 294625*x^7 + 2886875*x^8 + 28598035*x^9 + 285786575*x^10 + 2876602225*x^11 +...
where A(x)^2 = A(x^2/(1-10*x)).
RELATED SERIES.
A(x)^2 = x^2 + 10*x^3 + 105*x^4 + 1100*x^5 + 11540*x^6 + 121200*x^7 + 1274350*x^8 + 13414000*x^9 + 141353220*x^10 + 1491161000*x^11 + 15747360500*x^12 +...
A( x/(1+5*x) ) = x + 15*x^3 + 345*x^5 + 9000*x^7 + 251160*x^9 + 7328475*x^11 + 220880925*x^13 + 6824229750*x^15 + 214969962405*x^17 + 6877343600775*x^19 +...
A( x^2/(1-25*x^2) ) = x^2 + 30*x^4 + 915*x^6 + 28350*x^8 + 891345*x^10 + 28401750*x^12 + 915916500*x^14 + 29852415000*x^16 + 982068551160*x^18 +...
where A( x^2/(1-25*x^2) ) = A( x/(1+5*x) )^2.
Let B(x) = x/Series_Reversion(A(x)), then A(x) = x*B(A(x)), where
B(x) = 1 + 5*x + 15*x^2 - 105*x^4 + 1575*x^6 - 29190*x^8 + 603225*x^10 - 13352850*x^12 + 309605625*x^14 - 7422255645*x^16 +...+ A264415(n)*x^(2*n) +...
such that B(x) = F(x^2) + 5*x = F(x)^2 - 25*x and F(x) is the g.f. of A264415.

Prog

(PARI) {a(n) = my(A=x); for(i=1, n, A = sqrt( subst(A, x, x^2/(1-10*x +x*O(x^n))) ) ); polcoeff(A, n)}
for(n=1, 30, print1(a(n), ", "))

Crossrefs

Cf. A264415, A264224, A264225, A264226.

Sequence in context: A078846 A027259 A007036 * A052798 A137973 A371520

Adjacent sequences: A264224 A264225 A264226 * A264228 A264229 A264230

Keyword

nonn

Author

Paul D. Hanna, Nov 08 2015

Status

approved