A212392 G.f. satisfies: A(x) = (x + A(A(x)))^2 where g.f. A(x) = Sum_{n>=1} a(n)*x^(3*n-1).

1, 2, 9, 56, 400, 3096, 25256, 213832, 1861272, 16552320, 149737632, 1373597892, 12747475260, 119465392536, 1129016386080, 10747541655584, 102960795706704, 991886971036248, 9603034303017640, 93386133268757760, 911779906476551616, 8934398271363272642

Offset

1,2

Comments

Conjecture: n divides a(n); see A212391.

More generally, we have the conjecture:

If A(x) = ( x + A(A(x)) )^b

where A(x) = Sum_{n>=1} a(n) * x^((b^2-1)*(n-1)+b)

then ((b-1)*(n-1)+1) divides a(n).

Links

Paul D. Hanna, Table of n, a(n) for n = 1..200

Formula

a(n) = n*A212391(n).

Example

G.f.: A(x) = x^2 + 2*x^5 + 9*x^8 + 56*x^11 + 400*x^14 + 3096*x^17 + 25256*x^20 +...
such that A(x) = (x + A(A(x)))^2, where
A(A(x)) = x^4 + 4*x^7 + 24*x^10 + 168*x^13 + 1284*x^16 + 10384*x^19 + 87364*x^22 + 756808*x^25 + 6704968*x^28 + 60471040*x^31 +...+ A212277(n+1)*x^(3*n+1) +...
Note that sqrt(A(A(x))) = A(x) + A(A(A(x))), where
sqrt(A(A(x))) = x^2 + 2*x^5 + 10*x^8 + 64*x^11 + 464*x^14 + 3624*x^17 + 29746*x^20 + 252976*x^23 + 2209488*x^26 + 19701504*x^29 +...
A(A(A(x))) = x^8 + 8*x^11 + 64*x^14 + 528*x^17 + 4490*x^20 + 39144*x^23 + 348216*x^26 + 3149184*x^29 + 28872401*x^32 +...

Prog

(PARI) {a(n)=local(A=x^2+x^3); for(i=1, n, A=(x+subst(A, x, A+O(x^(3*n))))^2); polcoeff(A, 3*n-1)}
for(n=1, 30, print1(a(n), ", "))
(Maxima) A(n, k):= if  n<2*k then 0 else if n/2=k then 1 else sum(binomial(2*k, j)*sum(A(i, 2*k-j)*A(n-j, i), i, 2*k-j+1, n-j-1), j, 0, 2*k-1);
makelist(A(n, 1), n, 1, 17); [Vladimir Kruchinin, May 15 2012]

Crossrefs

Cf. A212391, A212277.

Sequence in context: A179405 A081004 A198953 * A186262 A138740 A276370

Adjacent sequences: A212389 A212390 A212391 * A212393 A212394 A212395

Keyword

nonn

Author

Paul D. Hanna, May 12 2012

Status

approved