A258378 O.g.f. satisfies A^3(z) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(z))) )^2.

1, 5, 37, 385, 5417, 99421, 2296077, 64510617, 2142013137, 82103710517, 3566271497845, 173005328363057, 9265752053418233, 542783129304580237, 34511577062800532573, 2366512551126709790793, 174056559606294111346593, 13666923859188010833522789, 1140970414332381380968275653

Offset: 0

Author

Peter Bala, May 28 2015

Keywords

• nonn
• easy

Comments

The binomial transform of an o.g.f. A(z) is given by BINOMIAL(A(z)) = 1/(1 - z)*A(z/(1 - z)).

For general remarks on a solution to the functional equation A^(N+1)(z) = 1/(1 - z)*(BINOMIAL(BINOMIAL(A(z))) )^N for integer N, and the connection with triangle A145901 see A258377 (case N = 1). This is the case N = 2.

From Peter Bala, Dec 06 2017: (Start)

a(n) appears to be of the form 4*m + 1. Calculation suggests that for k = 1,2,3,..., the sequence a(n) (mod 2^k) is purely periodic with period length a divisor of 2^(k-1). For example, a(n) (mod 8) = (1, 5, 5, 1, 1, 5, 5, 1,...) seems to be purely periodic with period length 4 and a(n) (mod 16) = (1, 5, 5, 1, 9, 13, 13, 9, 1, 5, 5, 1, 9, 13, 13, 9,...) seems to be purely periodic with period length 8 (both checked up to n = 1000). (End)

Links

• N. J. A. Sloane, Transforms.

Crossrefs

Cf. A019538, A145901, A258377 (N = 1), A258379 (N = 3), A258380 (N = 4), A258381 (N = 5).

Programs

Maple

#A258378
with(combinat):
#recursively define the row polynomials R(n,x) of A145901
R := proc (n, x) option remember; if n = 0 then 1 else 1 + x*add(binomial(n, i)*2^(n-i)*R(i,x), i = 0..n-1) end if; end proc:
#define a family of sequences depending on an integer parameter k
a := proc (n, k) option remember; if n = 0 then 1 else 1/n*add(R(i+1,k)*a(n-1-i,k), i = 0..n-1) end if; end proc:
# display the case k = 2
seq(a(n,2), n = 0..18);

Mathematica

R[n_, x_] := R[n, x] = If[n == 0, 1, 1 + x*Sum[Binomial[n, i]*2^(n - i)*R[i, x], {i, 0, n - 1}]];
a[n_, k_] := a[n, k] = If[n == 0, 1, 1/n*Sum[R[i + 1, k]*a[n - 1 - i, k], {i, 0, n - 1}]];
a[n_] := a[n, 2];
a /@ Range[0, 18] (* Jean-François Alcover, Oct 02 2019 *)

Formula

a(0) = 1 and for n >= 1, a(n) = 1/n*Sum_{i = 0..n-1} R(i+1,2)*a(n-1-i), where R(n,x) denotes the n-th row polynomial of A145901.

O.g.f.: A(z) = 1 + 5*z + 37*z^2 + 385*z^3 + 5417*z^4 + ... satisfies A^3(z) = 1/(1 - z)*1/(1 - 2*z)^2*A^2(z/(1 - 2*z)).

O.g.f.: A(z) = exp( Sum_{k >= 1} R(k,2)*z^k/k ).

a(n) ~ (n-1)! * 2^(n - 1/2) / (sqrt(3) * log(3/2)^(n+1)). - Vaclav Kotesovec, May 28 2025