A258381 O.g.f. satisfies A^6(z) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(z))) )^5.

1, 11, 181, 4191, 131241, 5360883, 275510493, 17223156423, 1272268864593, 108480982129883, 10481174173743109, 1130938869235448879, 134719322898080187129, 17552325198110327173059, 2482129971814696069384749, 378542038806168341351484567, 61920836368469049844434420897

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 = 5.

From Peter Bala, Dec 06 2017: (Start)

a(n) appears to be always odd. Calculation suggests that for k = 1,2,3,... the sequence a(n) (mod 2^k) is purely periodic with period 2^(k-1). For example, a(n) (mod 4) = (1,3,1,3,...) seems to be purely periodic with period 2 and a(n) (mod 8) = (1,3,5,7,1,3,5,7,...) seems to be purely periodic with period 4 (both checked up to n = 1000).

The sequences a(n) (mod k), for other values of k, appear to have interesting but more complicated patterns. An example is given below.

(End)

Examples

			a(n) (mod 5) begins [1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1,...]. - _Peter Bala_, Dec 06 2017
		

Links

• N. J. A. Sloane, Transforms.

Crossrefs

Cf. A019538, A145901, A258377 (N = 1), A258378 (N = 2), A258379 (N = 3), A258380 (N = 4).

Programs

Maple

#A258381
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 = 5
seq(a(n,5), n = 0..16);

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, 5];
a /@ Range[0, 16] (* 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+1,5)*a(n-1-i), where R(n,x) denotes the n-th row polynomial of A145901.

O.g.f.: A(z) = 1 + 11*z + 181*z^2 + 4191*z^3 + 131241*z^4 + ... satisfies A^6(z) = 1/(1 - z)*1/(1 - 2*z)^5*A^5(z/(1 - 2*z)).

O.g.f.: A(z) = exp( Sum_{k >= 1} R(k,5)*z^k/k ).

a(n) ~ (n-1)! * 2^(n - 1/2) / (sqrt(15) * log(6/5)^(n+1)). - Vaclav Kotesovec, May 28 2025