

A258379


O.g.f. satisfies A^4(z) = 1/(1  z)*( BINOMIAL(BINOMIAL(A(z))) )^3.


6



1, 7, 73, 1071, 21249, 549927, 17907177, 709326255, 33202983873, 1794040660359, 109844961440841, 7511188035994479, 567027585432472641, 46818521577433459239, 4195842793686119552361, 405529683304196611790703, 42039822952112350680798849, 4652599937163116610404900871
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OFFSET

0,2


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 = 3.


LINKS



FORMULA

a(0) = 1 and for n >= 1, a(n) = 1/n*Sum_{i = 0..n1} R(i+1,3)*a(n1i), where R(n,x) denotes the nth row polynomial of A145901.
O.g.f.: A(z) = 1 + 7*z + 73*z^2 + 1071*z^3 + 21249*z^4 + ... satisfies A^4(z) = 1/(1  z)*1/(1  2*z)^3*A^3(z/(1  2*z)).
O.g.f.: A(z) = exp( Sum_{k >= 1} R(k,3)*z^k/k ).
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 length a divisor of 2^(k1). For example, a(n) (mod 4) = (1,3,1,3,...) seems to be purely periodic with period 2, a(n) (mod 8) = (1,7,1,7,...) seems to be purely periodic also with period 2 while a(n) (mod 16) = (1,7,9,15,1,7,9,15,...) seems to be purely periodic with period 4 (all three 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)


EXAMPLE

a(n) (mod 3) = (1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0, 0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0, 0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0, 0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,...).  Peter Bala, Dec 06 2017


MAPLE

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^(ni)*R(i, x), i = 0..n1) 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(n1i, k), i = 0..n1) end if; end proc:
# display the case k = 3
seq(a(n, 3), n = 0..17);


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, 3];


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



